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pretraining/mathematica/linear_algebra/cross_product_w_steps/1063.txt

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+Problem:
+Find the cross product of the following vectors:
+$\left(
+\begin{array}{c}
+ -5 \\
+ -3 \\
+ -8 \\
+\end{array}
+\right)$ and
+$\left(
+\begin{array}{c}
+ 2 \\
+ -7 \\
+ -2 \\
+\end{array}
+\right)$
+Answer:
+\begin{array}{l}
+ 
+\begin{array}{l}
+ \text{Compute the following cross product}: \\
+ \, (-5,-3,-8)\, \times \, (2,-7,-2)\,  \\
+\end{array}
+ \\
+\hline
+ 
+\begin{array}{l}
+ \text{Construct }\text{a }\text{matrix }\text{where }\text{the }\text{first }\text{row }\text{contains }\text{unit }\text{vectors }\hat{\text{i}}, \hat{\text{j}}, \text{and }\hat{\text{k}}; \text{and }\text{the }\text{second }\text{and }\text{third }\text{rows }\text{are }\text{made }\text{of }\text{vectors }\, (-5,-3,-8)\,  \text{and }\, (2,-7,-2)\, : \\
+ \left(
+\begin{array}{ccc}
+ \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\
+ -5 & -3 & -8 \\
+ 2 & -7 & -2 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Take }\text{the }\text{determinant }\text{of }\text{this }\text{matrix}: \\
+ \left| 
+\begin{array}{ccc}
+ \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\
+ -5 & -3 & -8 \\
+ 2 & -7 & -2 \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Expand }\text{with }\text{respect }\text{to }\text{row }1: \\
+  \text{= }\left| 
+\begin{array}{ccc}
+ \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\
+ -5 & -3 & -8 \\
+ 2 & -7 & -2 \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{The }\text{determinant }\text{of }\text{the }\text{matrix }\left(
+\begin{array}{ccc}
+ \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\
+ -5 & -3 & -8 \\
+ 2 & -7 & -2 \\
+\end{array}
+\right) \text{is }\text{given }\text{by }\hat{\text{i}} \left| 
+\begin{array}{cc}
+ -3 & -8 \\
+ -7 & -2 \\
+\end{array}
+\right| +\left(-\hat{\text{j}}\right) \left| 
+\begin{array}{cc}
+ -5 & -8 \\
+ 2 & -2 \\
+\end{array}
+\right| +\hat{\text{k}} \left| 
+\begin{array}{cc}
+ -5 & -3 \\
+ 2 & -7 \\
+\end{array}
+\right| : \\
+  \text{= }\hat{\text{i}} \left| 
+\begin{array}{cc}
+ -3 & -8 \\
+ -7 & -2 \\
+\end{array}
+\right| +\left(-\hat{\text{j}}\right) \left| 
+\begin{array}{cc}
+ -5 & -8 \\
+ 2 & -2 \\
+\end{array}
+\right| +\hat{\text{k}} \left| 
+\begin{array}{cc}
+ -5 & -3 \\
+ 2 & -7 \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \hat{\text{i}} \left| 
+\begin{array}{cc}
+ -3 & -8 \\
+ -7 & -2 \\
+\end{array}
+\right| =\hat{\text{i}} ((-3)\, (-2)-(-8)\, (-7))=\hat{\text{i}} (-50)=\fbox{$-50 \hat{\text{i}}$}: \\
+  \text{= }\fbox{$-50 \hat{\text{i}}$}+\left(-\hat{\text{j}}\right) \left| 
+\begin{array}{cc}
+ -5 & -8 \\
+ 2 & -2 \\
+\end{array}
+\right| +\hat{\text{k}} \left| 
+\begin{array}{cc}
+ -5 & -3 \\
+ 2 & -7 \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ -\hat{\text{j}} \left| 
+\begin{array}{cc}
+ -5 & -8 \\
+ 2 & -2 \\
+\end{array}
+\right| =-\hat{\text{j}} ((-5)\, (-2)-(-8)\, \times \, 2)=-\hat{\text{j}} 26=\fbox{$-26 \hat{\text{j}}$}: \\
+  \text{= }-50 \hat{\text{i}}+\fbox{$-26 \hat{\text{j}}$}+\hat{\text{k}} \left| 
+\begin{array}{cc}
+ -5 & -3 \\
+ 2 & -7 \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \hat{\text{k}} \left| 
+\begin{array}{cc}
+ -5 & -3 \\
+ 2 & -7 \\
+\end{array}
+\right| =\hat{\text{k}} ((-5)\, (-7)-(-3)\, \times \, 2)=\hat{\text{k}} 41=\fbox{$41 \hat{\text{k}}$}: \\
+  \text{= }-50 \hat{\text{i}}-26 \hat{\text{j}}+\fbox{$41 \hat{\text{k}}$} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Order }\text{the }\text{terms }\text{in }\text{a }\text{more }\text{natural }\text{way}: \\
+  \text{= }-50 \hat{\text{i}}-26 \hat{\text{j}}+41 \hat{\text{k}} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ -50 \hat{\text{i}}-26 \hat{\text{j}}+41 \hat{\text{k}}=\, (-50,-26,41)\, : \\
+ \fbox{$
+\begin{array}{ll}
+ \text{Answer:} &  \\
+ \text{} & \, (-50,-26,41)\,  \\
+\end{array}
+$} \\
+\end{array}
+ \\
+\end{array}

pretraining/mathematica/linear_algebra/cross_product_w_steps/1146.txt

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+Problem:
+Find the cross product of the following vectors:
+$\left(
+\begin{array}{c}
+ 2 \\
+ 4 \\
+ 5 \\
+\end{array}
+\right)$ and
+$\left(
+\begin{array}{c}
+ -2 \\
+ 6 \\
+ -7 \\
+\end{array}
+\right)$
+Answer:
+\begin{array}{l}
+ 
+\begin{array}{l}
+ \text{Compute the following cross product}: \\
+ \, (2,4,5)\, \times \, (-2,6,-7)\,  \\
+\end{array}
+ \\
+\hline
+ 
+\begin{array}{l}
+ \text{Construct }\text{a }\text{matrix }\text{where }\text{the }\text{first }\text{row }\text{contains }\text{unit }\text{vectors }\hat{\text{i}}, \hat{\text{j}}, \text{and }\hat{\text{k}}; \text{and }\text{the }\text{second }\text{and }\text{third }\text{rows }\text{are }\text{made }\text{of }\text{vectors }\, (2,4,5)\,  \text{and }\, (-2,6,-7)\, : \\
+ \left(
+\begin{array}{ccc}
+ \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\
+ 2 & 4 & 5 \\
+ -2 & 6 & -7 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Take }\text{the }\text{determinant }\text{of }\text{this }\text{matrix}: \\
+ \left| 
+\begin{array}{ccc}
+ \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\
+ 2 & 4 & 5 \\
+ -2 & 6 & -7 \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Expand }\text{with }\text{respect }\text{to }\text{row }1: \\
+  \text{= }\left| 
+\begin{array}{ccc}
+ \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\
+ 2 & 4 & 5 \\
+ -2 & 6 & -7 \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{The }\text{determinant }\text{of }\text{the }\text{matrix }\left(
+\begin{array}{ccc}
+ \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\
+ 2 & 4 & 5 \\
+ -2 & 6 & -7 \\
+\end{array}
+\right) \text{is }\text{given }\text{by }\hat{\text{i}} \left| 
+\begin{array}{cc}
+ 4 & 5 \\
+ 6 & -7 \\
+\end{array}
+\right| +\left(-\hat{\text{j}}\right) \left| 
+\begin{array}{cc}
+ 2 & 5 \\
+ -2 & -7 \\
+\end{array}
+\right| +\hat{\text{k}} \left| 
+\begin{array}{cc}
+ 2 & 4 \\
+ -2 & 6 \\
+\end{array}
+\right| : \\
+  \text{= }\hat{\text{i}} \left| 
+\begin{array}{cc}
+ 4 & 5 \\
+ 6 & -7 \\
+\end{array}
+\right| +\left(-\hat{\text{j}}\right) \left| 
+\begin{array}{cc}
+ 2 & 5 \\
+ -2 & -7 \\
+\end{array}
+\right| +\hat{\text{k}} \left| 
+\begin{array}{cc}
+ 2 & 4 \\
+ -2 & 6 \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \hat{\text{i}} \left| 
+\begin{array}{cc}
+ 4 & 5 \\
+ 6 & -7 \\
+\end{array}
+\right| =\hat{\text{i}} (4 (-7)-5\ 6)=\hat{\text{i}} (-58)=\fbox{$-58 \hat{\text{i}}$}: \\
+  \text{= }\fbox{$-58 \hat{\text{i}}$}+\left(-\hat{\text{j}}\right) \left| 
+\begin{array}{cc}
+ 2 & 5 \\
+ -2 & -7 \\
+\end{array}
+\right| +\hat{\text{k}} \left| 
+\begin{array}{cc}
+ 2 & 4 \\
+ -2 & 6 \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ -\hat{\text{j}} \left| 
+\begin{array}{cc}
+ 2 & 5 \\
+ -2 & -7 \\
+\end{array}
+\right| =-\hat{\text{j}} (2 (-7)-5 (-2))=-\hat{\text{j}} (-4)=\fbox{$4 \hat{\text{j}}$}: \\
+  \text{= }-58 \hat{\text{i}}+\fbox{$4 \hat{\text{j}}$}+\hat{\text{k}} \left| 
+\begin{array}{cc}
+ 2 & 4 \\
+ -2 & 6 \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \hat{\text{k}} \left| 
+\begin{array}{cc}
+ 2 & 4 \\
+ -2 & 6 \\
+\end{array}
+\right| =\hat{\text{k}} (2\ 6-4 (-2))=\hat{\text{k}} 20=\fbox{$20 \hat{\text{k}}$}: \\
+  \text{= }-58 \hat{\text{i}}+4 \hat{\text{j}}+\fbox{$20 \hat{\text{k}}$} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Order }\text{the }\text{terms }\text{in }\text{a }\text{more }\text{natural }\text{way}: \\
+  \text{= }-58 \hat{\text{i}}+4 \hat{\text{j}}+20 \hat{\text{k}} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ -58 \hat{\text{i}}+4 \hat{\text{j}}+20 \hat{\text{k}}=\, (-58,4,20)\, : \\
+ \fbox{$
+\begin{array}{ll}
+ \text{Answer:} &  \\
+ \text{} & \, (-58,4,20)\,  \\
+\end{array}
+$} \\
+\end{array}
+ \\
+\end{array}

pretraining/mathematica/linear_algebra/cross_product_w_steps/117.txt

@@ -0,0 +1,169 @@
+Problem:
+Find the cross product of the following vectors:
+$\left(
+\begin{array}{c}
+ -10 \\
+ -2 \\
+ 1 \\
+\end{array}
+\right)$ and
+$\left(
+\begin{array}{c}
+ 9 \\
+ 4 \\
+ 2 \\
+\end{array}
+\right)$
+Answer:
+\begin{array}{l}
+ 
+\begin{array}{l}
+ \text{Compute the following cross product}: \\
+ \, (-10,-2,1)\, \times \, (9,4,2)\,  \\
+\end{array}
+ \\
+\hline
+ 
+\begin{array}{l}
+ \text{Construct }\text{a }\text{matrix }\text{where }\text{the }\text{first }\text{row }\text{contains }\text{unit }\text{vectors }\hat{\text{i}}, \hat{\text{j}}, \text{and }\hat{\text{k}}; \text{and }\text{the }\text{second }\text{and }\text{third }\text{rows }\text{are }\text{made }\text{of }\text{vectors }\, (-10,-2,1)\,  \text{and }\, (9,4,2)\, : \\
+ \left(
+\begin{array}{ccc}
+ \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\
+ -10 & -2 & 1 \\
+ 9 & 4 & 2 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Take }\text{the }\text{determinant }\text{of }\text{this }\text{matrix}: \\
+ \left| 
+\begin{array}{ccc}
+ \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\
+ -10 & -2 & 1 \\
+ 9 & 4 & 2 \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Expand }\text{with }\text{respect }\text{to }\text{row }1: \\
+  \text{= }\left| 
+\begin{array}{ccc}
+ \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\
+ -10 & -2 & 1 \\
+ 9 & 4 & 2 \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{The }\text{determinant }\text{of }\text{the }\text{matrix }\left(
+\begin{array}{ccc}
+ \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\
+ -10 & -2 & 1 \\
+ 9 & 4 & 2 \\
+\end{array}
+\right) \text{is }\text{given }\text{by }\hat{\text{i}} \left| 
+\begin{array}{cc}
+ -2 & 1 \\
+ 4 & 2 \\
+\end{array}
+\right| +\left(-\hat{\text{j}}\right) \left| 
+\begin{array}{cc}
+ -10 & 1 \\
+ 9 & 2 \\
+\end{array}
+\right| +\hat{\text{k}} \left| 
+\begin{array}{cc}
+ -10 & -2 \\
+ 9 & 4 \\
+\end{array}
+\right| : \\
+  \text{= }\hat{\text{i}} \left| 
+\begin{array}{cc}
+ -2 & 1 \\
+ 4 & 2 \\
+\end{array}
+\right| +\left(-\hat{\text{j}}\right) \left| 
+\begin{array}{cc}
+ -10 & 1 \\
+ 9 & 2 \\
+\end{array}
+\right| +\hat{\text{k}} \left| 
+\begin{array}{cc}
+ -10 & -2 \\
+ 9 & 4 \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \hat{\text{i}} \left| 
+\begin{array}{cc}
+ -2 & 1 \\
+ 4 & 2 \\
+\end{array}
+\right| =\hat{\text{i}} ((-2)\, \times \, 2-1\ 4)=\hat{\text{i}} (-8)=\fbox{$-8 \hat{\text{i}}$}: \\
+  \text{= }\fbox{$-8 \hat{\text{i}}$}+\left(-\hat{\text{j}}\right) \left| 
+\begin{array}{cc}
+ -10 & 1 \\
+ 9 & 2 \\
+\end{array}
+\right| +\hat{\text{k}} \left| 
+\begin{array}{cc}
+ -10 & -2 \\
+ 9 & 4 \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ -\hat{\text{j}} \left| 
+\begin{array}{cc}
+ -10 & 1 \\
+ 9 & 2 \\
+\end{array}
+\right| =-\hat{\text{j}} ((-10)\, \times \, 2-1\ 9)=-\hat{\text{j}} (-29)=\fbox{$29 \hat{\text{j}}$}: \\
+  \text{= }-8 \hat{\text{i}}+\fbox{$29 \hat{\text{j}}$}+\hat{\text{k}} \left| 
+\begin{array}{cc}
+ -10 & -2 \\
+ 9 & 4 \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \hat{\text{k}} \left| 
+\begin{array}{cc}
+ -10 & -2 \\
+ 9 & 4 \\
+\end{array}
+\right| =\hat{\text{k}} ((-10)\, \times \, 4-(-2)\, \times \, 9)=\hat{\text{k}} (-22)=\fbox{$-22 \hat{\text{k}}$}: \\
+  \text{= }-8 \hat{\text{i}}+29 \hat{\text{j}}+\fbox{$-22 \hat{\text{k}}$} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Order }\text{the }\text{terms }\text{in }\text{a }\text{more }\text{natural }\text{way}: \\
+  \text{= }-8 \hat{\text{i}}+29 \hat{\text{j}}-22 \hat{\text{k}} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ -8 \hat{\text{i}}+29 \hat{\text{j}}-22 \hat{\text{k}}=\, (-8,29,-22)\, : \\
+ \fbox{$
+\begin{array}{ll}
+ \text{Answer:} &  \\
+ \text{} & \, (-8,29,-22)\,  \\
+\end{array}
+$} \\
+\end{array}
+ \\
+\end{array}

pretraining/mathematica/linear_algebra/cross_product_w_steps/1191.txt

@@ -0,0 +1,169 @@
+Problem:
+Find the cross product of the following vectors:
+$\left(
+\begin{array}{c}
+ -\frac{35}{8} \\
+ \frac{117}{16} \\
+ -\frac{1}{4} \\
+\end{array}
+\right)$ and
+$\left(
+\begin{array}{c}
+ \frac{5}{8} \\
+ \frac{33}{4} \\
+ -\frac{129}{16} \\
+\end{array}
+\right)$
+Answer:
+\begin{array}{l}
+ 
+\begin{array}{l}
+ \text{Compute the following cross product}: \\
+ \, \left(-\frac{35}{8},\frac{117}{16},-\frac{1}{4}\right)\, \times \, \left(\frac{5}{8},\frac{33}{4},-\frac{129}{16}\right)\,  \\
+\end{array}
+ \\
+\hline
+ 
+\begin{array}{l}
+ \text{Construct }\text{a }\text{matrix }\text{where }\text{the }\text{first }\text{row }\text{contains }\text{unit }\text{vectors }\hat{\text{i}}, \hat{\text{j}}, \text{and }\hat{\text{k}}; \text{and }\text{the }\text{second }\text{and }\text{third }\text{rows }\text{are }\text{made }\text{of }\text{vectors }\, \left(-\frac{35}{8},\frac{117}{16},-\frac{1}{4}\right)\,  \text{and }\, \left(\frac{5}{8},\frac{33}{4},-\frac{129}{16}\right)\, : \\
+ \left(
+\begin{array}{ccc}
+ \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\
+ -\frac{35}{8} & \frac{117}{16} & -\frac{1}{4} \\
+ \frac{5}{8} & \frac{33}{4} & -\frac{129}{16} \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Take }\text{the }\text{determinant }\text{of }\text{this }\text{matrix}: \\
+ \left| 
+\begin{array}{ccc}
+ \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\
+ -\frac{35}{8} & \frac{117}{16} & -\frac{1}{4} \\
+ \frac{5}{8} & \frac{33}{4} & -\frac{129}{16} \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Expand }\text{with }\text{respect }\text{to }\text{row }1: \\
+  \text{= }\left| 
+\begin{array}{ccc}
+ \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\
+ -\frac{35}{8} & \frac{117}{16} & -\frac{1}{4} \\
+ \frac{5}{8} & \frac{33}{4} & -\frac{129}{16} \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{The }\text{determinant }\text{of }\text{the }\text{matrix }\left(
+\begin{array}{ccc}
+ \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\
+ -\frac{35}{8} & \frac{117}{16} & -\frac{1}{4} \\
+ \frac{5}{8} & \frac{33}{4} & -\frac{129}{16} \\
+\end{array}
+\right) \text{is }\text{given }\text{by }\hat{\text{i}} \left| 
+\begin{array}{cc}
+ \frac{117}{16} & -\frac{1}{4} \\
+ \frac{33}{4} & -\frac{129}{16} \\
+\end{array}
+\right| +\left(-\hat{\text{j}}\right) \left| 
+\begin{array}{cc}
+ -\frac{35}{8} & -\frac{1}{4} \\
+ \frac{5}{8} & -\frac{129}{16} \\
+\end{array}
+\right| +\hat{\text{k}} \left| 
+\begin{array}{cc}
+ -\frac{35}{8} & \frac{117}{16} \\
+ \frac{5}{8} & \frac{33}{4} \\
+\end{array}
+\right| : \\
+  \text{= }\hat{\text{i}} \left| 
+\begin{array}{cc}
+ \frac{117}{16} & -\frac{1}{4} \\
+ \frac{33}{4} & -\frac{129}{16} \\
+\end{array}
+\right| +\left(-\hat{\text{j}}\right) \left| 
+\begin{array}{cc}
+ -\frac{35}{8} & -\frac{1}{4} \\
+ \frac{5}{8} & -\frac{129}{16} \\
+\end{array}
+\right| +\hat{\text{k}} \left| 
+\begin{array}{cc}
+ -\frac{35}{8} & \frac{117}{16} \\
+ \frac{5}{8} & \frac{33}{4} \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \hat{\text{i}} \left| 
+\begin{array}{cc}
+ \frac{117}{16} & -\frac{1}{4} \\
+ \frac{33}{4} & -\frac{129}{16} \\
+\end{array}
+\right| =\hat{\text{i}} \left(\frac{117 (-129)}{16\ 16}-\frac{1}{4} \left(-\frac{33}{4}\right)\right)=\frac{\hat{\text{i}} (-14565)}{256}=\fbox{$-\frac{14565 \hat{\text{i}}}{256}$}: \\
+  \text{= }\fbox{$-\frac{14565 \hat{\text{i}}}{256}$}+\left(-\hat{\text{j}}\right) \left| 
+\begin{array}{cc}
+ -\frac{35}{8} & -\frac{1}{4} \\
+ \frac{5}{8} & -\frac{129}{16} \\
+\end{array}
+\right| +\hat{\text{k}} \left| 
+\begin{array}{cc}
+ -\frac{35}{8} & \frac{117}{16} \\
+ \frac{5}{8} & \frac{33}{4} \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ -\hat{\text{j}} \left| 
+\begin{array}{cc}
+ -\frac{35}{8} & -\frac{1}{4} \\
+ \frac{5}{8} & -\frac{129}{16} \\
+\end{array}
+\right| =-\hat{\text{j}} \left(\left(-\frac{35}{8}\right)\, \left(-\frac{129}{16}\right)-\frac{1}{4} \left(-\frac{5}{8}\right)\right)=\frac{-\hat{\text{j}} 4535}{128}=\fbox{$-\frac{4535 \hat{\text{j}}}{128}$}: \\
+  \text{= }\frac{-14565 \hat{\text{i}}}{256}+\fbox{$-\frac{4535 \hat{\text{j}}}{128}$}+\hat{\text{k}} \left| 
+\begin{array}{cc}
+ -\frac{35}{8} & \frac{117}{16} \\
+ \frac{5}{8} & \frac{33}{4} \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \hat{\text{k}} \left| 
+\begin{array}{cc}
+ -\frac{35}{8} & \frac{117}{16} \\
+ \frac{5}{8} & \frac{33}{4} \\
+\end{array}
+\right| =\hat{\text{k}} \left(\left(-\frac{35}{8}\right)\, \times \, \frac{33}{4}-\frac{117\ 5}{16\ 8}\right)=\frac{\hat{\text{k}} (-5205)}{128}=\fbox{$-\frac{5205 \hat{\text{k}}}{128}$}: \\
+  \text{= }\frac{-14565 \hat{\text{i}}}{256}-\frac{4535 \hat{\text{j}}}{128}+\fbox{$-\frac{5205 \hat{\text{k}}}{128}$} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Order }\text{the }\text{terms }\text{in }\text{a }\text{more }\text{natural }\text{way}: \\
+  \text{= }-\frac{14565 \hat{\text{i}}}{256}-\frac{4535 \hat{\text{j}}}{128}-\frac{5205 \hat{\text{k}}}{128} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ -\frac{14565 \hat{\text{i}}}{256}-\frac{4535 \hat{\text{j}}}{128}-\frac{5205 \hat{\text{k}}}{128}=\, \left(-\frac{14565}{256},-\frac{4535}{128},-\frac{5205}{128}\right)\, : \\
+ \fbox{$
+\begin{array}{ll}
+ \text{Answer:} &  \\
+ \text{} & \, \left(-\frac{14565}{256},-\frac{4535}{128},-\frac{5205}{128}\right)\,  \\
+\end{array}
+$} \\
+\end{array}
+ \\
+\end{array}

pretraining/mathematica/linear_algebra/cross_product_w_steps/1212.txt

@@ -0,0 +1,169 @@
+Problem:
+Find the cross product of the following vectors:
+$\left(
+\begin{array}{c}
+ -7 \\
+ 0 \\
+ 6 \\
+\end{array}
+\right)$ and
+$\left(
+\begin{array}{c}
+ 8 \\
+ 6 \\
+ 7 \\
+\end{array}
+\right)$
+Answer:
+\begin{array}{l}
+ 
+\begin{array}{l}
+ \text{Compute the following cross product}: \\
+ \, (-7,0,6)\, \times \, (8,6,7)\,  \\
+\end{array}
+ \\
+\hline
+ 
+\begin{array}{l}
+ \text{Construct }\text{a }\text{matrix }\text{where }\text{the }\text{first }\text{row }\text{contains }\text{unit }\text{vectors }\hat{\text{i}}, \hat{\text{j}}, \text{and }\hat{\text{k}}; \text{and }\text{the }\text{second }\text{and }\text{third }\text{rows }\text{are }\text{made }\text{of }\text{vectors }\, (-7,0,6)\,  \text{and }\, (8,6,7)\, : \\
+ \left(
+\begin{array}{ccc}
+ \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\
+ -7 & 0 & 6 \\
+ 8 & 6 & 7 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Take }\text{the }\text{determinant }\text{of }\text{this }\text{matrix}: \\
+ \left| 
+\begin{array}{ccc}
+ \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\
+ -7 & 0 & 6 \\
+ 8 & 6 & 7 \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Expand }\text{with }\text{respect }\text{to }\text{row }1: \\
+  \text{= }\left| 
+\begin{array}{ccc}
+ \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\
+ -7 & 0 & 6 \\
+ 8 & 6 & 7 \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{The }\text{determinant }\text{of }\text{the }\text{matrix }\left(
+\begin{array}{ccc}
+ \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\
+ -7 & 0 & 6 \\
+ 8 & 6 & 7 \\
+\end{array}
+\right) \text{is }\text{given }\text{by }\hat{\text{i}} \left| 
+\begin{array}{cc}
+ 0 & 6 \\
+ 6 & 7 \\
+\end{array}
+\right| +\left(-\hat{\text{j}}\right) \left| 
+\begin{array}{cc}
+ -7 & 6 \\
+ 8 & 7 \\
+\end{array}
+\right| +\hat{\text{k}} \left| 
+\begin{array}{cc}
+ -7 & 0 \\
+ 8 & 6 \\
+\end{array}
+\right| : \\
+  \text{= }\hat{\text{i}} \left| 
+\begin{array}{cc}
+ 0 & 6 \\
+ 6 & 7 \\
+\end{array}
+\right| +\left(-\hat{\text{j}}\right) \left| 
+\begin{array}{cc}
+ -7 & 6 \\
+ 8 & 7 \\
+\end{array}
+\right| +\hat{\text{k}} \left| 
+\begin{array}{cc}
+ -7 & 0 \\
+ 8 & 6 \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \hat{\text{i}} \left| 
+\begin{array}{cc}
+ 0 & 6 \\
+ 6 & 7 \\
+\end{array}
+\right| =\hat{\text{i}} (0\ 7-6\ 6)=\hat{\text{i}} (-36)=\fbox{$-36 \hat{\text{i}}$}: \\
+  \text{= }\fbox{$-36 \hat{\text{i}}$}+\left(-\hat{\text{j}}\right) \left| 
+\begin{array}{cc}
+ -7 & 6 \\
+ 8 & 7 \\
+\end{array}
+\right| +\hat{\text{k}} \left| 
+\begin{array}{cc}
+ -7 & 0 \\
+ 8 & 6 \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ -\hat{\text{j}} \left| 
+\begin{array}{cc}
+ -7 & 6 \\
+ 8 & 7 \\
+\end{array}
+\right| =-\hat{\text{j}} ((-7)\, \times \, 7-6\ 8)=-\hat{\text{j}} (-97)=\fbox{$97 \hat{\text{j}}$}: \\
+  \text{= }-36 \hat{\text{i}}+\fbox{$97 \hat{\text{j}}$}+\hat{\text{k}} \left| 
+\begin{array}{cc}
+ -7 & 0 \\
+ 8 & 6 \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \hat{\text{k}} \left| 
+\begin{array}{cc}
+ -7 & 0 \\
+ 8 & 6 \\
+\end{array}
+\right| =\hat{\text{k}} ((-7)\, \times \, 6-0\ 8)=\hat{\text{k}} (-42)=\fbox{$-42 \hat{\text{k}}$}: \\
+  \text{= }-36 \hat{\text{i}}+97 \hat{\text{j}}+\fbox{$-42 \hat{\text{k}}$} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Order }\text{the }\text{terms }\text{in }\text{a }\text{more }\text{natural }\text{way}: \\
+  \text{= }-36 \hat{\text{i}}+97 \hat{\text{j}}-42 \hat{\text{k}} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ -36 \hat{\text{i}}+97 \hat{\text{j}}-42 \hat{\text{k}}=\, (-36,97,-42)\, : \\
+ \fbox{$
+\begin{array}{ll}
+ \text{Answer:} &  \\
+ \text{} & \, (-36,97,-42)\,  \\
+\end{array}
+$} \\
+\end{array}
+ \\
+\end{array}

pretraining/mathematica/linear_algebra/cross_product_w_steps/1279.txt

@@ -0,0 +1,169 @@
+Problem:
+Find the cross product of the following vectors:
+$\left(
+\begin{array}{c}
+ -10 \\
+ 5 \\
+ 8 \\
+\end{array}
+\right)$ and
+$\left(
+\begin{array}{c}
+ 6 \\
+ -5 \\
+ 8 \\
+\end{array}
+\right)$
+Answer:
+\begin{array}{l}
+ 
+\begin{array}{l}
+ \text{Compute the following cross product}: \\
+ \, (-10,5,8)\, \times \, (6,-5,8)\,  \\
+\end{array}
+ \\
+\hline
+ 
+\begin{array}{l}
+ \text{Construct }\text{a }\text{matrix }\text{where }\text{the }\text{first }\text{row }\text{contains }\text{unit }\text{vectors }\hat{\text{i}}, \hat{\text{j}}, \text{and }\hat{\text{k}}; \text{and }\text{the }\text{second }\text{and }\text{third }\text{rows }\text{are }\text{made }\text{of }\text{vectors }\, (-10,5,8)\,  \text{and }\, (6,-5,8)\, : \\
+ \left(
+\begin{array}{ccc}
+ \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\
+ -10 & 5 & 8 \\
+ 6 & -5 & 8 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Take }\text{the }\text{determinant }\text{of }\text{this }\text{matrix}: \\
+ \left| 
+\begin{array}{ccc}
+ \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\
+ -10 & 5 & 8 \\
+ 6 & -5 & 8 \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Expand }\text{with }\text{respect }\text{to }\text{row }1: \\
+  \text{= }\left| 
+\begin{array}{ccc}
+ \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\
+ -10 & 5 & 8 \\
+ 6 & -5 & 8 \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{The }\text{determinant }\text{of }\text{the }\text{matrix }\left(
+\begin{array}{ccc}
+ \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\
+ -10 & 5 & 8 \\
+ 6 & -5 & 8 \\
+\end{array}
+\right) \text{is }\text{given }\text{by }\hat{\text{i}} \left| 
+\begin{array}{cc}
+ 5 & 8 \\
+ -5 & 8 \\
+\end{array}
+\right| +\left(-\hat{\text{j}}\right) \left| 
+\begin{array}{cc}
+ -10 & 8 \\
+ 6 & 8 \\
+\end{array}
+\right| +\hat{\text{k}} \left| 
+\begin{array}{cc}
+ -10 & 5 \\
+ 6 & -5 \\
+\end{array}
+\right| : \\
+  \text{= }\hat{\text{i}} \left| 
+\begin{array}{cc}
+ 5 & 8 \\
+ -5 & 8 \\
+\end{array}
+\right| +\left(-\hat{\text{j}}\right) \left| 
+\begin{array}{cc}
+ -10 & 8 \\
+ 6 & 8 \\
+\end{array}
+\right| +\hat{\text{k}} \left| 
+\begin{array}{cc}
+ -10 & 5 \\
+ 6 & -5 \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \hat{\text{i}} \left| 
+\begin{array}{cc}
+ 5 & 8 \\
+ -5 & 8 \\
+\end{array}
+\right| =\hat{\text{i}} (5\ 8-8 (-5))=\hat{\text{i}} 80=\fbox{$80 \hat{\text{i}}$}: \\
+  \text{= }\fbox{$80 \hat{\text{i}}$}+\left(-\hat{\text{j}}\right) \left| 
+\begin{array}{cc}
+ -10 & 8 \\
+ 6 & 8 \\
+\end{array}
+\right| +\hat{\text{k}} \left| 
+\begin{array}{cc}
+ -10 & 5 \\
+ 6 & -5 \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ -\hat{\text{j}} \left| 
+\begin{array}{cc}
+ -10 & 8 \\
+ 6 & 8 \\
+\end{array}
+\right| =-\hat{\text{j}} ((-10)\, \times \, 8-8\ 6)=-\hat{\text{j}} (-128)=\fbox{$128 \hat{\text{j}}$}: \\
+  \text{= }80 \hat{\text{i}}+\fbox{$128 \hat{\text{j}}$}+\hat{\text{k}} \left| 
+\begin{array}{cc}
+ -10 & 5 \\
+ 6 & -5 \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \hat{\text{k}} \left| 
+\begin{array}{cc}
+ -10 & 5 \\
+ 6 & -5 \\
+\end{array}
+\right| =\hat{\text{k}} ((-10)\, (-5)-5\ 6)=\hat{\text{k}} 20=\fbox{$20 \hat{\text{k}}$}: \\
+  \text{= }80 \hat{\text{i}}+128 \hat{\text{j}}+\fbox{$20 \hat{\text{k}}$} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Order }\text{the }\text{terms }\text{in }\text{a }\text{more }\text{natural }\text{way}: \\
+  \text{= }80 \hat{\text{i}}+128 \hat{\text{j}}+20 \hat{\text{k}} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ 80 \hat{\text{i}}+128 \hat{\text{j}}+20 \hat{\text{k}}=\, (80,128,20)\, : \\
+ \fbox{$
+\begin{array}{ll}
+ \text{Answer:} &  \\
+ \text{} & \, (80,128,20)\,  \\
+\end{array}
+$} \\
+\end{array}
+ \\
+\end{array}

pretraining/mathematica/linear_algebra/cross_product_w_steps/1392.txt

@@ -0,0 +1,169 @@
+Problem:
+Find the cross product of the following vectors:
+$\left(
+\begin{array}{c}
+ 9 \\
+ -5 \\
+ -7 \\
+\end{array}
+\right)$ and
+$\left(
+\begin{array}{c}
+ 10 \\
+ 3 \\
+ 2 \\
+\end{array}
+\right)$
+Answer:
+\begin{array}{l}
+ 
+\begin{array}{l}
+ \text{Compute the following cross product}: \\
+ \, (9,-5,-7)\, \times \, (10,3,2)\,  \\
+\end{array}
+ \\
+\hline
+ 
+\begin{array}{l}
+ \text{Construct }\text{a }\text{matrix }\text{where }\text{the }\text{first }\text{row }\text{contains }\text{unit }\text{vectors }\hat{\text{i}}, \hat{\text{j}}, \text{and }\hat{\text{k}}; \text{and }\text{the }\text{second }\text{and }\text{third }\text{rows }\text{are }\text{made }\text{of }\text{vectors }\, (9,-5,-7)\,  \text{and }\, (10,3,2)\, : \\
+ \left(
+\begin{array}{ccc}
+ \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\
+ 9 & -5 & -7 \\
+ 10 & 3 & 2 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Take }\text{the }\text{determinant }\text{of }\text{this }\text{matrix}: \\
+ \left| 
+\begin{array}{ccc}
+ \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\
+ 9 & -5 & -7 \\
+ 10 & 3 & 2 \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Expand }\text{with }\text{respect }\text{to }\text{row }1: \\
+  \text{= }\left| 
+\begin{array}{ccc}
+ \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\
+ 9 & -5 & -7 \\
+ 10 & 3 & 2 \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{The }\text{determinant }\text{of }\text{the }\text{matrix }\left(
+\begin{array}{ccc}
+ \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\
+ 9 & -5 & -7 \\
+ 10 & 3 & 2 \\
+\end{array}
+\right) \text{is }\text{given }\text{by }\hat{\text{i}} \left| 
+\begin{array}{cc}
+ -5 & -7 \\
+ 3 & 2 \\
+\end{array}
+\right| +\left(-\hat{\text{j}}\right) \left| 
+\begin{array}{cc}
+ 9 & -7 \\
+ 10 & 2 \\
+\end{array}
+\right| +\hat{\text{k}} \left| 
+\begin{array}{cc}
+ 9 & -5 \\
+ 10 & 3 \\
+\end{array}
+\right| : \\
+  \text{= }\hat{\text{i}} \left| 
+\begin{array}{cc}
+ -5 & -7 \\
+ 3 & 2 \\
+\end{array}
+\right| +\left(-\hat{\text{j}}\right) \left| 
+\begin{array}{cc}
+ 9 & -7 \\
+ 10 & 2 \\
+\end{array}
+\right| +\hat{\text{k}} \left| 
+\begin{array}{cc}
+ 9 & -5 \\
+ 10 & 3 \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \hat{\text{i}} \left| 
+\begin{array}{cc}
+ -5 & -7 \\
+ 3 & 2 \\
+\end{array}
+\right| =\hat{\text{i}} ((-5)\, \times \, 2-(-7)\, \times \, 3)=\hat{\text{i}} 11=\fbox{$11 \hat{\text{i}}$}: \\
+  \text{= }\fbox{$11 \hat{\text{i}}$}+\left(-\hat{\text{j}}\right) \left| 
+\begin{array}{cc}
+ 9 & -7 \\
+ 10 & 2 \\
+\end{array}
+\right| +\hat{\text{k}} \left| 
+\begin{array}{cc}
+ 9 & -5 \\
+ 10 & 3 \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ -\hat{\text{j}} \left| 
+\begin{array}{cc}
+ 9 & -7 \\
+ 10 & 2 \\
+\end{array}
+\right| =-\hat{\text{j}} (9\ 2-(-7)\, \times \, 10)=-\hat{\text{j}} 88=\fbox{$-88 \hat{\text{j}}$}: \\
+  \text{= }11 \hat{\text{i}}+\fbox{$-88 \hat{\text{j}}$}+\hat{\text{k}} \left| 
+\begin{array}{cc}
+ 9 & -5 \\
+ 10 & 3 \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \hat{\text{k}} \left| 
+\begin{array}{cc}
+ 9 & -5 \\
+ 10 & 3 \\
+\end{array}
+\right| =\hat{\text{k}} (9\ 3-(-5)\, \times \, 10)=\hat{\text{k}} 77=\fbox{$77 \hat{\text{k}}$}: \\
+  \text{= }11 \hat{\text{i}}-88 \hat{\text{j}}+\fbox{$77 \hat{\text{k}}$} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Order }\text{the }\text{terms }\text{in }\text{a }\text{more }\text{natural }\text{way}: \\
+  \text{= }11 \hat{\text{i}}-88 \hat{\text{j}}+77 \hat{\text{k}} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ 11 \hat{\text{i}}-88 \hat{\text{j}}+77 \hat{\text{k}}=\, (11,-88,77)\, : \\
+ \fbox{$
+\begin{array}{ll}
+ \text{Answer:} &  \\
+ \text{} & \, (11,-88,77)\,  \\
+\end{array}
+$} \\
+\end{array}
+ \\
+\end{array}

pretraining/mathematica/linear_algebra/cross_product_w_steps/1396.txt

@@ -0,0 +1,169 @@
+Problem:
+Find the cross product of the following vectors:
+$\left(
+\begin{array}{c}
+ 4 \\
+ -10 \\
+ 5 \\
+\end{array}
+\right)$ and
+$\left(
+\begin{array}{c}
+ -10 \\
+ -5 \\
+ 2 \\
+\end{array}
+\right)$
+Answer:
+\begin{array}{l}
+ 
+\begin{array}{l}
+ \text{Compute the following cross product}: \\
+ \, (4,-10,5)\, \times \, (-10,-5,2)\,  \\
+\end{array}
+ \\
+\hline
+ 
+\begin{array}{l}
+ \text{Construct }\text{a }\text{matrix }\text{where }\text{the }\text{first }\text{row }\text{contains }\text{unit }\text{vectors }\hat{\text{i}}, \hat{\text{j}}, \text{and }\hat{\text{k}}; \text{and }\text{the }\text{second }\text{and }\text{third }\text{rows }\text{are }\text{made }\text{of }\text{vectors }\, (4,-10,5)\,  \text{and }\, (-10,-5,2)\, : \\
+ \left(
+\begin{array}{ccc}
+ \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\
+ 4 & -10 & 5 \\
+ -10 & -5 & 2 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Take }\text{the }\text{determinant }\text{of }\text{this }\text{matrix}: \\
+ \left| 
+\begin{array}{ccc}
+ \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\
+ 4 & -10 & 5 \\
+ -10 & -5 & 2 \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Expand }\text{with }\text{respect }\text{to }\text{row }1: \\
+  \text{= }\left| 
+\begin{array}{ccc}
+ \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\
+ 4 & -10 & 5 \\
+ -10 & -5 & 2 \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{The }\text{determinant }\text{of }\text{the }\text{matrix }\left(
+\begin{array}{ccc}
+ \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\
+ 4 & -10 & 5 \\
+ -10 & -5 & 2 \\
+\end{array}
+\right) \text{is }\text{given }\text{by }\hat{\text{i}} \left| 
+\begin{array}{cc}
+ -10 & 5 \\
+ -5 & 2 \\
+\end{array}
+\right| +\left(-\hat{\text{j}}\right) \left| 
+\begin{array}{cc}
+ 4 & 5 \\
+ -10 & 2 \\
+\end{array}
+\right| +\hat{\text{k}} \left| 
+\begin{array}{cc}
+ 4 & -10 \\
+ -10 & -5 \\
+\end{array}
+\right| : \\
+  \text{= }\hat{\text{i}} \left| 
+\begin{array}{cc}
+ -10 & 5 \\
+ -5 & 2 \\
+\end{array}
+\right| +\left(-\hat{\text{j}}\right) \left| 
+\begin{array}{cc}
+ 4 & 5 \\
+ -10 & 2 \\
+\end{array}
+\right| +\hat{\text{k}} \left| 
+\begin{array}{cc}
+ 4 & -10 \\
+ -10 & -5 \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \hat{\text{i}} \left| 
+\begin{array}{cc}
+ -10 & 5 \\
+ -5 & 2 \\
+\end{array}
+\right| =\hat{\text{i}} ((-10)\, \times \, 2-5 (-5))=\hat{\text{i}} 5=\fbox{$5 \hat{\text{i}}$}: \\
+  \text{= }\fbox{$5 \hat{\text{i}}$}+\left(-\hat{\text{j}}\right) \left| 
+\begin{array}{cc}
+ 4 & 5 \\
+ -10 & 2 \\
+\end{array}
+\right| +\hat{\text{k}} \left| 
+\begin{array}{cc}
+ 4 & -10 \\
+ -10 & -5 \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ -\hat{\text{j}} \left| 
+\begin{array}{cc}
+ 4 & 5 \\
+ -10 & 2 \\
+\end{array}
+\right| =-\hat{\text{j}} (4\ 2-5 (-10))=-\hat{\text{j}} 58=\fbox{$-58 \hat{\text{j}}$}: \\
+  \text{= }5 \hat{\text{i}}+\fbox{$-58 \hat{\text{j}}$}+\hat{\text{k}} \left| 
+\begin{array}{cc}
+ 4 & -10 \\
+ -10 & -5 \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \hat{\text{k}} \left| 
+\begin{array}{cc}
+ 4 & -10 \\
+ -10 & -5 \\
+\end{array}
+\right| =\hat{\text{k}} (4 (-5)-(-10)\, (-10))=\hat{\text{k}} (-120)=\fbox{$-120 \hat{\text{k}}$}: \\
+  \text{= }5 \hat{\text{i}}-58 \hat{\text{j}}+\fbox{$-120 \hat{\text{k}}$} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Order }\text{the }\text{terms }\text{in }\text{a }\text{more }\text{natural }\text{way}: \\
+  \text{= }5 \hat{\text{i}}-58 \hat{\text{j}}-120 \hat{\text{k}} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ 5 \hat{\text{i}}-58 \hat{\text{j}}-120 \hat{\text{k}}=\, (5,-58,-120)\, : \\
+ \fbox{$
+\begin{array}{ll}
+ \text{Answer:} &  \\
+ \text{} & \, (5,-58,-120)\,  \\
+\end{array}
+$} \\
+\end{array}
+ \\
+\end{array}

pretraining/mathematica/linear_algebra/cross_product_w_steps/1416.txt

@@ -0,0 +1,169 @@
+Problem:
+Find the cross product of the following vectors:
+$\left(
+\begin{array}{c}
+ -1 \\
+ -8 \\
+ -2 \\
+\end{array}
+\right)$ and
+$\left(
+\begin{array}{c}
+ -5 \\
+ 8 \\
+ 4 \\
+\end{array}
+\right)$
+Answer:
+\begin{array}{l}
+ 
+\begin{array}{l}
+ \text{Compute the following cross product}: \\
+ \, (-1,-8,-2)\, \times \, (-5,8,4)\,  \\
+\end{array}
+ \\
+\hline
+ 
+\begin{array}{l}
+ \text{Construct }\text{a }\text{matrix }\text{where }\text{the }\text{first }\text{row }\text{contains }\text{unit }\text{vectors }\hat{\text{i}}, \hat{\text{j}}, \text{and }\hat{\text{k}}; \text{and }\text{the }\text{second }\text{and }\text{third }\text{rows }\text{are }\text{made }\text{of }\text{vectors }\, (-1,-8,-2)\,  \text{and }\, (-5,8,4)\, : \\
+ \left(
+\begin{array}{ccc}
+ \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\
+ -1 & -8 & -2 \\
+ -5 & 8 & 4 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Take }\text{the }\text{determinant }\text{of }\text{this }\text{matrix}: \\
+ \left| 
+\begin{array}{ccc}
+ \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\
+ -1 & -8 & -2 \\
+ -5 & 8 & 4 \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Expand }\text{with }\text{respect }\text{to }\text{row }1: \\
+  \text{= }\left| 
+\begin{array}{ccc}
+ \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\
+ -1 & -8 & -2 \\
+ -5 & 8 & 4 \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{The }\text{determinant }\text{of }\text{the }\text{matrix }\left(
+\begin{array}{ccc}
+ \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\
+ -1 & -8 & -2 \\
+ -5 & 8 & 4 \\
+\end{array}
+\right) \text{is }\text{given }\text{by }\hat{\text{i}} \left| 
+\begin{array}{cc}
+ -8 & -2 \\
+ 8 & 4 \\
+\end{array}
+\right| +\left(-\hat{\text{j}}\right) \left| 
+\begin{array}{cc}
+ -1 & -2 \\
+ -5 & 4 \\
+\end{array}
+\right| +\hat{\text{k}} \left| 
+\begin{array}{cc}
+ -1 & -8 \\
+ -5 & 8 \\
+\end{array}
+\right| : \\
+  \text{= }\hat{\text{i}} \left| 
+\begin{array}{cc}
+ -8 & -2 \\
+ 8 & 4 \\
+\end{array}
+\right| +\left(-\hat{\text{j}}\right) \left| 
+\begin{array}{cc}
+ -1 & -2 \\
+ -5 & 4 \\
+\end{array}
+\right| +\hat{\text{k}} \left| 
+\begin{array}{cc}
+ -1 & -8 \\
+ -5 & 8 \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \hat{\text{i}} \left| 
+\begin{array}{cc}
+ -8 & -2 \\
+ 8 & 4 \\
+\end{array}
+\right| =\hat{\text{i}} ((-8)\, \times \, 4-(-2)\, \times \, 8)=\hat{\text{i}} (-16)=\fbox{$-16 \hat{\text{i}}$}: \\
+  \text{= }\fbox{$-16 \hat{\text{i}}$}+\left(-\hat{\text{j}}\right) \left| 
+\begin{array}{cc}
+ -1 & -2 \\
+ -5 & 4 \\
+\end{array}
+\right| +\hat{\text{k}} \left| 
+\begin{array}{cc}
+ -1 & -8 \\
+ -5 & 8 \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ -\hat{\text{j}} \left| 
+\begin{array}{cc}
+ -1 & -2 \\
+ -5 & 4 \\
+\end{array}
+\right| =-\hat{\text{j}} ((-1)\, \times \, 4-(-2)\, (-5))=-\hat{\text{j}} (-14)=\fbox{$14 \hat{\text{j}}$}: \\
+  \text{= }-16 \hat{\text{i}}+\fbox{$14 \hat{\text{j}}$}+\hat{\text{k}} \left| 
+\begin{array}{cc}
+ -1 & -8 \\
+ -5 & 8 \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \hat{\text{k}} \left| 
+\begin{array}{cc}
+ -1 & -8 \\
+ -5 & 8 \\
+\end{array}
+\right| =\hat{\text{k}} ((-1)\, \times \, 8-(-8)\, (-5))=\hat{\text{k}} (-48)=\fbox{$-48 \hat{\text{k}}$}: \\
+  \text{= }-16 \hat{\text{i}}+14 \hat{\text{j}}+\fbox{$-48 \hat{\text{k}}$} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Order }\text{the }\text{terms }\text{in }\text{a }\text{more }\text{natural }\text{way}: \\
+  \text{= }-16 \hat{\text{i}}+14 \hat{\text{j}}-48 \hat{\text{k}} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ -16 \hat{\text{i}}+14 \hat{\text{j}}-48 \hat{\text{k}}=\, (-16,14,-48)\, : \\
+ \fbox{$
+\begin{array}{ll}
+ \text{Answer:} &  \\
+ \text{} & \, (-16,14,-48)\,  \\
+\end{array}
+$} \\
+\end{array}
+ \\
+\end{array}

pretraining/mathematica/linear_algebra/cross_product_w_steps/1535.txt

@@ -0,0 +1,169 @@
+Problem:
+Find the cross product of the following vectors:
+$\left(
+\begin{array}{c}
+ -9 \\
+ -\frac{8}{5} \\
+ -\frac{3}{5} \\
+\end{array}
+\right)$ and
+$\left(
+\begin{array}{c}
+ -\frac{42}{5} \\
+ \frac{87}{10} \\
+ \frac{51}{10} \\
+\end{array}
+\right)$
+Answer:
+\begin{array}{l}
+ 
+\begin{array}{l}
+ \text{Compute the following cross product}: \\
+ \, \left(-9,-\frac{8}{5},-\frac{3}{5}\right)\, \times \, \left(-\frac{42}{5},\frac{87}{10},\frac{51}{10}\right)\,  \\
+\end{array}
+ \\
+\hline
+ 
+\begin{array}{l}
+ \text{Construct }\text{a }\text{matrix }\text{where }\text{the }\text{first }\text{row }\text{contains }\text{unit }\text{vectors }\hat{\text{i}}, \hat{\text{j}}, \text{and }\hat{\text{k}}; \text{and }\text{the }\text{second }\text{and }\text{third }\text{rows }\text{are }\text{made }\text{of }\text{vectors }\, \left(-9,-\frac{8}{5},-\frac{3}{5}\right)\,  \text{and }\, \left(-\frac{42}{5},\frac{87}{10},\frac{51}{10}\right)\, : \\
+ \left(
+\begin{array}{ccc}
+ \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\
+ -9 & -\frac{8}{5} & -\frac{3}{5} \\
+ -\frac{42}{5} & \frac{87}{10} & \frac{51}{10} \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Take }\text{the }\text{determinant }\text{of }\text{this }\text{matrix}: \\
+ \left| 
+\begin{array}{ccc}
+ \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\
+ -9 & -\frac{8}{5} & -\frac{3}{5} \\
+ -\frac{42}{5} & \frac{87}{10} & \frac{51}{10} \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Expand }\text{with }\text{respect }\text{to }\text{row }1: \\
+  \text{= }\left| 
+\begin{array}{ccc}
+ \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\
+ -9 & -\frac{8}{5} & -\frac{3}{5} \\
+ -\frac{42}{5} & \frac{87}{10} & \frac{51}{10} \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{The }\text{determinant }\text{of }\text{the }\text{matrix }\left(
+\begin{array}{ccc}
+ \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\
+ -9 & -\frac{8}{5} & -\frac{3}{5} \\
+ -\frac{42}{5} & \frac{87}{10} & \frac{51}{10} \\
+\end{array}
+\right) \text{is }\text{given }\text{by }\hat{\text{i}} \left| 
+\begin{array}{cc}
+ -\frac{8}{5} & -\frac{3}{5} \\
+ \frac{87}{10} & \frac{51}{10} \\
+\end{array}
+\right| +\left(-\hat{\text{j}}\right) \left| 
+\begin{array}{cc}
+ -9 & -\frac{3}{5} \\
+ -\frac{42}{5} & \frac{51}{10} \\
+\end{array}
+\right| +\hat{\text{k}} \left| 
+\begin{array}{cc}
+ -9 & -\frac{8}{5} \\
+ -\frac{42}{5} & \frac{87}{10} \\
+\end{array}
+\right| : \\
+  \text{= }\hat{\text{i}} \left| 
+\begin{array}{cc}
+ -\frac{8}{5} & -\frac{3}{5} \\
+ \frac{87}{10} & \frac{51}{10} \\
+\end{array}
+\right| +\left(-\hat{\text{j}}\right) \left| 
+\begin{array}{cc}
+ -9 & -\frac{3}{5} \\
+ -\frac{42}{5} & \frac{51}{10} \\
+\end{array}
+\right| +\hat{\text{k}} \left| 
+\begin{array}{cc}
+ -9 & -\frac{8}{5} \\
+ -\frac{42}{5} & \frac{87}{10} \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \hat{\text{i}} \left| 
+\begin{array}{cc}
+ -\frac{8}{5} & -\frac{3}{5} \\
+ \frac{87}{10} & \frac{51}{10} \\
+\end{array}
+\right| =\hat{\text{i}} \left(\left(-\frac{8}{5}\right)\, \times \, \frac{51}{10}-\left(-\frac{3}{5}\right)\, \times \, \frac{87}{10}\right)=\frac{\hat{\text{i}} (-147)}{50}=\fbox{$-\frac{147 \hat{\text{i}}}{50}$}: \\
+  \text{= }\fbox{$-\frac{147 \hat{\text{i}}}{50}$}+\left(-\hat{\text{j}}\right) \left| 
+\begin{array}{cc}
+ -9 & -\frac{3}{5} \\
+ -\frac{42}{5} & \frac{51}{10} \\
+\end{array}
+\right| +\hat{\text{k}} \left| 
+\begin{array}{cc}
+ -9 & -\frac{8}{5} \\
+ -\frac{42}{5} & \frac{87}{10} \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ -\hat{\text{j}} \left| 
+\begin{array}{cc}
+ -9 & -\frac{3}{5} \\
+ -\frac{42}{5} & \frac{51}{10} \\
+\end{array}
+\right| =-\hat{\text{j}} \left((-9)\, \times \, \frac{51}{10}-\left(-\frac{3}{5}\right)\, \left(-\frac{42}{5}\right)\right)=\frac{-\hat{\text{j}} (-2547)}{50}=\fbox{$\frac{2547 \hat{\text{j}}}{50}$}: \\
+  \text{= }\frac{-147 \hat{\text{i}}}{50}+\fbox{$\frac{2547 \hat{\text{j}}}{50}$}+\hat{\text{k}} \left| 
+\begin{array}{cc}
+ -9 & -\frac{8}{5} \\
+ -\frac{42}{5} & \frac{87}{10} \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \hat{\text{k}} \left| 
+\begin{array}{cc}
+ -9 & -\frac{8}{5} \\
+ -\frac{42}{5} & \frac{87}{10} \\
+\end{array}
+\right| =\hat{\text{k}} \left((-9)\, \times \, \frac{87}{10}-\left(-\frac{8}{5}\right)\, \left(-\frac{42}{5}\right)\right)=\frac{\hat{\text{k}} (-4587)}{50}=\fbox{$-\frac{4587 \hat{\text{k}}}{50}$}: \\
+  \text{= }\frac{-147 \hat{\text{i}}}{50}+\frac{2547 \hat{\text{j}}}{50}+\fbox{$-\frac{4587 \hat{\text{k}}}{50}$} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Order }\text{the }\text{terms }\text{in }\text{a }\text{more }\text{natural }\text{way}: \\
+  \text{= }-\frac{147 \hat{\text{i}}}{50}+\frac{2547 \hat{\text{j}}}{50}-\frac{4587 \hat{\text{k}}}{50} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ -\frac{147 \hat{\text{i}}}{50}+\frac{2547 \hat{\text{j}}}{50}-\frac{4587 \hat{\text{k}}}{50}=\, \left(-\frac{147}{50},\frac{2547}{50},-\frac{4587}{50}\right)\, : \\
+ \fbox{$
+\begin{array}{ll}
+ \text{Answer:} &  \\
+ \text{} & \, \left(-\frac{147}{50},\frac{2547}{50},-\frac{4587}{50}\right)\,  \\
+\end{array}
+$} \\
+\end{array}
+ \\
+\end{array}

pretraining/mathematica/linear_algebra/cross_product_w_steps/1605.txt

@@ -0,0 +1,169 @@
+Problem:
+Find the cross product of the following vectors:
+$\left(
+\begin{array}{c}
+ -\frac{39}{4} \\
+ -\frac{13}{4} \\
+ \frac{35}{4} \\
+\end{array}
+\right)$ and
+$\left(
+\begin{array}{c}
+ \frac{21}{4} \\
+ -\frac{7}{2} \\
+ -\frac{3}{4} \\
+\end{array}
+\right)$
+Answer:
+\begin{array}{l}
+ 
+\begin{array}{l}
+ \text{Compute the following cross product}: \\
+ \, \left(-\frac{39}{4},-\frac{13}{4},\frac{35}{4}\right)\, \times \, \left(\frac{21}{4},-\frac{7}{2},-\frac{3}{4}\right)\,  \\
+\end{array}
+ \\
+\hline
+ 
+\begin{array}{l}
+ \text{Construct }\text{a }\text{matrix }\text{where }\text{the }\text{first }\text{row }\text{contains }\text{unit }\text{vectors }\hat{\text{i}}, \hat{\text{j}}, \text{and }\hat{\text{k}}; \text{and }\text{the }\text{second }\text{and }\text{third }\text{rows }\text{are }\text{made }\text{of }\text{vectors }\, \left(-\frac{39}{4},-\frac{13}{4},\frac{35}{4}\right)\,  \text{and }\, \left(\frac{21}{4},-\frac{7}{2},-\frac{3}{4}\right)\, : \\
+ \left(
+\begin{array}{ccc}
+ \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\
+ -\frac{39}{4} & -\frac{13}{4} & \frac{35}{4} \\
+ \frac{21}{4} & -\frac{7}{2} & -\frac{3}{4} \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Take }\text{the }\text{determinant }\text{of }\text{this }\text{matrix}: \\
+ \left| 
+\begin{array}{ccc}
+ \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\
+ -\frac{39}{4} & -\frac{13}{4} & \frac{35}{4} \\
+ \frac{21}{4} & -\frac{7}{2} & -\frac{3}{4} \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Expand }\text{with }\text{respect }\text{to }\text{row }1: \\
+  \text{= }\left| 
+\begin{array}{ccc}
+ \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\
+ -\frac{39}{4} & -\frac{13}{4} & \frac{35}{4} \\
+ \frac{21}{4} & -\frac{7}{2} & -\frac{3}{4} \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{The }\text{determinant }\text{of }\text{the }\text{matrix }\left(
+\begin{array}{ccc}
+ \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\
+ -\frac{39}{4} & -\frac{13}{4} & \frac{35}{4} \\
+ \frac{21}{4} & -\frac{7}{2} & -\frac{3}{4} \\
+\end{array}
+\right) \text{is }\text{given }\text{by }\hat{\text{i}} \left| 
+\begin{array}{cc}
+ -\frac{13}{4} & \frac{35}{4} \\
+ -\frac{7}{2} & -\frac{3}{4} \\
+\end{array}
+\right| +\left(-\hat{\text{j}}\right) \left| 
+\begin{array}{cc}
+ -\frac{39}{4} & \frac{35}{4} \\
+ \frac{21}{4} & -\frac{3}{4} \\
+\end{array}
+\right| +\hat{\text{k}} \left| 
+\begin{array}{cc}
+ -\frac{39}{4} & -\frac{13}{4} \\
+ \frac{21}{4} & -\frac{7}{2} \\
+\end{array}
+\right| : \\
+  \text{= }\hat{\text{i}} \left| 
+\begin{array}{cc}
+ -\frac{13}{4} & \frac{35}{4} \\
+ -\frac{7}{2} & -\frac{3}{4} \\
+\end{array}
+\right| +\left(-\hat{\text{j}}\right) \left| 
+\begin{array}{cc}
+ -\frac{39}{4} & \frac{35}{4} \\
+ \frac{21}{4} & -\frac{3}{4} \\
+\end{array}
+\right| +\hat{\text{k}} \left| 
+\begin{array}{cc}
+ -\frac{39}{4} & -\frac{13}{4} \\
+ \frac{21}{4} & -\frac{7}{2} \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \hat{\text{i}} \left| 
+\begin{array}{cc}
+ -\frac{13}{4} & \frac{35}{4} \\
+ -\frac{7}{2} & -\frac{3}{4} \\
+\end{array}
+\right| =\hat{\text{i}} \left(\left(-\frac{13}{4}\right)\, \left(-\frac{3}{4}\right)-\frac{35 (-7)}{4\ 2}\right)=\frac{\hat{\text{i}} 529}{16}=\fbox{$\frac{529 \hat{\text{i}}}{16}$}: \\
+  \text{= }\fbox{$\frac{529 \hat{\text{i}}}{16}$}+\left(-\hat{\text{j}}\right) \left| 
+\begin{array}{cc}
+ -\frac{39}{4} & \frac{35}{4} \\
+ \frac{21}{4} & -\frac{3}{4} \\
+\end{array}
+\right| +\hat{\text{k}} \left| 
+\begin{array}{cc}
+ -\frac{39}{4} & -\frac{13}{4} \\
+ \frac{21}{4} & -\frac{7}{2} \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ -\hat{\text{j}} \left| 
+\begin{array}{cc}
+ -\frac{39}{4} & \frac{35}{4} \\
+ \frac{21}{4} & -\frac{3}{4} \\
+\end{array}
+\right| =-\hat{\text{j}} \left(\left(-\frac{39}{4}\right)\, \left(-\frac{3}{4}\right)-\frac{35\ 21}{4\ 4}\right)=\frac{-\hat{\text{j}} (-309)}{8}=\fbox{$\frac{309 \hat{\text{j}}}{8}$}: \\
+  \text{= }\frac{529 \hat{\text{i}}}{16}+\fbox{$\frac{309 \hat{\text{j}}}{8}$}+\hat{\text{k}} \left| 
+\begin{array}{cc}
+ -\frac{39}{4} & -\frac{13}{4} \\
+ \frac{21}{4} & -\frac{7}{2} \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \hat{\text{k}} \left| 
+\begin{array}{cc}
+ -\frac{39}{4} & -\frac{13}{4} \\
+ \frac{21}{4} & -\frac{7}{2} \\
+\end{array}
+\right| =\hat{\text{k}} \left(\left(-\frac{39}{4}\right)\, \left(-\frac{7}{2}\right)-\left(-\frac{13}{4}\right)\, \times \, \frac{21}{4}\right)=\frac{\hat{\text{k}} 819}{16}=\fbox{$\frac{819 \hat{\text{k}}}{16}$}: \\
+  \text{= }\frac{529 \hat{\text{i}}}{16}+\frac{309 \hat{\text{j}}}{8}+\fbox{$\frac{819 \hat{\text{k}}}{16}$} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Order }\text{the }\text{terms }\text{in }\text{a }\text{more }\text{natural }\text{way}: \\
+  \text{= }\frac{529 \hat{\text{i}}}{16}+\frac{309 \hat{\text{j}}}{8}+\frac{819 \hat{\text{k}}}{16} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \frac{529 \hat{\text{i}}}{16}+\frac{309 \hat{\text{j}}}{8}+\frac{819 \hat{\text{k}}}{16}=\, \left(\frac{529}{16},\frac{309}{8},\frac{819}{16}\right)\, : \\
+ \fbox{$
+\begin{array}{ll}
+ \text{Answer:} &  \\
+ \text{} & \, \left(\frac{529}{16},\frac{309}{8},\frac{819}{16}\right)\,  \\
+\end{array}
+$} \\
+\end{array}
+ \\
+\end{array}

pretraining/mathematica/linear_algebra/cross_product_w_steps/1647.txt

@@ -0,0 +1,169 @@
+Problem:
+Find the cross product of the following vectors:
+$\left(
+\begin{array}{c}
+ -3 \\
+ -4 \\
+ \frac{23}{3} \\
+\end{array}
+\right)$ and
+$\left(
+\begin{array}{c}
+ -\frac{25}{3} \\
+ \frac{29}{6} \\
+ -\frac{17}{2} \\
+\end{array}
+\right)$
+Answer:
+\begin{array}{l}
+ 
+\begin{array}{l}
+ \text{Compute the following cross product}: \\
+ \, \left(-3,-4,\frac{23}{3}\right)\, \times \, \left(-\frac{25}{3},\frac{29}{6},-\frac{17}{2}\right)\,  \\
+\end{array}
+ \\
+\hline
+ 
+\begin{array}{l}
+ \text{Construct }\text{a }\text{matrix }\text{where }\text{the }\text{first }\text{row }\text{contains }\text{unit }\text{vectors }\hat{\text{i}}, \hat{\text{j}}, \text{and }\hat{\text{k}}; \text{and }\text{the }\text{second }\text{and }\text{third }\text{rows }\text{are }\text{made }\text{of }\text{vectors }\, \left(-3,-4,\frac{23}{3}\right)\,  \text{and }\, \left(-\frac{25}{3},\frac{29}{6},-\frac{17}{2}\right)\, : \\
+ \left(
+\begin{array}{ccc}
+ \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\
+ -3 & -4 & \frac{23}{3} \\
+ -\frac{25}{3} & \frac{29}{6} & -\frac{17}{2} \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Take }\text{the }\text{determinant }\text{of }\text{this }\text{matrix}: \\
+ \left| 
+\begin{array}{ccc}
+ \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\
+ -3 & -4 & \frac{23}{3} \\
+ -\frac{25}{3} & \frac{29}{6} & -\frac{17}{2} \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Expand }\text{with }\text{respect }\text{to }\text{row }1: \\
+  \text{= }\left| 
+\begin{array}{ccc}
+ \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\
+ -3 & -4 & \frac{23}{3} \\
+ -\frac{25}{3} & \frac{29}{6} & -\frac{17}{2} \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{The }\text{determinant }\text{of }\text{the }\text{matrix }\left(
+\begin{array}{ccc}
+ \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\
+ -3 & -4 & \frac{23}{3} \\
+ -\frac{25}{3} & \frac{29}{6} & -\frac{17}{2} \\
+\end{array}
+\right) \text{is }\text{given }\text{by }\hat{\text{i}} \left| 
+\begin{array}{cc}
+ -4 & \frac{23}{3} \\
+ \frac{29}{6} & -\frac{17}{2} \\
+\end{array}
+\right| +\left(-\hat{\text{j}}\right) \left| 
+\begin{array}{cc}
+ -3 & \frac{23}{3} \\
+ -\frac{25}{3} & -\frac{17}{2} \\
+\end{array}
+\right| +\hat{\text{k}} \left| 
+\begin{array}{cc}
+ -3 & -4 \\
+ -\frac{25}{3} & \frac{29}{6} \\
+\end{array}
+\right| : \\
+  \text{= }\hat{\text{i}} \left| 
+\begin{array}{cc}
+ -4 & \frac{23}{3} \\
+ \frac{29}{6} & -\frac{17}{2} \\
+\end{array}
+\right| +\left(-\hat{\text{j}}\right) \left| 
+\begin{array}{cc}
+ -3 & \frac{23}{3} \\
+ -\frac{25}{3} & -\frac{17}{2} \\
+\end{array}
+\right| +\hat{\text{k}} \left| 
+\begin{array}{cc}
+ -3 & -4 \\
+ -\frac{25}{3} & \frac{29}{6} \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \hat{\text{i}} \left| 
+\begin{array}{cc}
+ -4 & \frac{23}{3} \\
+ \frac{29}{6} & -\frac{17}{2} \\
+\end{array}
+\right| =\hat{\text{i}} \left((-4)\, \left(-\frac{17}{2}\right)-\frac{23\ 29}{3\ 6}\right)=\frac{\hat{\text{i}} (-55)}{18}=\fbox{$-\frac{55 \hat{\text{i}}}{18}$}: \\
+  \text{= }\fbox{$-\frac{55 \hat{\text{i}}}{18}$}+\left(-\hat{\text{j}}\right) \left| 
+\begin{array}{cc}
+ -3 & \frac{23}{3} \\
+ -\frac{25}{3} & -\frac{17}{2} \\
+\end{array}
+\right| +\hat{\text{k}} \left| 
+\begin{array}{cc}
+ -3 & -4 \\
+ -\frac{25}{3} & \frac{29}{6} \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ -\hat{\text{j}} \left| 
+\begin{array}{cc}
+ -3 & \frac{23}{3} \\
+ -\frac{25}{3} & -\frac{17}{2} \\
+\end{array}
+\right| =-\hat{\text{j}} \left((-3)\, \left(-\frac{17}{2}\right)-\frac{23 (-25)}{3\ 3}\right)=\frac{-\hat{\text{j}} 1609}{18}=\fbox{$-\frac{1609 \hat{\text{j}}}{18}$}: \\
+  \text{= }\frac{-55 \hat{\text{i}}}{18}+\fbox{$-\frac{1609 \hat{\text{j}}}{18}$}+\hat{\text{k}} \left| 
+\begin{array}{cc}
+ -3 & -4 \\
+ -\frac{25}{3} & \frac{29}{6} \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \hat{\text{k}} \left| 
+\begin{array}{cc}
+ -3 & -4 \\
+ -\frac{25}{3} & \frac{29}{6} \\
+\end{array}
+\right| =\hat{\text{k}} \left((-3)\, \times \, \frac{29}{6}-(-4)\, \left(-\frac{25}{3}\right)\right)=\frac{\hat{\text{k}} (-287)}{6}=\fbox{$-\frac{287 \hat{\text{k}}}{6}$}: \\
+  \text{= }\frac{-55 \hat{\text{i}}}{18}-\frac{1609 \hat{\text{j}}}{18}+\fbox{$-\frac{287 \hat{\text{k}}}{6}$} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Order }\text{the }\text{terms }\text{in }\text{a }\text{more }\text{natural }\text{way}: \\
+  \text{= }-\frac{55 \hat{\text{i}}}{18}-\frac{1609 \hat{\text{j}}}{18}-\frac{287 \hat{\text{k}}}{6} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ -\frac{55 \hat{\text{i}}}{18}-\frac{1609 \hat{\text{j}}}{18}-\frac{287 \hat{\text{k}}}{6}=\, \left(-\frac{55}{18},-\frac{1609}{18},-\frac{287}{6}\right)\, : \\
+ \fbox{$
+\begin{array}{ll}
+ \text{Answer:} &  \\
+ \text{} & \, \left(-\frac{55}{18},-\frac{1609}{18},-\frac{287}{6}\right)\,  \\
+\end{array}
+$} \\
+\end{array}
+ \\
+\end{array}

pretraining/mathematica/linear_algebra/cross_product_w_steps/1724.txt

@@ -0,0 +1,169 @@
+Problem:
+Find the cross product of the following vectors:
+$\left(
+\begin{array}{c}
+ \frac{11}{3} \\
+ \frac{23}{3} \\
+ -5 \\
+\end{array}
+\right)$ and
+$\left(
+\begin{array}{c}
+ \frac{17}{3} \\
+ 8 \\
+ \frac{26}{3} \\
+\end{array}
+\right)$
+Answer:
+\begin{array}{l}
+ 
+\begin{array}{l}
+ \text{Compute the following cross product}: \\
+ \, \left(\frac{11}{3},\frac{23}{3},-5\right)\, \times \, \left(\frac{17}{3},8,\frac{26}{3}\right)\,  \\
+\end{array}
+ \\
+\hline
+ 
+\begin{array}{l}
+ \text{Construct }\text{a }\text{matrix }\text{where }\text{the }\text{first }\text{row }\text{contains }\text{unit }\text{vectors }\hat{\text{i}}, \hat{\text{j}}, \text{and }\hat{\text{k}}; \text{and }\text{the }\text{second }\text{and }\text{third }\text{rows }\text{are }\text{made }\text{of }\text{vectors }\, \left(\frac{11}{3},\frac{23}{3},-5\right)\,  \text{and }\, \left(\frac{17}{3},8,\frac{26}{3}\right)\, : \\
+ \left(
+\begin{array}{ccc}
+ \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\
+ \frac{11}{3} & \frac{23}{3} & -5 \\
+ \frac{17}{3} & 8 & \frac{26}{3} \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Take }\text{the }\text{determinant }\text{of }\text{this }\text{matrix}: \\
+ \left| 
+\begin{array}{ccc}
+ \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\
+ \frac{11}{3} & \frac{23}{3} & -5 \\
+ \frac{17}{3} & 8 & \frac{26}{3} \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Expand }\text{with }\text{respect }\text{to }\text{row }1: \\
+  \text{= }\left| 
+\begin{array}{ccc}
+ \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\
+ \frac{11}{3} & \frac{23}{3} & -5 \\
+ \frac{17}{3} & 8 & \frac{26}{3} \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{The }\text{determinant }\text{of }\text{the }\text{matrix }\left(
+\begin{array}{ccc}
+ \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\
+ \frac{11}{3} & \frac{23}{3} & -5 \\
+ \frac{17}{3} & 8 & \frac{26}{3} \\
+\end{array}
+\right) \text{is }\text{given }\text{by }\hat{\text{i}} \left| 
+\begin{array}{cc}
+ \frac{23}{3} & -5 \\
+ 8 & \frac{26}{3} \\
+\end{array}
+\right| +\left(-\hat{\text{j}}\right) \left| 
+\begin{array}{cc}
+ \frac{11}{3} & -5 \\
+ \frac{17}{3} & \frac{26}{3} \\
+\end{array}
+\right| +\hat{\text{k}} \left| 
+\begin{array}{cc}
+ \frac{11}{3} & \frac{23}{3} \\
+ \frac{17}{3} & 8 \\
+\end{array}
+\right| : \\
+  \text{= }\hat{\text{i}} \left| 
+\begin{array}{cc}
+ \frac{23}{3} & -5 \\
+ 8 & \frac{26}{3} \\
+\end{array}
+\right| +\left(-\hat{\text{j}}\right) \left| 
+\begin{array}{cc}
+ \frac{11}{3} & -5 \\
+ \frac{17}{3} & \frac{26}{3} \\
+\end{array}
+\right| +\hat{\text{k}} \left| 
+\begin{array}{cc}
+ \frac{11}{3} & \frac{23}{3} \\
+ \frac{17}{3} & 8 \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \hat{\text{i}} \left| 
+\begin{array}{cc}
+ \frac{23}{3} & -5 \\
+ 8 & \frac{26}{3} \\
+\end{array}
+\right| =\hat{\text{i}} \left(\frac{23\ 26}{3\ 3}-(-5)\, \times \, 8\right)=\frac{\hat{\text{i}} 958}{9}=\fbox{$\frac{958 \hat{\text{i}}}{9}$}: \\
+  \text{= }\fbox{$\frac{958 \hat{\text{i}}}{9}$}+\left(-\hat{\text{j}}\right) \left| 
+\begin{array}{cc}
+ \frac{11}{3} & -5 \\
+ \frac{17}{3} & \frac{26}{3} \\
+\end{array}
+\right| +\hat{\text{k}} \left| 
+\begin{array}{cc}
+ \frac{11}{3} & \frac{23}{3} \\
+ \frac{17}{3} & 8 \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ -\hat{\text{j}} \left| 
+\begin{array}{cc}
+ \frac{11}{3} & -5 \\
+ \frac{17}{3} & \frac{26}{3} \\
+\end{array}
+\right| =-\hat{\text{j}} \left(\frac{11\ 26}{3\ 3}-(-5)\, \times \, \frac{17}{3}\right)=\frac{-\hat{\text{j}} 541}{9}=\fbox{$-\frac{541 \hat{\text{j}}}{9}$}: \\
+  \text{= }\frac{958 \hat{\text{i}}}{9}+\fbox{$-\frac{541 \hat{\text{j}}}{9}$}+\hat{\text{k}} \left| 
+\begin{array}{cc}
+ \frac{11}{3} & \frac{23}{3} \\
+ \frac{17}{3} & 8 \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \hat{\text{k}} \left| 
+\begin{array}{cc}
+ \frac{11}{3} & \frac{23}{3} \\
+ \frac{17}{3} & 8 \\
+\end{array}
+\right| =\hat{\text{k}} \left(\frac{11\ 8}{3}-\frac{23\ 17}{3\ 3}\right)=\frac{\hat{\text{k}} (-127)}{9}=\fbox{$-\frac{127 \hat{\text{k}}}{9}$}: \\
+  \text{= }\frac{958 \hat{\text{i}}}{9}-\frac{541 \hat{\text{j}}}{9}+\fbox{$-\frac{127 \hat{\text{k}}}{9}$} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Order }\text{the }\text{terms }\text{in }\text{a }\text{more }\text{natural }\text{way}: \\
+  \text{= }\frac{958 \hat{\text{i}}}{9}-\frac{541 \hat{\text{j}}}{9}-\frac{127 \hat{\text{k}}}{9} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \frac{958 \hat{\text{i}}}{9}-\frac{541 \hat{\text{j}}}{9}-\frac{127 \hat{\text{k}}}{9}=\, \left(\frac{958}{9},-\frac{541}{9},-\frac{127}{9}\right)\, : \\
+ \fbox{$
+\begin{array}{ll}
+ \text{Answer:} &  \\
+ \text{} & \, \left(\frac{958}{9},-\frac{541}{9},-\frac{127}{9}\right)\,  \\
+\end{array}
+$} \\
+\end{array}
+ \\
+\end{array}

pretraining/mathematica/linear_algebra/cross_product_w_steps/1729.txt

@@ -0,0 +1,169 @@
+Problem:
+Find the cross product of the following vectors:
+$\left(
+\begin{array}{c}
+ 1 \\
+ 3 \\
+ 1 \\
+\end{array}
+\right)$ and
+$\left(
+\begin{array}{c}
+ -3 \\
+ -9 \\
+ 6 \\
+\end{array}
+\right)$
+Answer:
+\begin{array}{l}
+ 
+\begin{array}{l}
+ \text{Compute the following cross product}: \\
+ \, (1,3,1)\, \times \, (-3,-9,6)\,  \\
+\end{array}
+ \\
+\hline
+ 
+\begin{array}{l}
+ \text{Construct }\text{a }\text{matrix }\text{where }\text{the }\text{first }\text{row }\text{contains }\text{unit }\text{vectors }\hat{\text{i}}, \hat{\text{j}}, \text{and }\hat{\text{k}}; \text{and }\text{the }\text{second }\text{and }\text{third }\text{rows }\text{are }\text{made }\text{of }\text{vectors }\, (1,3,1)\,  \text{and }\, (-3,-9,6)\, : \\
+ \left(
+\begin{array}{ccc}
+ \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\
+ 1 & 3 & 1 \\
+ -3 & -9 & 6 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Take }\text{the }\text{determinant }\text{of }\text{this }\text{matrix}: \\
+ \left| 
+\begin{array}{ccc}
+ \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\
+ 1 & 3 & 1 \\
+ -3 & -9 & 6 \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Expand }\text{with }\text{respect }\text{to }\text{row }1: \\
+  \text{= }\left| 
+\begin{array}{ccc}
+ \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\
+ 1 & 3 & 1 \\
+ -3 & -9 & 6 \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{The }\text{determinant }\text{of }\text{the }\text{matrix }\left(
+\begin{array}{ccc}
+ \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\
+ 1 & 3 & 1 \\
+ -3 & -9 & 6 \\
+\end{array}
+\right) \text{is }\text{given }\text{by }\hat{\text{i}} \left| 
+\begin{array}{cc}
+ 3 & 1 \\
+ -9 & 6 \\
+\end{array}
+\right| +\left(-\hat{\text{j}}\right) \left| 
+\begin{array}{cc}
+ 1 & 1 \\
+ -3 & 6 \\
+\end{array}
+\right| +\hat{\text{k}} \left| 
+\begin{array}{cc}
+ 1 & 3 \\
+ -3 & -9 \\
+\end{array}
+\right| : \\
+  \text{= }\hat{\text{i}} \left| 
+\begin{array}{cc}
+ 3 & 1 \\
+ -9 & 6 \\
+\end{array}
+\right| +\left(-\hat{\text{j}}\right) \left| 
+\begin{array}{cc}
+ 1 & 1 \\
+ -3 & 6 \\
+\end{array}
+\right| +\hat{\text{k}} \left| 
+\begin{array}{cc}
+ 1 & 3 \\
+ -3 & -9 \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \hat{\text{i}} \left| 
+\begin{array}{cc}
+ 3 & 1 \\
+ -9 & 6 \\
+\end{array}
+\right| =\hat{\text{i}} (3\ 6-1 (-9))=\hat{\text{i}} 27=\fbox{$27 \hat{\text{i}}$}: \\
+  \text{= }\fbox{$27 \hat{\text{i}}$}+\left(-\hat{\text{j}}\right) \left| 
+\begin{array}{cc}
+ 1 & 1 \\
+ -3 & 6 \\
+\end{array}
+\right| +\hat{\text{k}} \left| 
+\begin{array}{cc}
+ 1 & 3 \\
+ -3 & -9 \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ -\hat{\text{j}} \left| 
+\begin{array}{cc}
+ 1 & 1 \\
+ -3 & 6 \\
+\end{array}
+\right| =-\hat{\text{j}} (1\ 6-1 (-3))=-\hat{\text{j}} 9=\fbox{$-9 \hat{\text{j}}$}: \\
+  \text{= }27 \hat{\text{i}}+\fbox{$-9 \hat{\text{j}}$}+\hat{\text{k}} \left| 
+\begin{array}{cc}
+ 1 & 3 \\
+ -3 & -9 \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \hat{\text{k}} \left| 
+\begin{array}{cc}
+ 1 & 3 \\
+ -3 & -9 \\
+\end{array}
+\right| =\hat{\text{k}} (1 (-9)-3 (-3))=\hat{\text{k}} 0=\fbox{$0$}: \\
+  \text{= }27 \hat{\text{i}}-9 \hat{\text{j}}+\fbox{$0$} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Order }\text{the }\text{terms }\text{in }\text{a }\text{more }\text{natural }\text{way}: \\
+  \text{= }27 \hat{\text{i}}-9 \hat{\text{j}} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ 27 \hat{\text{i}}-9 \hat{\text{j}}=\, (27,-9,0)\, : \\
+ \fbox{$
+\begin{array}{ll}
+ \text{Answer:} &  \\
+ \text{} & \, (27,-9,0)\,  \\
+\end{array}
+$} \\
+\end{array}
+ \\
+\end{array}

pretraining/mathematica/linear_algebra/cross_product_w_steps/1730.txt

@@ -0,0 +1,169 @@
+Problem:
+Find the cross product of the following vectors:
+$\left(
+\begin{array}{c}
+ 2 \\
+ \frac{13}{3} \\
+ -\frac{1}{3} \\
+\end{array}
+\right)$ and
+$\left(
+\begin{array}{c}
+ 1 \\
+ -\frac{5}{3} \\
+ \frac{2}{3} \\
+\end{array}
+\right)$
+Answer:
+\begin{array}{l}
+ 
+\begin{array}{l}
+ \text{Compute the following cross product}: \\
+ \, \left(2,\frac{13}{3},-\frac{1}{3}\right)\, \times \, \left(1,-\frac{5}{3},\frac{2}{3}\right)\,  \\
+\end{array}
+ \\
+\hline
+ 
+\begin{array}{l}
+ \text{Construct }\text{a }\text{matrix }\text{where }\text{the }\text{first }\text{row }\text{contains }\text{unit }\text{vectors }\hat{\text{i}}, \hat{\text{j}}, \text{and }\hat{\text{k}}; \text{and }\text{the }\text{second }\text{and }\text{third }\text{rows }\text{are }\text{made }\text{of }\text{vectors }\, \left(2,\frac{13}{3},-\frac{1}{3}\right)\,  \text{and }\, \left(1,-\frac{5}{3},\frac{2}{3}\right)\, : \\
+ \left(
+\begin{array}{ccc}
+ \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\
+ 2 & \frac{13}{3} & -\frac{1}{3} \\
+ 1 & -\frac{5}{3} & \frac{2}{3} \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Take }\text{the }\text{determinant }\text{of }\text{this }\text{matrix}: \\
+ \left| 
+\begin{array}{ccc}
+ \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\
+ 2 & \frac{13}{3} & -\frac{1}{3} \\
+ 1 & -\frac{5}{3} & \frac{2}{3} \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Expand }\text{with }\text{respect }\text{to }\text{row }1: \\
+  \text{= }\left| 
+\begin{array}{ccc}
+ \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\
+ 2 & \frac{13}{3} & -\frac{1}{3} \\
+ 1 & -\frac{5}{3} & \frac{2}{3} \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{The }\text{determinant }\text{of }\text{the }\text{matrix }\left(
+\begin{array}{ccc}
+ \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\
+ 2 & \frac{13}{3} & -\frac{1}{3} \\
+ 1 & -\frac{5}{3} & \frac{2}{3} \\
+\end{array}
+\right) \text{is }\text{given }\text{by }\hat{\text{i}} \left| 
+\begin{array}{cc}
+ \frac{13}{3} & -\frac{1}{3} \\
+ -\frac{5}{3} & \frac{2}{3} \\
+\end{array}
+\right| +\left(-\hat{\text{j}}\right) \left| 
+\begin{array}{cc}
+ 2 & -\frac{1}{3} \\
+ 1 & \frac{2}{3} \\
+\end{array}
+\right| +\hat{\text{k}} \left| 
+\begin{array}{cc}
+ 2 & \frac{13}{3} \\
+ 1 & -\frac{5}{3} \\
+\end{array}
+\right| : \\
+  \text{= }\hat{\text{i}} \left| 
+\begin{array}{cc}
+ \frac{13}{3} & -\frac{1}{3} \\
+ -\frac{5}{3} & \frac{2}{3} \\
+\end{array}
+\right| +\left(-\hat{\text{j}}\right) \left| 
+\begin{array}{cc}
+ 2 & -\frac{1}{3} \\
+ 1 & \frac{2}{3} \\
+\end{array}
+\right| +\hat{\text{k}} \left| 
+\begin{array}{cc}
+ 2 & \frac{13}{3} \\
+ 1 & -\frac{5}{3} \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \hat{\text{i}} \left| 
+\begin{array}{cc}
+ \frac{13}{3} & -\frac{1}{3} \\
+ -\frac{5}{3} & \frac{2}{3} \\
+\end{array}
+\right| =\hat{\text{i}} \left(\frac{13\ 2}{3\ 3}-\frac{5}{3\ 3}\right)=\frac{\hat{\text{i}} 7}{3}=\fbox{$\frac{7 \hat{\text{i}}}{3}$}: \\
+  \text{= }\fbox{$\frac{7 \hat{\text{i}}}{3}$}+\left(-\hat{\text{j}}\right) \left| 
+\begin{array}{cc}
+ 2 & -\frac{1}{3} \\
+ 1 & \frac{2}{3} \\
+\end{array}
+\right| +\hat{\text{k}} \left| 
+\begin{array}{cc}
+ 2 & \frac{13}{3} \\
+ 1 & -\frac{5}{3} \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ -\hat{\text{j}} \left| 
+\begin{array}{cc}
+ 2 & -\frac{1}{3} \\
+ 1 & \frac{2}{3} \\
+\end{array}
+\right| =-\hat{\text{j}} \left(\frac{2\ 2}{3}-\frac{1}{3} (-1)\right)=\frac{-\hat{\text{j}} 5}{3}=\fbox{$-\frac{5 \hat{\text{j}}}{3}$}: \\
+  \text{= }\frac{7 \hat{\text{i}}}{3}+\fbox{$-\frac{5 \hat{\text{j}}}{3}$}+\hat{\text{k}} \left| 
+\begin{array}{cc}
+ 2 & \frac{13}{3} \\
+ 1 & -\frac{5}{3} \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \hat{\text{k}} \left| 
+\begin{array}{cc}
+ 2 & \frac{13}{3} \\
+ 1 & -\frac{5}{3} \\
+\end{array}
+\right| =\hat{\text{k}} \left(\frac{2 (-5)}{3}-\frac{13}{3}\right)=\frac{\hat{\text{k}} (-23)}{3}=\fbox{$-\frac{23 \hat{\text{k}}}{3}$}: \\
+  \text{= }\frac{7 \hat{\text{i}}}{3}-\frac{5 \hat{\text{j}}}{3}+\fbox{$-\frac{23 \hat{\text{k}}}{3}$} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Order }\text{the }\text{terms }\text{in }\text{a }\text{more }\text{natural }\text{way}: \\
+  \text{= }\frac{7 \hat{\text{i}}}{3}-\frac{5 \hat{\text{j}}}{3}-\frac{23 \hat{\text{k}}}{3} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \frac{7 \hat{\text{i}}}{3}-\frac{5 \hat{\text{j}}}{3}-\frac{23 \hat{\text{k}}}{3}=\, \left(\frac{7}{3},-\frac{5}{3},-\frac{23}{3}\right)\, : \\
+ \fbox{$
+\begin{array}{ll}
+ \text{Answer:} &  \\
+ \text{} & \, \left(\frac{7}{3},-\frac{5}{3},-\frac{23}{3}\right)\,  \\
+\end{array}
+$} \\
+\end{array}
+ \\
+\end{array}

pretraining/mathematica/linear_algebra/cross_product_w_steps/1735.txt

@@ -0,0 +1,169 @@
+Problem:
+Find the cross product of the following vectors:
+$\left(
+\begin{array}{c}
+ -8 \\
+ 1 \\
+ -6 \\
+\end{array}
+\right)$ and
+$\left(
+\begin{array}{c}
+ -9 \\
+ -3 \\
+ -6 \\
+\end{array}
+\right)$
+Answer:
+\begin{array}{l}
+ 
+\begin{array}{l}
+ \text{Compute the following cross product}: \\
+ \, (-8,1,-6)\, \times \, (-9,-3,-6)\,  \\
+\end{array}
+ \\
+\hline
+ 
+\begin{array}{l}
+ \text{Construct }\text{a }\text{matrix }\text{where }\text{the }\text{first }\text{row }\text{contains }\text{unit }\text{vectors }\hat{\text{i}}, \hat{\text{j}}, \text{and }\hat{\text{k}}; \text{and }\text{the }\text{second }\text{and }\text{third }\text{rows }\text{are }\text{made }\text{of }\text{vectors }\, (-8,1,-6)\,  \text{and }\, (-9,-3,-6)\, : \\
+ \left(
+\begin{array}{ccc}
+ \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\
+ -8 & 1 & -6 \\
+ -9 & -3 & -6 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Take }\text{the }\text{determinant }\text{of }\text{this }\text{matrix}: \\
+ \left| 
+\begin{array}{ccc}
+ \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\
+ -8 & 1 & -6 \\
+ -9 & -3 & -6 \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Expand }\text{with }\text{respect }\text{to }\text{row }1: \\
+  \text{= }\left| 
+\begin{array}{ccc}
+ \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\
+ -8 & 1 & -6 \\
+ -9 & -3 & -6 \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{The }\text{determinant }\text{of }\text{the }\text{matrix }\left(
+\begin{array}{ccc}
+ \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\
+ -8 & 1 & -6 \\
+ -9 & -3 & -6 \\
+\end{array}
+\right) \text{is }\text{given }\text{by }\hat{\text{i}} \left| 
+\begin{array}{cc}
+ 1 & -6 \\
+ -3 & -6 \\
+\end{array}
+\right| +\left(-\hat{\text{j}}\right) \left| 
+\begin{array}{cc}
+ -8 & -6 \\
+ -9 & -6 \\
+\end{array}
+\right| +\hat{\text{k}} \left| 
+\begin{array}{cc}
+ -8 & 1 \\
+ -9 & -3 \\
+\end{array}
+\right| : \\
+  \text{= }\hat{\text{i}} \left| 
+\begin{array}{cc}
+ 1 & -6 \\
+ -3 & -6 \\
+\end{array}
+\right| +\left(-\hat{\text{j}}\right) \left| 
+\begin{array}{cc}
+ -8 & -6 \\
+ -9 & -6 \\
+\end{array}
+\right| +\hat{\text{k}} \left| 
+\begin{array}{cc}
+ -8 & 1 \\
+ -9 & -3 \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \hat{\text{i}} \left| 
+\begin{array}{cc}
+ 1 & -6 \\
+ -3 & -6 \\
+\end{array}
+\right| =\hat{\text{i}} (1 (-6)-(-6)\, (-3))=\hat{\text{i}} (-24)=\fbox{$-24 \hat{\text{i}}$}: \\
+  \text{= }\fbox{$-24 \hat{\text{i}}$}+\left(-\hat{\text{j}}\right) \left| 
+\begin{array}{cc}
+ -8 & -6 \\
+ -9 & -6 \\
+\end{array}
+\right| +\hat{\text{k}} \left| 
+\begin{array}{cc}
+ -8 & 1 \\
+ -9 & -3 \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ -\hat{\text{j}} \left| 
+\begin{array}{cc}
+ -8 & -6 \\
+ -9 & -6 \\
+\end{array}
+\right| =-\hat{\text{j}} ((-8)\, (-6)-(-6)\, (-9))=-\hat{\text{j}} (-6)=\fbox{$6 \hat{\text{j}}$}: \\
+  \text{= }-24 \hat{\text{i}}+\fbox{$6 \hat{\text{j}}$}+\hat{\text{k}} \left| 
+\begin{array}{cc}
+ -8 & 1 \\
+ -9 & -3 \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \hat{\text{k}} \left| 
+\begin{array}{cc}
+ -8 & 1 \\
+ -9 & -3 \\
+\end{array}
+\right| =\hat{\text{k}} ((-8)\, (-3)-1 (-9))=\hat{\text{k}} 33=\fbox{$33 \hat{\text{k}}$}: \\
+  \text{= }-24 \hat{\text{i}}+6 \hat{\text{j}}+\fbox{$33 \hat{\text{k}}$} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Order }\text{the }\text{terms }\text{in }\text{a }\text{more }\text{natural }\text{way}: \\
+  \text{= }-24 \hat{\text{i}}+6 \hat{\text{j}}+33 \hat{\text{k}} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ -24 \hat{\text{i}}+6 \hat{\text{j}}+33 \hat{\text{k}}=\, (-24,6,33)\, : \\
+ \fbox{$
+\begin{array}{ll}
+ \text{Answer:} &  \\
+ \text{} & \, (-24,6,33)\,  \\
+\end{array}
+$} \\
+\end{array}
+ \\
+\end{array}

pretraining/mathematica/linear_algebra/cross_product_w_steps/1777.txt

@@ -0,0 +1,169 @@
+Problem:
+Find the cross product of the following vectors:
+$\left(
+\begin{array}{c}
+ 9 \\
+ -6 \\
+ 8 \\
+\end{array}
+\right)$ and
+$\left(
+\begin{array}{c}
+ 6 \\
+ -6 \\
+ -7 \\
+\end{array}
+\right)$
+Answer:
+\begin{array}{l}
+ 
+\begin{array}{l}
+ \text{Compute the following cross product}: \\
+ \, (9,-6,8)\, \times \, (6,-6,-7)\,  \\
+\end{array}
+ \\
+\hline
+ 
+\begin{array}{l}
+ \text{Construct }\text{a }\text{matrix }\text{where }\text{the }\text{first }\text{row }\text{contains }\text{unit }\text{vectors }\hat{\text{i}}, \hat{\text{j}}, \text{and }\hat{\text{k}}; \text{and }\text{the }\text{second }\text{and }\text{third }\text{rows }\text{are }\text{made }\text{of }\text{vectors }\, (9,-6,8)\,  \text{and }\, (6,-6,-7)\, : \\
+ \left(
+\begin{array}{ccc}
+ \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\
+ 9 & -6 & 8 \\
+ 6 & -6 & -7 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Take }\text{the }\text{determinant }\text{of }\text{this }\text{matrix}: \\
+ \left| 
+\begin{array}{ccc}
+ \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\
+ 9 & -6 & 8 \\
+ 6 & -6 & -7 \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Expand }\text{with }\text{respect }\text{to }\text{row }1: \\
+  \text{= }\left| 
+\begin{array}{ccc}
+ \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\
+ 9 & -6 & 8 \\
+ 6 & -6 & -7 \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{The }\text{determinant }\text{of }\text{the }\text{matrix }\left(
+\begin{array}{ccc}
+ \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\
+ 9 & -6 & 8 \\
+ 6 & -6 & -7 \\
+\end{array}
+\right) \text{is }\text{given }\text{by }\hat{\text{i}} \left| 
+\begin{array}{cc}
+ -6 & 8 \\
+ -6 & -7 \\
+\end{array}
+\right| +\left(-\hat{\text{j}}\right) \left| 
+\begin{array}{cc}
+ 9 & 8 \\
+ 6 & -7 \\
+\end{array}
+\right| +\hat{\text{k}} \left| 
+\begin{array}{cc}
+ 9 & -6 \\
+ 6 & -6 \\
+\end{array}
+\right| : \\
+  \text{= }\hat{\text{i}} \left| 
+\begin{array}{cc}
+ -6 & 8 \\
+ -6 & -7 \\
+\end{array}
+\right| +\left(-\hat{\text{j}}\right) \left| 
+\begin{array}{cc}
+ 9 & 8 \\
+ 6 & -7 \\
+\end{array}
+\right| +\hat{\text{k}} \left| 
+\begin{array}{cc}
+ 9 & -6 \\
+ 6 & -6 \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \hat{\text{i}} \left| 
+\begin{array}{cc}
+ -6 & 8 \\
+ -6 & -7 \\
+\end{array}
+\right| =\hat{\text{i}} ((-6)\, (-7)-8 (-6))=\hat{\text{i}} 90=\fbox{$90 \hat{\text{i}}$}: \\
+  \text{= }\fbox{$90 \hat{\text{i}}$}+\left(-\hat{\text{j}}\right) \left| 
+\begin{array}{cc}
+ 9 & 8 \\
+ 6 & -7 \\
+\end{array}
+\right| +\hat{\text{k}} \left| 
+\begin{array}{cc}
+ 9 & -6 \\
+ 6 & -6 \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ -\hat{\text{j}} \left| 
+\begin{array}{cc}
+ 9 & 8 \\
+ 6 & -7 \\
+\end{array}
+\right| =-\hat{\text{j}} (9 (-7)-8\ 6)=-\hat{\text{j}} (-111)=\fbox{$111 \hat{\text{j}}$}: \\
+  \text{= }90 \hat{\text{i}}+\fbox{$111 \hat{\text{j}}$}+\hat{\text{k}} \left| 
+\begin{array}{cc}
+ 9 & -6 \\
+ 6 & -6 \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \hat{\text{k}} \left| 
+\begin{array}{cc}
+ 9 & -6 \\
+ 6 & -6 \\
+\end{array}
+\right| =\hat{\text{k}} (9 (-6)-(-6)\, \times \, 6)=\hat{\text{k}} (-18)=\fbox{$-18 \hat{\text{k}}$}: \\
+  \text{= }90 \hat{\text{i}}+111 \hat{\text{j}}+\fbox{$-18 \hat{\text{k}}$} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Order }\text{the }\text{terms }\text{in }\text{a }\text{more }\text{natural }\text{way}: \\
+  \text{= }90 \hat{\text{i}}+111 \hat{\text{j}}-18 \hat{\text{k}} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ 90 \hat{\text{i}}+111 \hat{\text{j}}-18 \hat{\text{k}}=\, (90,111,-18)\, : \\
+ \fbox{$
+\begin{array}{ll}
+ \text{Answer:} &  \\
+ \text{} & \, (90,111,-18)\,  \\
+\end{array}
+$} \\
+\end{array}
+ \\
+\end{array}

pretraining/mathematica/linear_algebra/cross_product_w_steps/1796.txt

@@ -0,0 +1,169 @@
+Problem:
+Find the cross product of the following vectors:
+$\left(
+\begin{array}{c}
+ -9 \\
+ -2 \\
+ 6 \\
+\end{array}
+\right)$ and
+$\left(
+\begin{array}{c}
+ -2 \\
+ -6 \\
+ -10 \\
+\end{array}
+\right)$
+Answer:
+\begin{array}{l}
+ 
+\begin{array}{l}
+ \text{Compute the following cross product}: \\
+ \, (-9,-2,6)\, \times \, (-2,-6,-10)\,  \\
+\end{array}
+ \\
+\hline
+ 
+\begin{array}{l}
+ \text{Construct }\text{a }\text{matrix }\text{where }\text{the }\text{first }\text{row }\text{contains }\text{unit }\text{vectors }\hat{\text{i}}, \hat{\text{j}}, \text{and }\hat{\text{k}}; \text{and }\text{the }\text{second }\text{and }\text{third }\text{rows }\text{are }\text{made }\text{of }\text{vectors }\, (-9,-2,6)\,  \text{and }\, (-2,-6,-10)\, : \\
+ \left(
+\begin{array}{ccc}
+ \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\
+ -9 & -2 & 6 \\
+ -2 & -6 & -10 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Take }\text{the }\text{determinant }\text{of }\text{this }\text{matrix}: \\
+ \left| 
+\begin{array}{ccc}
+ \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\
+ -9 & -2 & 6 \\
+ -2 & -6 & -10 \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Expand }\text{with }\text{respect }\text{to }\text{row }1: \\
+  \text{= }\left| 
+\begin{array}{ccc}
+ \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\
+ -9 & -2 & 6 \\
+ -2 & -6 & -10 \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{The }\text{determinant }\text{of }\text{the }\text{matrix }\left(
+\begin{array}{ccc}
+ \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\
+ -9 & -2 & 6 \\
+ -2 & -6 & -10 \\
+\end{array}
+\right) \text{is }\text{given }\text{by }\hat{\text{i}} \left| 
+\begin{array}{cc}
+ -2 & 6 \\
+ -6 & -10 \\
+\end{array}
+\right| +\left(-\hat{\text{j}}\right) \left| 
+\begin{array}{cc}
+ -9 & 6 \\
+ -2 & -10 \\
+\end{array}
+\right| +\hat{\text{k}} \left| 
+\begin{array}{cc}
+ -9 & -2 \\
+ -2 & -6 \\
+\end{array}
+\right| : \\
+  \text{= }\hat{\text{i}} \left| 
+\begin{array}{cc}
+ -2 & 6 \\
+ -6 & -10 \\
+\end{array}
+\right| +\left(-\hat{\text{j}}\right) \left| 
+\begin{array}{cc}
+ -9 & 6 \\
+ -2 & -10 \\
+\end{array}
+\right| +\hat{\text{k}} \left| 
+\begin{array}{cc}
+ -9 & -2 \\
+ -2 & -6 \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \hat{\text{i}} \left| 
+\begin{array}{cc}
+ -2 & 6 \\
+ -6 & -10 \\
+\end{array}
+\right| =\hat{\text{i}} ((-2)\, (-10)-6 (-6))=\hat{\text{i}} 56=\fbox{$56 \hat{\text{i}}$}: \\
+  \text{= }\fbox{$56 \hat{\text{i}}$}+\left(-\hat{\text{j}}\right) \left| 
+\begin{array}{cc}
+ -9 & 6 \\
+ -2 & -10 \\
+\end{array}
+\right| +\hat{\text{k}} \left| 
+\begin{array}{cc}
+ -9 & -2 \\
+ -2 & -6 \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ -\hat{\text{j}} \left| 
+\begin{array}{cc}
+ -9 & 6 \\
+ -2 & -10 \\
+\end{array}
+\right| =-\hat{\text{j}} ((-9)\, (-10)-6 (-2))=-\hat{\text{j}} 102=\fbox{$-102 \hat{\text{j}}$}: \\
+  \text{= }56 \hat{\text{i}}+\fbox{$-102 \hat{\text{j}}$}+\hat{\text{k}} \left| 
+\begin{array}{cc}
+ -9 & -2 \\
+ -2 & -6 \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \hat{\text{k}} \left| 
+\begin{array}{cc}
+ -9 & -2 \\
+ -2 & -6 \\
+\end{array}
+\right| =\hat{\text{k}} ((-9)\, (-6)-(-2)\, (-2))=\hat{\text{k}} 50=\fbox{$50 \hat{\text{k}}$}: \\
+  \text{= }56 \hat{\text{i}}-102 \hat{\text{j}}+\fbox{$50 \hat{\text{k}}$} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Order }\text{the }\text{terms }\text{in }\text{a }\text{more }\text{natural }\text{way}: \\
+  \text{= }56 \hat{\text{i}}-102 \hat{\text{j}}+50 \hat{\text{k}} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ 56 \hat{\text{i}}-102 \hat{\text{j}}+50 \hat{\text{k}}=\, (56,-102,50)\, : \\
+ \fbox{$
+\begin{array}{ll}
+ \text{Answer:} &  \\
+ \text{} & \, (56,-102,50)\,  \\
+\end{array}
+$} \\
+\end{array}
+ \\
+\end{array}

pretraining/mathematica/linear_algebra/cross_product_w_steps/1816.txt

@@ -0,0 +1,169 @@
+Problem:
+Find the cross product of the following vectors:
+$\left(
+\begin{array}{c}
+ -\frac{13}{8} \\
+ -\frac{5}{8} \\
+ \frac{21}{4} \\
+\end{array}
+\right)$ and
+$\left(
+\begin{array}{c}
+ -\frac{137}{16} \\
+ -\frac{13}{2} \\
+ \frac{47}{16} \\
+\end{array}
+\right)$
+Answer:
+\begin{array}{l}
+ 
+\begin{array}{l}
+ \text{Compute the following cross product}: \\
+ \, \left(-\frac{13}{8},-\frac{5}{8},\frac{21}{4}\right)\, \times \, \left(-\frac{137}{16},-\frac{13}{2},\frac{47}{16}\right)\,  \\
+\end{array}
+ \\
+\hline
+ 
+\begin{array}{l}
+ \text{Construct }\text{a }\text{matrix }\text{where }\text{the }\text{first }\text{row }\text{contains }\text{unit }\text{vectors }\hat{\text{i}}, \hat{\text{j}}, \text{and }\hat{\text{k}}; \text{and }\text{the }\text{second }\text{and }\text{third }\text{rows }\text{are }\text{made }\text{of }\text{vectors }\, \left(-\frac{13}{8},-\frac{5}{8},\frac{21}{4}\right)\,  \text{and }\, \left(-\frac{137}{16},-\frac{13}{2},\frac{47}{16}\right)\, : \\
+ \left(
+\begin{array}{ccc}
+ \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\
+ -\frac{13}{8} & -\frac{5}{8} & \frac{21}{4} \\
+ -\frac{137}{16} & -\frac{13}{2} & \frac{47}{16} \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Take }\text{the }\text{determinant }\text{of }\text{this }\text{matrix}: \\
+ \left| 
+\begin{array}{ccc}
+ \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\
+ -\frac{13}{8} & -\frac{5}{8} & \frac{21}{4} \\
+ -\frac{137}{16} & -\frac{13}{2} & \frac{47}{16} \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Expand }\text{with }\text{respect }\text{to }\text{row }1: \\
+  \text{= }\left| 
+\begin{array}{ccc}
+ \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\
+ -\frac{13}{8} & -\frac{5}{8} & \frac{21}{4} \\
+ -\frac{137}{16} & -\frac{13}{2} & \frac{47}{16} \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{The }\text{determinant }\text{of }\text{the }\text{matrix }\left(
+\begin{array}{ccc}
+ \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\
+ -\frac{13}{8} & -\frac{5}{8} & \frac{21}{4} \\
+ -\frac{137}{16} & -\frac{13}{2} & \frac{47}{16} \\
+\end{array}
+\right) \text{is }\text{given }\text{by }\hat{\text{i}} \left| 
+\begin{array}{cc}
+ -\frac{5}{8} & \frac{21}{4} \\
+ -\frac{13}{2} & \frac{47}{16} \\
+\end{array}
+\right| +\left(-\hat{\text{j}}\right) \left| 
+\begin{array}{cc}
+ -\frac{13}{8} & \frac{21}{4} \\
+ -\frac{137}{16} & \frac{47}{16} \\
+\end{array}
+\right| +\hat{\text{k}} \left| 
+\begin{array}{cc}
+ -\frac{13}{8} & -\frac{5}{8} \\
+ -\frac{137}{16} & -\frac{13}{2} \\
+\end{array}
+\right| : \\
+  \text{= }\hat{\text{i}} \left| 
+\begin{array}{cc}
+ -\frac{5}{8} & \frac{21}{4} \\
+ -\frac{13}{2} & \frac{47}{16} \\
+\end{array}
+\right| +\left(-\hat{\text{j}}\right) \left| 
+\begin{array}{cc}
+ -\frac{13}{8} & \frac{21}{4} \\
+ -\frac{137}{16} & \frac{47}{16} \\
+\end{array}
+\right| +\hat{\text{k}} \left| 
+\begin{array}{cc}
+ -\frac{13}{8} & -\frac{5}{8} \\
+ -\frac{137}{16} & -\frac{13}{2} \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \hat{\text{i}} \left| 
+\begin{array}{cc}
+ -\frac{5}{8} & \frac{21}{4} \\
+ -\frac{13}{2} & \frac{47}{16} \\
+\end{array}
+\right| =\hat{\text{i}} \left(\left(-\frac{5}{8}\right)\, \times \, \frac{47}{16}-\frac{21 (-13)}{4\ 2}\right)=\frac{\hat{\text{i}} 4133}{128}=\fbox{$\frac{4133 \hat{\text{i}}}{128}$}: \\
+  \text{= }\fbox{$\frac{4133 \hat{\text{i}}}{128}$}+\left(-\hat{\text{j}}\right) \left| 
+\begin{array}{cc}
+ -\frac{13}{8} & \frac{21}{4} \\
+ -\frac{137}{16} & \frac{47}{16} \\
+\end{array}
+\right| +\hat{\text{k}} \left| 
+\begin{array}{cc}
+ -\frac{13}{8} & -\frac{5}{8} \\
+ -\frac{137}{16} & -\frac{13}{2} \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ -\hat{\text{j}} \left| 
+\begin{array}{cc}
+ -\frac{13}{8} & \frac{21}{4} \\
+ -\frac{137}{16} & \frac{47}{16} \\
+\end{array}
+\right| =-\hat{\text{j}} \left(\left(-\frac{13}{8}\right)\, \times \, \frac{47}{16}-\frac{21 (-137)}{4\ 16}\right)=\frac{-\hat{\text{j}} 5143}{128}=\fbox{$-\frac{5143 \hat{\text{j}}}{128}$}: \\
+  \text{= }\frac{4133 \hat{\text{i}}}{128}+\fbox{$-\frac{5143 \hat{\text{j}}}{128}$}+\hat{\text{k}} \left| 
+\begin{array}{cc}
+ -\frac{13}{8} & -\frac{5}{8} \\
+ -\frac{137}{16} & -\frac{13}{2} \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \hat{\text{k}} \left| 
+\begin{array}{cc}
+ -\frac{13}{8} & -\frac{5}{8} \\
+ -\frac{137}{16} & -\frac{13}{2} \\
+\end{array}
+\right| =\hat{\text{k}} \left(\left(-\frac{13}{8}\right)\, \left(-\frac{13}{2}\right)-\left(-\frac{5}{8}\right)\, \left(-\frac{137}{16}\right)\right)=\frac{\hat{\text{k}} 667}{128}=\fbox{$\frac{667 \hat{\text{k}}}{128}$}: \\
+  \text{= }\frac{4133 \hat{\text{i}}}{128}-\frac{5143 \hat{\text{j}}}{128}+\fbox{$\frac{667 \hat{\text{k}}}{128}$} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Order }\text{the }\text{terms }\text{in }\text{a }\text{more }\text{natural }\text{way}: \\
+  \text{= }\frac{4133 \hat{\text{i}}}{128}-\frac{5143 \hat{\text{j}}}{128}+\frac{667 \hat{\text{k}}}{128} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \frac{4133 \hat{\text{i}}}{128}-\frac{5143 \hat{\text{j}}}{128}+\frac{667 \hat{\text{k}}}{128}=\, \left(\frac{4133}{128},-\frac{5143}{128},\frac{667}{128}\right)\, : \\
+ \fbox{$
+\begin{array}{ll}
+ \text{Answer:} &  \\
+ \text{} & \, \left(\frac{4133}{128},-\frac{5143}{128},\frac{667}{128}\right)\,  \\
+\end{array}
+$} \\
+\end{array}
+ \\
+\end{array}

pretraining/mathematica/linear_algebra/cross_product_w_steps/1865.txt

@@ -0,0 +1,169 @@
+Problem:
+Find the cross product of the following vectors:
+$\left(
+\begin{array}{c}
+ 2 \\
+ \frac{17}{2} \\
+ 3 \\
+\end{array}
+\right)$ and
+$\left(
+\begin{array}{c}
+ 4 \\
+ -4 \\
+ -\frac{1}{2} \\
+\end{array}
+\right)$
+Answer:
+\begin{array}{l}
+ 
+\begin{array}{l}
+ \text{Compute the following cross product}: \\
+ \, \left(2,\frac{17}{2},3\right)\, \times \, \left(4,-4,-\frac{1}{2}\right)\,  \\
+\end{array}
+ \\
+\hline
+ 
+\begin{array}{l}
+ \text{Construct }\text{a }\text{matrix }\text{where }\text{the }\text{first }\text{row }\text{contains }\text{unit }\text{vectors }\hat{\text{i}}, \hat{\text{j}}, \text{and }\hat{\text{k}}; \text{and }\text{the }\text{second }\text{and }\text{third }\text{rows }\text{are }\text{made }\text{of }\text{vectors }\, \left(2,\frac{17}{2},3\right)\,  \text{and }\, \left(4,-4,-\frac{1}{2}\right)\, : \\
+ \left(
+\begin{array}{ccc}
+ \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\
+ 2 & \frac{17}{2} & 3 \\
+ 4 & -4 & -\frac{1}{2} \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Take }\text{the }\text{determinant }\text{of }\text{this }\text{matrix}: \\
+ \left| 
+\begin{array}{ccc}
+ \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\
+ 2 & \frac{17}{2} & 3 \\
+ 4 & -4 & -\frac{1}{2} \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Expand }\text{with }\text{respect }\text{to }\text{row }1: \\
+  \text{= }\left| 
+\begin{array}{ccc}
+ \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\
+ 2 & \frac{17}{2} & 3 \\
+ 4 & -4 & -\frac{1}{2} \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{The }\text{determinant }\text{of }\text{the }\text{matrix }\left(
+\begin{array}{ccc}
+ \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\
+ 2 & \frac{17}{2} & 3 \\
+ 4 & -4 & -\frac{1}{2} \\
+\end{array}
+\right) \text{is }\text{given }\text{by }\hat{\text{i}} \left| 
+\begin{array}{cc}
+ \frac{17}{2} & 3 \\
+ -4 & -\frac{1}{2} \\
+\end{array}
+\right| +\left(-\hat{\text{j}}\right) \left| 
+\begin{array}{cc}
+ 2 & 3 \\
+ 4 & -\frac{1}{2} \\
+\end{array}
+\right| +\hat{\text{k}} \left| 
+\begin{array}{cc}
+ 2 & \frac{17}{2} \\
+ 4 & -4 \\
+\end{array}
+\right| : \\
+  \text{= }\hat{\text{i}} \left| 
+\begin{array}{cc}
+ \frac{17}{2} & 3 \\
+ -4 & -\frac{1}{2} \\
+\end{array}
+\right| +\left(-\hat{\text{j}}\right) \left| 
+\begin{array}{cc}
+ 2 & 3 \\
+ 4 & -\frac{1}{2} \\
+\end{array}
+\right| +\hat{\text{k}} \left| 
+\begin{array}{cc}
+ 2 & \frac{17}{2} \\
+ 4 & -4 \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \hat{\text{i}} \left| 
+\begin{array}{cc}
+ \frac{17}{2} & 3 \\
+ -4 & -\frac{1}{2} \\
+\end{array}
+\right| =\hat{\text{i}} \left(\frac{1}{2} \left(-\frac{17}{2}\right)-3 (-4)\right)=\frac{\hat{\text{i}} 31}{4}=\fbox{$\frac{31 \hat{\text{i}}}{4}$}: \\
+  \text{= }\fbox{$\frac{31 \hat{\text{i}}}{4}$}+\left(-\hat{\text{j}}\right) \left| 
+\begin{array}{cc}
+ 2 & 3 \\
+ 4 & -\frac{1}{2} \\
+\end{array}
+\right| +\hat{\text{k}} \left| 
+\begin{array}{cc}
+ 2 & \frac{17}{2} \\
+ 4 & -4 \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ -\hat{\text{j}} \left| 
+\begin{array}{cc}
+ 2 & 3 \\
+ 4 & -\frac{1}{2} \\
+\end{array}
+\right| =-\hat{\text{j}} \left(\frac{1}{2} (-2)-3\ 4\right)=-\hat{\text{j}} (-13)=\fbox{$13 \hat{\text{j}}$}: \\
+  \text{= }\frac{31 \hat{\text{i}}}{4}+\fbox{$13 \hat{\text{j}}$}+\hat{\text{k}} \left| 
+\begin{array}{cc}
+ 2 & \frac{17}{2} \\
+ 4 & -4 \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \hat{\text{k}} \left| 
+\begin{array}{cc}
+ 2 & \frac{17}{2} \\
+ 4 & -4 \\
+\end{array}
+\right| =\hat{\text{k}} \left(2 (-4)-\frac{17\ 4}{2}\right)=\hat{\text{k}} (-42)=\fbox{$-42 \hat{\text{k}}$}: \\
+  \text{= }\frac{31 \hat{\text{i}}}{4}+13 \hat{\text{j}}+\fbox{$-42 \hat{\text{k}}$} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Order }\text{the }\text{terms }\text{in }\text{a }\text{more }\text{natural }\text{way}: \\
+  \text{= }\frac{31 \hat{\text{i}}}{4}+13 \hat{\text{j}}-42 \hat{\text{k}} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \frac{31 \hat{\text{i}}}{4}+13 \hat{\text{j}}-42 \hat{\text{k}}=\, \left(\frac{31}{4},13,-42\right)\, : \\
+ \fbox{$
+\begin{array}{ll}
+ \text{Answer:} &  \\
+ \text{} & \, \left(\frac{31}{4},13,-42\right)\,  \\
+\end{array}
+$} \\
+\end{array}
+ \\
+\end{array}

pretraining/mathematica/linear_algebra/cross_product_w_steps/1924.txt

@@ -0,0 +1,169 @@
+Problem:
+Find the cross product of the following vectors:
+$\left(
+\begin{array}{c}
+ \frac{46}{9} \\
+ \frac{59}{9} \\
+ \frac{26}{9} \\
+\end{array}
+\right)$ and
+$\left(
+\begin{array}{c}
+ -\frac{25}{3} \\
+ \frac{74}{9} \\
+ \frac{65}{9} \\
+\end{array}
+\right)$
+Answer:
+\begin{array}{l}
+ 
+\begin{array}{l}
+ \text{Compute the following cross product}: \\
+ \, \left(\frac{46}{9},\frac{59}{9},\frac{26}{9}\right)\, \times \, \left(-\frac{25}{3},\frac{74}{9},\frac{65}{9}\right)\,  \\
+\end{array}
+ \\
+\hline
+ 
+\begin{array}{l}
+ \text{Construct }\text{a }\text{matrix }\text{where }\text{the }\text{first }\text{row }\text{contains }\text{unit }\text{vectors }\hat{\text{i}}, \hat{\text{j}}, \text{and }\hat{\text{k}}; \text{and }\text{the }\text{second }\text{and }\text{third }\text{rows }\text{are }\text{made }\text{of }\text{vectors }\, \left(\frac{46}{9},\frac{59}{9},\frac{26}{9}\right)\,  \text{and }\, \left(-\frac{25}{3},\frac{74}{9},\frac{65}{9}\right)\, : \\
+ \left(
+\begin{array}{ccc}
+ \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\
+ \frac{46}{9} & \frac{59}{9} & \frac{26}{9} \\
+ -\frac{25}{3} & \frac{74}{9} & \frac{65}{9} \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Take }\text{the }\text{determinant }\text{of }\text{this }\text{matrix}: \\
+ \left| 
+\begin{array}{ccc}
+ \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\
+ \frac{46}{9} & \frac{59}{9} & \frac{26}{9} \\
+ -\frac{25}{3} & \frac{74}{9} & \frac{65}{9} \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Expand }\text{with }\text{respect }\text{to }\text{row }1: \\
+  \text{= }\left| 
+\begin{array}{ccc}
+ \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\
+ \frac{46}{9} & \frac{59}{9} & \frac{26}{9} \\
+ -\frac{25}{3} & \frac{74}{9} & \frac{65}{9} \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{The }\text{determinant }\text{of }\text{the }\text{matrix }\left(
+\begin{array}{ccc}
+ \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\
+ \frac{46}{9} & \frac{59}{9} & \frac{26}{9} \\
+ -\frac{25}{3} & \frac{74}{9} & \frac{65}{9} \\
+\end{array}
+\right) \text{is }\text{given }\text{by }\hat{\text{i}} \left| 
+\begin{array}{cc}
+ \frac{59}{9} & \frac{26}{9} \\
+ \frac{74}{9} & \frac{65}{9} \\
+\end{array}
+\right| +\left(-\hat{\text{j}}\right) \left| 
+\begin{array}{cc}
+ \frac{46}{9} & \frac{26}{9} \\
+ -\frac{25}{3} & \frac{65}{9} \\
+\end{array}
+\right| +\hat{\text{k}} \left| 
+\begin{array}{cc}
+ \frac{46}{9} & \frac{59}{9} \\
+ -\frac{25}{3} & \frac{74}{9} \\
+\end{array}
+\right| : \\
+  \text{= }\hat{\text{i}} \left| 
+\begin{array}{cc}
+ \frac{59}{9} & \frac{26}{9} \\
+ \frac{74}{9} & \frac{65}{9} \\
+\end{array}
+\right| +\left(-\hat{\text{j}}\right) \left| 
+\begin{array}{cc}
+ \frac{46}{9} & \frac{26}{9} \\
+ -\frac{25}{3} & \frac{65}{9} \\
+\end{array}
+\right| +\hat{\text{k}} \left| 
+\begin{array}{cc}
+ \frac{46}{9} & \frac{59}{9} \\
+ -\frac{25}{3} & \frac{74}{9} \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \hat{\text{i}} \left| 
+\begin{array}{cc}
+ \frac{59}{9} & \frac{26}{9} \\
+ \frac{74}{9} & \frac{65}{9} \\
+\end{array}
+\right| =\hat{\text{i}} \left(\frac{59\ 65}{9\ 9}-\frac{26\ 74}{9\ 9}\right)=\frac{\hat{\text{i}} 637}{27}=\fbox{$\frac{637 \hat{\text{i}}}{27}$}: \\
+  \text{= }\fbox{$\frac{637 \hat{\text{i}}}{27}$}+\left(-\hat{\text{j}}\right) \left| 
+\begin{array}{cc}
+ \frac{46}{9} & \frac{26}{9} \\
+ -\frac{25}{3} & \frac{65}{9} \\
+\end{array}
+\right| +\hat{\text{k}} \left| 
+\begin{array}{cc}
+ \frac{46}{9} & \frac{59}{9} \\
+ -\frac{25}{3} & \frac{74}{9} \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ -\hat{\text{j}} \left| 
+\begin{array}{cc}
+ \frac{46}{9} & \frac{26}{9} \\
+ -\frac{25}{3} & \frac{65}{9} \\
+\end{array}
+\right| =-\hat{\text{j}} \left(\frac{46\ 65}{9\ 9}-\frac{26 (-25)}{9\ 3}\right)=\frac{-\hat{\text{j}} 4940}{81}=\fbox{$-\frac{4940 \hat{\text{j}}}{81}$}: \\
+  \text{= }\frac{637 \hat{\text{i}}}{27}+\fbox{$-\frac{4940 \hat{\text{j}}}{81}$}+\hat{\text{k}} \left| 
+\begin{array}{cc}
+ \frac{46}{9} & \frac{59}{9} \\
+ -\frac{25}{3} & \frac{74}{9} \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \hat{\text{k}} \left| 
+\begin{array}{cc}
+ \frac{46}{9} & \frac{59}{9} \\
+ -\frac{25}{3} & \frac{74}{9} \\
+\end{array}
+\right| =\hat{\text{k}} \left(\frac{46\ 74}{9\ 9}-\frac{59 (-25)}{9\ 3}\right)=\frac{\hat{\text{k}} 7829}{81}=\fbox{$\frac{7829 \hat{\text{k}}}{81}$}: \\
+  \text{= }\frac{637 \hat{\text{i}}}{27}-\frac{4940 \hat{\text{j}}}{81}+\fbox{$\frac{7829 \hat{\text{k}}}{81}$} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Order }\text{the }\text{terms }\text{in }\text{a }\text{more }\text{natural }\text{way}: \\
+  \text{= }\frac{637 \hat{\text{i}}}{27}-\frac{4940 \hat{\text{j}}}{81}+\frac{7829 \hat{\text{k}}}{81} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \frac{637 \hat{\text{i}}}{27}-\frac{4940 \hat{\text{j}}}{81}+\frac{7829 \hat{\text{k}}}{81}=\, \left(\frac{637}{27},-\frac{4940}{81},\frac{7829}{81}\right)\, : \\
+ \fbox{$
+\begin{array}{ll}
+ \text{Answer:} &  \\
+ \text{} & \, \left(\frac{637}{27},-\frac{4940}{81},\frac{7829}{81}\right)\,  \\
+\end{array}
+$} \\
+\end{array}
+ \\
+\end{array}

pretraining/mathematica/linear_algebra/cross_product_w_steps/1925.txt

@@ -0,0 +1,169 @@
+Problem:
+Find the cross product of the following vectors:
+$\left(
+\begin{array}{c}
+ -\frac{5}{2} \\
+ 2 \\
+ -\frac{13}{2} \\
+\end{array}
+\right)$ and
+$\left(
+\begin{array}{c}
+ -4 \\
+ \frac{7}{2} \\
+ -\frac{13}{2} \\
+\end{array}
+\right)$
+Answer:
+\begin{array}{l}
+ 
+\begin{array}{l}
+ \text{Compute the following cross product}: \\
+ \, \left(-\frac{5}{2},2,-\frac{13}{2}\right)\, \times \, \left(-4,\frac{7}{2},-\frac{13}{2}\right)\,  \\
+\end{array}
+ \\
+\hline
+ 
+\begin{array}{l}
+ \text{Construct }\text{a }\text{matrix }\text{where }\text{the }\text{first }\text{row }\text{contains }\text{unit }\text{vectors }\hat{\text{i}}, \hat{\text{j}}, \text{and }\hat{\text{k}}; \text{and }\text{the }\text{second }\text{and }\text{third }\text{rows }\text{are }\text{made }\text{of }\text{vectors }\, \left(-\frac{5}{2},2,-\frac{13}{2}\right)\,  \text{and }\, \left(-4,\frac{7}{2},-\frac{13}{2}\right)\, : \\
+ \left(
+\begin{array}{ccc}
+ \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\
+ -\frac{5}{2} & 2 & -\frac{13}{2} \\
+ -4 & \frac{7}{2} & -\frac{13}{2} \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Take }\text{the }\text{determinant }\text{of }\text{this }\text{matrix}: \\
+ \left| 
+\begin{array}{ccc}
+ \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\
+ -\frac{5}{2} & 2 & -\frac{13}{2} \\
+ -4 & \frac{7}{2} & -\frac{13}{2} \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Expand }\text{with }\text{respect }\text{to }\text{row }1: \\
+  \text{= }\left| 
+\begin{array}{ccc}
+ \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\
+ -\frac{5}{2} & 2 & -\frac{13}{2} \\
+ -4 & \frac{7}{2} & -\frac{13}{2} \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{The }\text{determinant }\text{of }\text{the }\text{matrix }\left(
+\begin{array}{ccc}
+ \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\
+ -\frac{5}{2} & 2 & -\frac{13}{2} \\
+ -4 & \frac{7}{2} & -\frac{13}{2} \\
+\end{array}
+\right) \text{is }\text{given }\text{by }\hat{\text{i}} \left| 
+\begin{array}{cc}
+ 2 & -\frac{13}{2} \\
+ \frac{7}{2} & -\frac{13}{2} \\
+\end{array}
+\right| +\left(-\hat{\text{j}}\right) \left| 
+\begin{array}{cc}
+ -\frac{5}{2} & -\frac{13}{2} \\
+ -4 & -\frac{13}{2} \\
+\end{array}
+\right| +\hat{\text{k}} \left| 
+\begin{array}{cc}
+ -\frac{5}{2} & 2 \\
+ -4 & \frac{7}{2} \\
+\end{array}
+\right| : \\
+  \text{= }\hat{\text{i}} \left| 
+\begin{array}{cc}
+ 2 & -\frac{13}{2} \\
+ \frac{7}{2} & -\frac{13}{2} \\
+\end{array}
+\right| +\left(-\hat{\text{j}}\right) \left| 
+\begin{array}{cc}
+ -\frac{5}{2} & -\frac{13}{2} \\
+ -4 & -\frac{13}{2} \\
+\end{array}
+\right| +\hat{\text{k}} \left| 
+\begin{array}{cc}
+ -\frac{5}{2} & 2 \\
+ -4 & \frac{7}{2} \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \hat{\text{i}} \left| 
+\begin{array}{cc}
+ 2 & -\frac{13}{2} \\
+ \frac{7}{2} & -\frac{13}{2} \\
+\end{array}
+\right| =\hat{\text{i}} \left(\frac{2 (-13)}{2}-\left(-\frac{13}{2}\right)\, \times \, \frac{7}{2}\right)=\frac{\hat{\text{i}} 39}{4}=\fbox{$\frac{39 \hat{\text{i}}}{4}$}: \\
+  \text{= }\fbox{$\frac{39 \hat{\text{i}}}{4}$}+\left(-\hat{\text{j}}\right) \left| 
+\begin{array}{cc}
+ -\frac{5}{2} & -\frac{13}{2} \\
+ -4 & -\frac{13}{2} \\
+\end{array}
+\right| +\hat{\text{k}} \left| 
+\begin{array}{cc}
+ -\frac{5}{2} & 2 \\
+ -4 & \frac{7}{2} \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ -\hat{\text{j}} \left| 
+\begin{array}{cc}
+ -\frac{5}{2} & -\frac{13}{2} \\
+ -4 & -\frac{13}{2} \\
+\end{array}
+\right| =-\hat{\text{j}} \left(\left(-\frac{5}{2}\right)\, \left(-\frac{13}{2}\right)-\left(-\frac{13}{2}\right)\, (-4)\right)=\frac{-\hat{\text{j}} (-39)}{4}=\fbox{$\frac{39 \hat{\text{j}}}{4}$}: \\
+  \text{= }\frac{39 \hat{\text{i}}}{4}+\fbox{$\frac{39 \hat{\text{j}}}{4}$}+\hat{\text{k}} \left| 
+\begin{array}{cc}
+ -\frac{5}{2} & 2 \\
+ -4 & \frac{7}{2} \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \hat{\text{k}} \left| 
+\begin{array}{cc}
+ -\frac{5}{2} & 2 \\
+ -4 & \frac{7}{2} \\
+\end{array}
+\right| =\hat{\text{k}} \left(\left(-\frac{5}{2}\right)\, \times \, \frac{7}{2}-2 (-4)\right)=\frac{\hat{\text{k}} (-3)}{4}=\fbox{$-\frac{3 \hat{\text{k}}}{4}$}: \\
+  \text{= }\frac{39 \hat{\text{i}}}{4}+\frac{39 \hat{\text{j}}}{4}+\fbox{$-\frac{3 \hat{\text{k}}}{4}$} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Order }\text{the }\text{terms }\text{in }\text{a }\text{more }\text{natural }\text{way}: \\
+  \text{= }\frac{39 \hat{\text{i}}}{4}+\frac{39 \hat{\text{j}}}{4}-\frac{3 \hat{\text{k}}}{4} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \frac{39 \hat{\text{i}}}{4}+\frac{39 \hat{\text{j}}}{4}-\frac{3 \hat{\text{k}}}{4}=\, \left(\frac{39}{4},\frac{39}{4},-\frac{3}{4}\right)\, : \\
+ \fbox{$
+\begin{array}{ll}
+ \text{Answer:} &  \\
+ \text{} & \, \left(\frac{39}{4},\frac{39}{4},-\frac{3}{4}\right)\,  \\
+\end{array}
+$} \\
+\end{array}
+ \\
+\end{array}

pretraining/mathematica/linear_algebra/cross_product_w_steps/1932.txt

@@ -0,0 +1,169 @@
+Problem:
+Find the cross product of the following vectors:
+$\left(
+\begin{array}{c}
+ 6 \\
+ 3 \\
+ 0 \\
+\end{array}
+\right)$ and
+$\left(
+\begin{array}{c}
+ 3 \\
+ 10 \\
+ -4 \\
+\end{array}
+\right)$
+Answer:
+\begin{array}{l}
+ 
+\begin{array}{l}
+ \text{Compute the following cross product}: \\
+ \, (6,3,0)\, \times \, (3,10,-4)\,  \\
+\end{array}
+ \\
+\hline
+ 
+\begin{array}{l}
+ \text{Construct }\text{a }\text{matrix }\text{where }\text{the }\text{first }\text{row }\text{contains }\text{unit }\text{vectors }\hat{\text{i}}, \hat{\text{j}}, \text{and }\hat{\text{k}}; \text{and }\text{the }\text{second }\text{and }\text{third }\text{rows }\text{are }\text{made }\text{of }\text{vectors }\, (6,3,0)\,  \text{and }\, (3,10,-4)\, : \\
+ \left(
+\begin{array}{ccc}
+ \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\
+ 6 & 3 & 0 \\
+ 3 & 10 & -4 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Take }\text{the }\text{determinant }\text{of }\text{this }\text{matrix}: \\
+ \left| 
+\begin{array}{ccc}
+ \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\
+ 6 & 3 & 0 \\
+ 3 & 10 & -4 \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Expand }\text{with }\text{respect }\text{to }\text{row }1: \\
+  \text{= }\left| 
+\begin{array}{ccc}
+ \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\
+ 6 & 3 & 0 \\
+ 3 & 10 & -4 \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{The }\text{determinant }\text{of }\text{the }\text{matrix }\left(
+\begin{array}{ccc}
+ \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\
+ 6 & 3 & 0 \\
+ 3 & 10 & -4 \\
+\end{array}
+\right) \text{is }\text{given }\text{by }\hat{\text{i}} \left| 
+\begin{array}{cc}
+ 3 & 0 \\
+ 10 & -4 \\
+\end{array}
+\right| +\left(-\hat{\text{j}}\right) \left| 
+\begin{array}{cc}
+ 6 & 0 \\
+ 3 & -4 \\
+\end{array}
+\right| +\hat{\text{k}} \left| 
+\begin{array}{cc}
+ 6 & 3 \\
+ 3 & 10 \\
+\end{array}
+\right| : \\
+  \text{= }\hat{\text{i}} \left| 
+\begin{array}{cc}
+ 3 & 0 \\
+ 10 & -4 \\
+\end{array}
+\right| +\left(-\hat{\text{j}}\right) \left| 
+\begin{array}{cc}
+ 6 & 0 \\
+ 3 & -4 \\
+\end{array}
+\right| +\hat{\text{k}} \left| 
+\begin{array}{cc}
+ 6 & 3 \\
+ 3 & 10 \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \hat{\text{i}} \left| 
+\begin{array}{cc}
+ 3 & 0 \\
+ 10 & -4 \\
+\end{array}
+\right| =\hat{\text{i}} (3 (-4)-0\ 10)=\hat{\text{i}} (-12)=\fbox{$-12 \hat{\text{i}}$}: \\
+  \text{= }\fbox{$-12 \hat{\text{i}}$}+\left(-\hat{\text{j}}\right) \left| 
+\begin{array}{cc}
+ 6 & 0 \\
+ 3 & -4 \\
+\end{array}
+\right| +\hat{\text{k}} \left| 
+\begin{array}{cc}
+ 6 & 3 \\
+ 3 & 10 \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ -\hat{\text{j}} \left| 
+\begin{array}{cc}
+ 6 & 0 \\
+ 3 & -4 \\
+\end{array}
+\right| =-\hat{\text{j}} (6 (-4)-0\ 3)=-\hat{\text{j}} (-24)=\fbox{$24 \hat{\text{j}}$}: \\
+  \text{= }-12 \hat{\text{i}}+\fbox{$24 \hat{\text{j}}$}+\hat{\text{k}} \left| 
+\begin{array}{cc}
+ 6 & 3 \\
+ 3 & 10 \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \hat{\text{k}} \left| 
+\begin{array}{cc}
+ 6 & 3 \\
+ 3 & 10 \\
+\end{array}
+\right| =\hat{\text{k}} (6\ 10-3\ 3)=\hat{\text{k}} 51=\fbox{$51 \hat{\text{k}}$}: \\
+  \text{= }-12 \hat{\text{i}}+24 \hat{\text{j}}+\fbox{$51 \hat{\text{k}}$} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Order }\text{the }\text{terms }\text{in }\text{a }\text{more }\text{natural }\text{way}: \\
+  \text{= }-12 \hat{\text{i}}+24 \hat{\text{j}}+51 \hat{\text{k}} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ -12 \hat{\text{i}}+24 \hat{\text{j}}+51 \hat{\text{k}}=\, (-12,24,51)\, : \\
+ \fbox{$
+\begin{array}{ll}
+ \text{Answer:} &  \\
+ \text{} & \, (-12,24,51)\,  \\
+\end{array}
+$} \\
+\end{array}
+ \\
+\end{array}

pretraining/mathematica/linear_algebra/cross_product_w_steps/2021.txt

@@ -0,0 +1,169 @@
+Problem:
+Find the cross product of the following vectors:
+$\left(
+\begin{array}{c}
+ -\frac{71}{8} \\
+ -\frac{51}{8} \\
+ -\frac{39}{8} \\
+\end{array}
+\right)$ and
+$\left(
+\begin{array}{c}
+ -\frac{11}{2} \\
+ -\frac{25}{4} \\
+ -\frac{69}{8} \\
+\end{array}
+\right)$
+Answer:
+\begin{array}{l}
+ 
+\begin{array}{l}
+ \text{Compute the following cross product}: \\
+ \, \left(-\frac{71}{8},-\frac{51}{8},-\frac{39}{8}\right)\, \times \, \left(-\frac{11}{2},-\frac{25}{4},-\frac{69}{8}\right)\,  \\
+\end{array}
+ \\
+\hline
+ 
+\begin{array}{l}
+ \text{Construct }\text{a }\text{matrix }\text{where }\text{the }\text{first }\text{row }\text{contains }\text{unit }\text{vectors }\hat{\text{i}}, \hat{\text{j}}, \text{and }\hat{\text{k}}; \text{and }\text{the }\text{second }\text{and }\text{third }\text{rows }\text{are }\text{made }\text{of }\text{vectors }\, \left(-\frac{71}{8},-\frac{51}{8},-\frac{39}{8}\right)\,  \text{and }\, \left(-\frac{11}{2},-\frac{25}{4},-\frac{69}{8}\right)\, : \\
+ \left(
+\begin{array}{ccc}
+ \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\
+ -\frac{71}{8} & -\frac{51}{8} & -\frac{39}{8} \\
+ -\frac{11}{2} & -\frac{25}{4} & -\frac{69}{8} \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Take }\text{the }\text{determinant }\text{of }\text{this }\text{matrix}: \\
+ \left| 
+\begin{array}{ccc}
+ \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\
+ -\frac{71}{8} & -\frac{51}{8} & -\frac{39}{8} \\
+ -\frac{11}{2} & -\frac{25}{4} & -\frac{69}{8} \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Expand }\text{with }\text{respect }\text{to }\text{row }1: \\
+  \text{= }\left| 
+\begin{array}{ccc}
+ \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\
+ -\frac{71}{8} & -\frac{51}{8} & -\frac{39}{8} \\
+ -\frac{11}{2} & -\frac{25}{4} & -\frac{69}{8} \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{The }\text{determinant }\text{of }\text{the }\text{matrix }\left(
+\begin{array}{ccc}
+ \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\
+ -\frac{71}{8} & -\frac{51}{8} & -\frac{39}{8} \\
+ -\frac{11}{2} & -\frac{25}{4} & -\frac{69}{8} \\
+\end{array}
+\right) \text{is }\text{given }\text{by }\hat{\text{i}} \left| 
+\begin{array}{cc}
+ -\frac{51}{8} & -\frac{39}{8} \\
+ -\frac{25}{4} & -\frac{69}{8} \\
+\end{array}
+\right| +\left(-\hat{\text{j}}\right) \left| 
+\begin{array}{cc}
+ -\frac{71}{8} & -\frac{39}{8} \\
+ -\frac{11}{2} & -\frac{69}{8} \\
+\end{array}
+\right| +\hat{\text{k}} \left| 
+\begin{array}{cc}
+ -\frac{71}{8} & -\frac{51}{8} \\
+ -\frac{11}{2} & -\frac{25}{4} \\
+\end{array}
+\right| : \\
+  \text{= }\hat{\text{i}} \left| 
+\begin{array}{cc}
+ -\frac{51}{8} & -\frac{39}{8} \\
+ -\frac{25}{4} & -\frac{69}{8} \\
+\end{array}
+\right| +\left(-\hat{\text{j}}\right) \left| 
+\begin{array}{cc}
+ -\frac{71}{8} & -\frac{39}{8} \\
+ -\frac{11}{2} & -\frac{69}{8} \\
+\end{array}
+\right| +\hat{\text{k}} \left| 
+\begin{array}{cc}
+ -\frac{71}{8} & -\frac{51}{8} \\
+ -\frac{11}{2} & -\frac{25}{4} \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \hat{\text{i}} \left| 
+\begin{array}{cc}
+ -\frac{51}{8} & -\frac{39}{8} \\
+ -\frac{25}{4} & -\frac{69}{8} \\
+\end{array}
+\right| =\hat{\text{i}} \left(\left(-\frac{51}{8}\right)\, \left(-\frac{69}{8}\right)-\left(-\frac{39}{8}\right)\, \left(-\frac{25}{4}\right)\right)=\frac{\hat{\text{i}} 1569}{64}=\fbox{$\frac{1569 \hat{\text{i}}}{64}$}: \\
+  \text{= }\fbox{$\frac{1569 \hat{\text{i}}}{64}$}+\left(-\hat{\text{j}}\right) \left| 
+\begin{array}{cc}
+ -\frac{71}{8} & -\frac{39}{8} \\
+ -\frac{11}{2} & -\frac{69}{8} \\
+\end{array}
+\right| +\hat{\text{k}} \left| 
+\begin{array}{cc}
+ -\frac{71}{8} & -\frac{51}{8} \\
+ -\frac{11}{2} & -\frac{25}{4} \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ -\hat{\text{j}} \left| 
+\begin{array}{cc}
+ -\frac{71}{8} & -\frac{39}{8} \\
+ -\frac{11}{2} & -\frac{69}{8} \\
+\end{array}
+\right| =-\hat{\text{j}} \left(\left(-\frac{71}{8}\right)\, \left(-\frac{69}{8}\right)-\left(-\frac{39}{8}\right)\, \left(-\frac{11}{2}\right)\right)=\frac{-\hat{\text{j}} 3183}{64}=\fbox{$-\frac{3183 \hat{\text{j}}}{64}$}: \\
+  \text{= }\frac{1569 \hat{\text{i}}}{64}+\fbox{$-\frac{3183 \hat{\text{j}}}{64}$}+\hat{\text{k}} \left| 
+\begin{array}{cc}
+ -\frac{71}{8} & -\frac{51}{8} \\
+ -\frac{11}{2} & -\frac{25}{4} \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \hat{\text{k}} \left| 
+\begin{array}{cc}
+ -\frac{71}{8} & -\frac{51}{8} \\
+ -\frac{11}{2} & -\frac{25}{4} \\
+\end{array}
+\right| =\hat{\text{k}} \left(\left(-\frac{71}{8}\right)\, \left(-\frac{25}{4}\right)-\left(-\frac{51}{8}\right)\, \left(-\frac{11}{2}\right)\right)=\frac{\hat{\text{k}} 653}{32}=\fbox{$\frac{653 \hat{\text{k}}}{32}$}: \\
+  \text{= }\frac{1569 \hat{\text{i}}}{64}-\frac{3183 \hat{\text{j}}}{64}+\fbox{$\frac{653 \hat{\text{k}}}{32}$} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Order }\text{the }\text{terms }\text{in }\text{a }\text{more }\text{natural }\text{way}: \\
+  \text{= }\frac{1569 \hat{\text{i}}}{64}-\frac{3183 \hat{\text{j}}}{64}+\frac{653 \hat{\text{k}}}{32} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \frac{1569 \hat{\text{i}}}{64}-\frac{3183 \hat{\text{j}}}{64}+\frac{653 \hat{\text{k}}}{32}=\, \left(\frac{1569}{64},-\frac{3183}{64},\frac{653}{32}\right)\, : \\
+ \fbox{$
+\begin{array}{ll}
+ \text{Answer:} &  \\
+ \text{} & \, \left(\frac{1569}{64},-\frac{3183}{64},\frac{653}{32}\right)\,  \\
+\end{array}
+$} \\
+\end{array}
+ \\
+\end{array}

pretraining/mathematica/linear_algebra/cross_product_w_steps/2044.txt

@@ -0,0 +1,169 @@
+Problem:
+Find the cross product of the following vectors:
+$\left(
+\begin{array}{c}
+ -3 \\
+ 8 \\
+ -6 \\
+\end{array}
+\right)$ and
+$\left(
+\begin{array}{c}
+ 4 \\
+ 9 \\
+ -8 \\
+\end{array}
+\right)$
+Answer:
+\begin{array}{l}
+ 
+\begin{array}{l}
+ \text{Compute the following cross product}: \\
+ \, (-3,8,-6)\, \times \, (4,9,-8)\,  \\
+\end{array}
+ \\
+\hline
+ 
+\begin{array}{l}
+ \text{Construct }\text{a }\text{matrix }\text{where }\text{the }\text{first }\text{row }\text{contains }\text{unit }\text{vectors }\hat{\text{i}}, \hat{\text{j}}, \text{and }\hat{\text{k}}; \text{and }\text{the }\text{second }\text{and }\text{third }\text{rows }\text{are }\text{made }\text{of }\text{vectors }\, (-3,8,-6)\,  \text{and }\, (4,9,-8)\, : \\
+ \left(
+\begin{array}{ccc}
+ \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\
+ -3 & 8 & -6 \\
+ 4 & 9 & -8 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Take }\text{the }\text{determinant }\text{of }\text{this }\text{matrix}: \\
+ \left| 
+\begin{array}{ccc}
+ \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\
+ -3 & 8 & -6 \\
+ 4 & 9 & -8 \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Expand }\text{with }\text{respect }\text{to }\text{row }1: \\
+  \text{= }\left| 
+\begin{array}{ccc}
+ \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\
+ -3 & 8 & -6 \\
+ 4 & 9 & -8 \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{The }\text{determinant }\text{of }\text{the }\text{matrix }\left(
+\begin{array}{ccc}
+ \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\
+ -3 & 8 & -6 \\
+ 4 & 9 & -8 \\
+\end{array}
+\right) \text{is }\text{given }\text{by }\hat{\text{i}} \left| 
+\begin{array}{cc}
+ 8 & -6 \\
+ 9 & -8 \\
+\end{array}
+\right| +\left(-\hat{\text{j}}\right) \left| 
+\begin{array}{cc}
+ -3 & -6 \\
+ 4 & -8 \\
+\end{array}
+\right| +\hat{\text{k}} \left| 
+\begin{array}{cc}
+ -3 & 8 \\
+ 4 & 9 \\
+\end{array}
+\right| : \\
+  \text{= }\hat{\text{i}} \left| 
+\begin{array}{cc}
+ 8 & -6 \\
+ 9 & -8 \\
+\end{array}
+\right| +\left(-\hat{\text{j}}\right) \left| 
+\begin{array}{cc}
+ -3 & -6 \\
+ 4 & -8 \\
+\end{array}
+\right| +\hat{\text{k}} \left| 
+\begin{array}{cc}
+ -3 & 8 \\
+ 4 & 9 \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \hat{\text{i}} \left| 
+\begin{array}{cc}
+ 8 & -6 \\
+ 9 & -8 \\
+\end{array}
+\right| =\hat{\text{i}} (8 (-8)-(-6)\, \times \, 9)=\hat{\text{i}} (-10)=\fbox{$-10 \hat{\text{i}}$}: \\
+  \text{= }\fbox{$-10 \hat{\text{i}}$}+\left(-\hat{\text{j}}\right) \left| 
+\begin{array}{cc}
+ -3 & -6 \\
+ 4 & -8 \\
+\end{array}
+\right| +\hat{\text{k}} \left| 
+\begin{array}{cc}
+ -3 & 8 \\
+ 4 & 9 \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ -\hat{\text{j}} \left| 
+\begin{array}{cc}
+ -3 & -6 \\
+ 4 & -8 \\
+\end{array}
+\right| =-\hat{\text{j}} ((-3)\, (-8)-(-6)\, \times \, 4)=-\hat{\text{j}} 48=\fbox{$-48 \hat{\text{j}}$}: \\
+  \text{= }-10 \hat{\text{i}}+\fbox{$-48 \hat{\text{j}}$}+\hat{\text{k}} \left| 
+\begin{array}{cc}
+ -3 & 8 \\
+ 4 & 9 \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \hat{\text{k}} \left| 
+\begin{array}{cc}
+ -3 & 8 \\
+ 4 & 9 \\
+\end{array}
+\right| =\hat{\text{k}} ((-3)\, \times \, 9-8\ 4)=\hat{\text{k}} (-59)=\fbox{$-59 \hat{\text{k}}$}: \\
+  \text{= }-10 \hat{\text{i}}-48 \hat{\text{j}}+\fbox{$-59 \hat{\text{k}}$} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Order }\text{the }\text{terms }\text{in }\text{a }\text{more }\text{natural }\text{way}: \\
+  \text{= }-10 \hat{\text{i}}-48 \hat{\text{j}}-59 \hat{\text{k}} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ -10 \hat{\text{i}}-48 \hat{\text{j}}-59 \hat{\text{k}}=\, (-10,-48,-59)\, : \\
+ \fbox{$
+\begin{array}{ll}
+ \text{Answer:} &  \\
+ \text{} & \, (-10,-48,-59)\,  \\
+\end{array}
+$} \\
+\end{array}
+ \\
+\end{array}

pretraining/mathematica/linear_algebra/cross_product_w_steps/2061.txt

@@ -0,0 +1,169 @@
+Problem:
+Find the cross product of the following vectors:
+$\left(
+\begin{array}{c}
+ -3 \\
+ 0 \\
+ 4 \\
+\end{array}
+\right)$ and
+$\left(
+\begin{array}{c}
+ 3 \\
+ 4 \\
+ -1 \\
+\end{array}
+\right)$
+Answer:
+\begin{array}{l}
+ 
+\begin{array}{l}
+ \text{Compute the following cross product}: \\
+ \, (-3,0,4)\, \times \, (3,4,-1)\,  \\
+\end{array}
+ \\
+\hline
+ 
+\begin{array}{l}
+ \text{Construct }\text{a }\text{matrix }\text{where }\text{the }\text{first }\text{row }\text{contains }\text{unit }\text{vectors }\hat{\text{i}}, \hat{\text{j}}, \text{and }\hat{\text{k}}; \text{and }\text{the }\text{second }\text{and }\text{third }\text{rows }\text{are }\text{made }\text{of }\text{vectors }\, (-3,0,4)\,  \text{and }\, (3,4,-1)\, : \\
+ \left(
+\begin{array}{ccc}
+ \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\
+ -3 & 0 & 4 \\
+ 3 & 4 & -1 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Take }\text{the }\text{determinant }\text{of }\text{this }\text{matrix}: \\
+ \left| 
+\begin{array}{ccc}
+ \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\
+ -3 & 0 & 4 \\
+ 3 & 4 & -1 \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Expand }\text{with }\text{respect }\text{to }\text{row }1: \\
+  \text{= }\left| 
+\begin{array}{ccc}
+ \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\
+ -3 & 0 & 4 \\
+ 3 & 4 & -1 \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{The }\text{determinant }\text{of }\text{the }\text{matrix }\left(
+\begin{array}{ccc}
+ \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\
+ -3 & 0 & 4 \\
+ 3 & 4 & -1 \\
+\end{array}
+\right) \text{is }\text{given }\text{by }\hat{\text{i}} \left| 
+\begin{array}{cc}
+ 0 & 4 \\
+ 4 & -1 \\
+\end{array}
+\right| +\left(-\hat{\text{j}}\right) \left| 
+\begin{array}{cc}
+ -3 & 4 \\
+ 3 & -1 \\
+\end{array}
+\right| +\hat{\text{k}} \left| 
+\begin{array}{cc}
+ -3 & 0 \\
+ 3 & 4 \\
+\end{array}
+\right| : \\
+  \text{= }\hat{\text{i}} \left| 
+\begin{array}{cc}
+ 0 & 4 \\
+ 4 & -1 \\
+\end{array}
+\right| +\left(-\hat{\text{j}}\right) \left| 
+\begin{array}{cc}
+ -3 & 4 \\
+ 3 & -1 \\
+\end{array}
+\right| +\hat{\text{k}} \left| 
+\begin{array}{cc}
+ -3 & 0 \\
+ 3 & 4 \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \hat{\text{i}} \left| 
+\begin{array}{cc}
+ 0 & 4 \\
+ 4 & -1 \\
+\end{array}
+\right| =\hat{\text{i}} (0 (-1)-4\ 4)=\hat{\text{i}} (-16)=\fbox{$-16 \hat{\text{i}}$}: \\
+  \text{= }\fbox{$-16 \hat{\text{i}}$}+\left(-\hat{\text{j}}\right) \left| 
+\begin{array}{cc}
+ -3 & 4 \\
+ 3 & -1 \\
+\end{array}
+\right| +\hat{\text{k}} \left| 
+\begin{array}{cc}
+ -3 & 0 \\
+ 3 & 4 \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ -\hat{\text{j}} \left| 
+\begin{array}{cc}
+ -3 & 4 \\
+ 3 & -1 \\
+\end{array}
+\right| =-\hat{\text{j}} ((-3)\, (-1)-4\ 3)=-\hat{\text{j}} (-9)=\fbox{$9 \hat{\text{j}}$}: \\
+  \text{= }-16 \hat{\text{i}}+\fbox{$9 \hat{\text{j}}$}+\hat{\text{k}} \left| 
+\begin{array}{cc}
+ -3 & 0 \\
+ 3 & 4 \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \hat{\text{k}} \left| 
+\begin{array}{cc}
+ -3 & 0 \\
+ 3 & 4 \\
+\end{array}
+\right| =\hat{\text{k}} ((-3)\, \times \, 4-0\ 3)=\hat{\text{k}} (-12)=\fbox{$-12 \hat{\text{k}}$}: \\
+  \text{= }-16 \hat{\text{i}}+9 \hat{\text{j}}+\fbox{$-12 \hat{\text{k}}$} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Order }\text{the }\text{terms }\text{in }\text{a }\text{more }\text{natural }\text{way}: \\
+  \text{= }-16 \hat{\text{i}}+9 \hat{\text{j}}-12 \hat{\text{k}} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ -16 \hat{\text{i}}+9 \hat{\text{j}}-12 \hat{\text{k}}=\, (-16,9,-12)\, : \\
+ \fbox{$
+\begin{array}{ll}
+ \text{Answer:} &  \\
+ \text{} & \, (-16,9,-12)\,  \\
+\end{array}
+$} \\
+\end{array}
+ \\
+\end{array}

pretraining/mathematica/linear_algebra/cross_product_w_steps/2205.txt

@@ -0,0 +1,169 @@
+Problem:
+Find the cross product of the following vectors:
+$\left(
+\begin{array}{c}
+ -\frac{83}{10} \\
+ \frac{46}{5} \\
+ \frac{1}{10} \\
+\end{array}
+\right)$ and
+$\left(
+\begin{array}{c}
+ -\frac{33}{10} \\
+ -\frac{28}{5} \\
+ 7 \\
+\end{array}
+\right)$
+Answer:
+\begin{array}{l}
+ 
+\begin{array}{l}
+ \text{Compute the following cross product}: \\
+ \, \left(-\frac{83}{10},\frac{46}{5},\frac{1}{10}\right)\, \times \, \left(-\frac{33}{10},-\frac{28}{5},7\right)\,  \\
+\end{array}
+ \\
+\hline
+ 
+\begin{array}{l}
+ \text{Construct }\text{a }\text{matrix }\text{where }\text{the }\text{first }\text{row }\text{contains }\text{unit }\text{vectors }\hat{\text{i}}, \hat{\text{j}}, \text{and }\hat{\text{k}}; \text{and }\text{the }\text{second }\text{and }\text{third }\text{rows }\text{are }\text{made }\text{of }\text{vectors }\, \left(-\frac{83}{10},\frac{46}{5},\frac{1}{10}\right)\,  \text{and }\, \left(-\frac{33}{10},-\frac{28}{5},7\right)\, : \\
+ \left(
+\begin{array}{ccc}
+ \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\
+ -\frac{83}{10} & \frac{46}{5} & \frac{1}{10} \\
+ -\frac{33}{10} & -\frac{28}{5} & 7 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Take }\text{the }\text{determinant }\text{of }\text{this }\text{matrix}: \\
+ \left| 
+\begin{array}{ccc}
+ \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\
+ -\frac{83}{10} & \frac{46}{5} & \frac{1}{10} \\
+ -\frac{33}{10} & -\frac{28}{5} & 7 \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Expand }\text{with }\text{respect }\text{to }\text{row }1: \\
+  \text{= }\left| 
+\begin{array}{ccc}
+ \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\
+ -\frac{83}{10} & \frac{46}{5} & \frac{1}{10} \\
+ -\frac{33}{10} & -\frac{28}{5} & 7 \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{The }\text{determinant }\text{of }\text{the }\text{matrix }\left(
+\begin{array}{ccc}
+ \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\
+ -\frac{83}{10} & \frac{46}{5} & \frac{1}{10} \\
+ -\frac{33}{10} & -\frac{28}{5} & 7 \\
+\end{array}
+\right) \text{is }\text{given }\text{by }\hat{\text{i}} \left| 
+\begin{array}{cc}
+ \frac{46}{5} & \frac{1}{10} \\
+ -\frac{28}{5} & 7 \\
+\end{array}
+\right| +\left(-\hat{\text{j}}\right) \left| 
+\begin{array}{cc}
+ -\frac{83}{10} & \frac{1}{10} \\
+ -\frac{33}{10} & 7 \\
+\end{array}
+\right| +\hat{\text{k}} \left| 
+\begin{array}{cc}
+ -\frac{83}{10} & \frac{46}{5} \\
+ -\frac{33}{10} & -\frac{28}{5} \\
+\end{array}
+\right| : \\
+  \text{= }\hat{\text{i}} \left| 
+\begin{array}{cc}
+ \frac{46}{5} & \frac{1}{10} \\
+ -\frac{28}{5} & 7 \\
+\end{array}
+\right| +\left(-\hat{\text{j}}\right) \left| 
+\begin{array}{cc}
+ -\frac{83}{10} & \frac{1}{10} \\
+ -\frac{33}{10} & 7 \\
+\end{array}
+\right| +\hat{\text{k}} \left| 
+\begin{array}{cc}
+ -\frac{83}{10} & \frac{46}{5} \\
+ -\frac{33}{10} & -\frac{28}{5} \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \hat{\text{i}} \left| 
+\begin{array}{cc}
+ \frac{46}{5} & \frac{1}{10} \\
+ -\frac{28}{5} & 7 \\
+\end{array}
+\right| =\hat{\text{i}} \left(\frac{46\ 7}{5}--\frac{28}{10\ 5}\right)=\frac{\hat{\text{i}} 1624}{25}=\fbox{$\frac{1624 \hat{\text{i}}}{25}$}: \\
+  \text{= }\fbox{$\frac{1624 \hat{\text{i}}}{25}$}+\left(-\hat{\text{j}}\right) \left| 
+\begin{array}{cc}
+ -\frac{83}{10} & \frac{1}{10} \\
+ -\frac{33}{10} & 7 \\
+\end{array}
+\right| +\hat{\text{k}} \left| 
+\begin{array}{cc}
+ -\frac{83}{10} & \frac{46}{5} \\
+ -\frac{33}{10} & -\frac{28}{5} \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ -\hat{\text{j}} \left| 
+\begin{array}{cc}
+ -\frac{83}{10} & \frac{1}{10} \\
+ -\frac{33}{10} & 7 \\
+\end{array}
+\right| =-\hat{\text{j}} \left(\left(-\frac{83}{10}\right)\, \times \, 7--\frac{33}{10\ 10}\right)=\frac{-\hat{\text{j}} (-5777)}{100}=\fbox{$\frac{5777 \hat{\text{j}}}{100}$}: \\
+  \text{= }\frac{1624 \hat{\text{i}}}{25}+\fbox{$\frac{5777 \hat{\text{j}}}{100}$}+\hat{\text{k}} \left| 
+\begin{array}{cc}
+ -\frac{83}{10} & \frac{46}{5} \\
+ -\frac{33}{10} & -\frac{28}{5} \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \hat{\text{k}} \left| 
+\begin{array}{cc}
+ -\frac{83}{10} & \frac{46}{5} \\
+ -\frac{33}{10} & -\frac{28}{5} \\
+\end{array}
+\right| =\hat{\text{k}} \left(\left(-\frac{83}{10}\right)\, \left(-\frac{28}{5}\right)-\frac{46 (-33)}{5\ 10}\right)=\frac{\hat{\text{k}} 1921}{25}=\fbox{$\frac{1921 \hat{\text{k}}}{25}$}: \\
+  \text{= }\frac{1624 \hat{\text{i}}}{25}+\frac{5777 \hat{\text{j}}}{100}+\fbox{$\frac{1921 \hat{\text{k}}}{25}$} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Order }\text{the }\text{terms }\text{in }\text{a }\text{more }\text{natural }\text{way}: \\
+  \text{= }\frac{1624 \hat{\text{i}}}{25}+\frac{5777 \hat{\text{j}}}{100}+\frac{1921 \hat{\text{k}}}{25} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \frac{1624 \hat{\text{i}}}{25}+\frac{5777 \hat{\text{j}}}{100}+\frac{1921 \hat{\text{k}}}{25}=\, \left(\frac{1624}{25},\frac{5777}{100},\frac{1921}{25}\right)\, : \\
+ \fbox{$
+\begin{array}{ll}
+ \text{Answer:} &  \\
+ \text{} & \, \left(\frac{1624}{25},\frac{5777}{100},\frac{1921}{25}\right)\,  \\
+\end{array}
+$} \\
+\end{array}
+ \\
+\end{array}

pretraining/mathematica/linear_algebra/cross_product_w_steps/2208.txt

@@ -0,0 +1,169 @@
+Problem:
+Find the cross product of the following vectors:
+$\left(
+\begin{array}{c}
+ -3 \\
+ -\frac{37}{8} \\
+ \frac{69}{8} \\
+\end{array}
+\right)$ and
+$\left(
+\begin{array}{c}
+ -\frac{19}{2} \\
+ -\frac{35}{4} \\
+ -\frac{25}{4} \\
+\end{array}
+\right)$
+Answer:
+\begin{array}{l}
+ 
+\begin{array}{l}
+ \text{Compute the following cross product}: \\
+ \, \left(-3,-\frac{37}{8},\frac{69}{8}\right)\, \times \, \left(-\frac{19}{2},-\frac{35}{4},-\frac{25}{4}\right)\,  \\
+\end{array}
+ \\
+\hline
+ 
+\begin{array}{l}
+ \text{Construct }\text{a }\text{matrix }\text{where }\text{the }\text{first }\text{row }\text{contains }\text{unit }\text{vectors }\hat{\text{i}}, \hat{\text{j}}, \text{and }\hat{\text{k}}; \text{and }\text{the }\text{second }\text{and }\text{third }\text{rows }\text{are }\text{made }\text{of }\text{vectors }\, \left(-3,-\frac{37}{8},\frac{69}{8}\right)\,  \text{and }\, \left(-\frac{19}{2},-\frac{35}{4},-\frac{25}{4}\right)\, : \\
+ \left(
+\begin{array}{ccc}
+ \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\
+ -3 & -\frac{37}{8} & \frac{69}{8} \\
+ -\frac{19}{2} & -\frac{35}{4} & -\frac{25}{4} \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Take }\text{the }\text{determinant }\text{of }\text{this }\text{matrix}: \\
+ \left| 
+\begin{array}{ccc}
+ \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\
+ -3 & -\frac{37}{8} & \frac{69}{8} \\
+ -\frac{19}{2} & -\frac{35}{4} & -\frac{25}{4} \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Expand }\text{with }\text{respect }\text{to }\text{row }1: \\
+  \text{= }\left| 
+\begin{array}{ccc}
+ \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\
+ -3 & -\frac{37}{8} & \frac{69}{8} \\
+ -\frac{19}{2} & -\frac{35}{4} & -\frac{25}{4} \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{The }\text{determinant }\text{of }\text{the }\text{matrix }\left(
+\begin{array}{ccc}
+ \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\
+ -3 & -\frac{37}{8} & \frac{69}{8} \\
+ -\frac{19}{2} & -\frac{35}{4} & -\frac{25}{4} \\
+\end{array}
+\right) \text{is }\text{given }\text{by }\hat{\text{i}} \left| 
+\begin{array}{cc}
+ -\frac{37}{8} & \frac{69}{8} \\
+ -\frac{35}{4} & -\frac{25}{4} \\
+\end{array}
+\right| +\left(-\hat{\text{j}}\right) \left| 
+\begin{array}{cc}
+ -3 & \frac{69}{8} \\
+ -\frac{19}{2} & -\frac{25}{4} \\
+\end{array}
+\right| +\hat{\text{k}} \left| 
+\begin{array}{cc}
+ -3 & -\frac{37}{8} \\
+ -\frac{19}{2} & -\frac{35}{4} \\
+\end{array}
+\right| : \\
+  \text{= }\hat{\text{i}} \left| 
+\begin{array}{cc}
+ -\frac{37}{8} & \frac{69}{8} \\
+ -\frac{35}{4} & -\frac{25}{4} \\
+\end{array}
+\right| +\left(-\hat{\text{j}}\right) \left| 
+\begin{array}{cc}
+ -3 & \frac{69}{8} \\
+ -\frac{19}{2} & -\frac{25}{4} \\
+\end{array}
+\right| +\hat{\text{k}} \left| 
+\begin{array}{cc}
+ -3 & -\frac{37}{8} \\
+ -\frac{19}{2} & -\frac{35}{4} \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \hat{\text{i}} \left| 
+\begin{array}{cc}
+ -\frac{37}{8} & \frac{69}{8} \\
+ -\frac{35}{4} & -\frac{25}{4} \\
+\end{array}
+\right| =\hat{\text{i}} \left(\left(-\frac{37}{8}\right)\, \left(-\frac{25}{4}\right)-\frac{69 (-35)}{8\ 4}\right)=\frac{\hat{\text{i}} 835}{8}=\fbox{$\frac{835 \hat{\text{i}}}{8}$}: \\
+  \text{= }\fbox{$\frac{835 \hat{\text{i}}}{8}$}+\left(-\hat{\text{j}}\right) \left| 
+\begin{array}{cc}
+ -3 & \frac{69}{8} \\
+ -\frac{19}{2} & -\frac{25}{4} \\
+\end{array}
+\right| +\hat{\text{k}} \left| 
+\begin{array}{cc}
+ -3 & -\frac{37}{8} \\
+ -\frac{19}{2} & -\frac{35}{4} \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ -\hat{\text{j}} \left| 
+\begin{array}{cc}
+ -3 & \frac{69}{8} \\
+ -\frac{19}{2} & -\frac{25}{4} \\
+\end{array}
+\right| =-\hat{\text{j}} \left((-3)\, \left(-\frac{25}{4}\right)-\frac{69 (-19)}{8\ 2}\right)=\frac{-\hat{\text{j}} 1611}{16}=\fbox{$-\frac{1611 \hat{\text{j}}}{16}$}: \\
+  \text{= }\frac{835 \hat{\text{i}}}{8}+\fbox{$-\frac{1611 \hat{\text{j}}}{16}$}+\hat{\text{k}} \left| 
+\begin{array}{cc}
+ -3 & -\frac{37}{8} \\
+ -\frac{19}{2} & -\frac{35}{4} \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \hat{\text{k}} \left| 
+\begin{array}{cc}
+ -3 & -\frac{37}{8} \\
+ -\frac{19}{2} & -\frac{35}{4} \\
+\end{array}
+\right| =\hat{\text{k}} \left((-3)\, \left(-\frac{35}{4}\right)-\left(-\frac{37}{8}\right)\, \left(-\frac{19}{2}\right)\right)=\frac{\hat{\text{k}} (-283)}{16}=\fbox{$-\frac{283 \hat{\text{k}}}{16}$}: \\
+  \text{= }\frac{835 \hat{\text{i}}}{8}-\frac{1611 \hat{\text{j}}}{16}+\fbox{$-\frac{283 \hat{\text{k}}}{16}$} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Order }\text{the }\text{terms }\text{in }\text{a }\text{more }\text{natural }\text{way}: \\
+  \text{= }\frac{835 \hat{\text{i}}}{8}-\frac{1611 \hat{\text{j}}}{16}-\frac{283 \hat{\text{k}}}{16} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \frac{835 \hat{\text{i}}}{8}-\frac{1611 \hat{\text{j}}}{16}-\frac{283 \hat{\text{k}}}{16}=\, \left(\frac{835}{8},-\frac{1611}{16},-\frac{283}{16}\right)\, : \\
+ \fbox{$
+\begin{array}{ll}
+ \text{Answer:} &  \\
+ \text{} & \, \left(\frac{835}{8},-\frac{1611}{16},-\frac{283}{16}\right)\,  \\
+\end{array}
+$} \\
+\end{array}
+ \\
+\end{array}

pretraining/mathematica/linear_algebra/cross_product_w_steps/2280.txt

@@ -0,0 +1,169 @@
+Problem:
+Find the cross product of the following vectors:
+$\left(
+\begin{array}{c}
+ -2 \\
+ \frac{18}{5} \\
+ 2 \\
+\end{array}
+\right)$ and
+$\left(
+\begin{array}{c}
+ \frac{42}{5} \\
+ -6 \\
+ -6 \\
+\end{array}
+\right)$
+Answer:
+\begin{array}{l}
+ 
+\begin{array}{l}
+ \text{Compute the following cross product}: \\
+ \, \left(-2,\frac{18}{5},2\right)\, \times \, \left(\frac{42}{5},-6,-6\right)\,  \\
+\end{array}
+ \\
+\hline
+ 
+\begin{array}{l}
+ \text{Construct }\text{a }\text{matrix }\text{where }\text{the }\text{first }\text{row }\text{contains }\text{unit }\text{vectors }\hat{\text{i}}, \hat{\text{j}}, \text{and }\hat{\text{k}}; \text{and }\text{the }\text{second }\text{and }\text{third }\text{rows }\text{are }\text{made }\text{of }\text{vectors }\, \left(-2,\frac{18}{5},2\right)\,  \text{and }\, \left(\frac{42}{5},-6,-6\right)\, : \\
+ \left(
+\begin{array}{ccc}
+ \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\
+ -2 & \frac{18}{5} & 2 \\
+ \frac{42}{5} & -6 & -6 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Take }\text{the }\text{determinant }\text{of }\text{this }\text{matrix}: \\
+ \left| 
+\begin{array}{ccc}
+ \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\
+ -2 & \frac{18}{5} & 2 \\
+ \frac{42}{5} & -6 & -6 \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Expand }\text{with }\text{respect }\text{to }\text{row }1: \\
+  \text{= }\left| 
+\begin{array}{ccc}
+ \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\
+ -2 & \frac{18}{5} & 2 \\
+ \frac{42}{5} & -6 & -6 \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{The }\text{determinant }\text{of }\text{the }\text{matrix }\left(
+\begin{array}{ccc}
+ \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\
+ -2 & \frac{18}{5} & 2 \\
+ \frac{42}{5} & -6 & -6 \\
+\end{array}
+\right) \text{is }\text{given }\text{by }\hat{\text{i}} \left| 
+\begin{array}{cc}
+ \frac{18}{5} & 2 \\
+ -6 & -6 \\
+\end{array}
+\right| +\left(-\hat{\text{j}}\right) \left| 
+\begin{array}{cc}
+ -2 & 2 \\
+ \frac{42}{5} & -6 \\
+\end{array}
+\right| +\hat{\text{k}} \left| 
+\begin{array}{cc}
+ -2 & \frac{18}{5} \\
+ \frac{42}{5} & -6 \\
+\end{array}
+\right| : \\
+  \text{= }\hat{\text{i}} \left| 
+\begin{array}{cc}
+ \frac{18}{5} & 2 \\
+ -6 & -6 \\
+\end{array}
+\right| +\left(-\hat{\text{j}}\right) \left| 
+\begin{array}{cc}
+ -2 & 2 \\
+ \frac{42}{5} & -6 \\
+\end{array}
+\right| +\hat{\text{k}} \left| 
+\begin{array}{cc}
+ -2 & \frac{18}{5} \\
+ \frac{42}{5} & -6 \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \hat{\text{i}} \left| 
+\begin{array}{cc}
+ \frac{18}{5} & 2 \\
+ -6 & -6 \\
+\end{array}
+\right| =\hat{\text{i}} \left(\frac{18 (-6)}{5}-2 (-6)\right)=\frac{\hat{\text{i}} (-48)}{5}=\fbox{$-\frac{48 \hat{\text{i}}}{5}$}: \\
+  \text{= }\fbox{$-\frac{48 \hat{\text{i}}}{5}$}+\left(-\hat{\text{j}}\right) \left| 
+\begin{array}{cc}
+ -2 & 2 \\
+ \frac{42}{5} & -6 \\
+\end{array}
+\right| +\hat{\text{k}} \left| 
+\begin{array}{cc}
+ -2 & \frac{18}{5} \\
+ \frac{42}{5} & -6 \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ -\hat{\text{j}} \left| 
+\begin{array}{cc}
+ -2 & 2 \\
+ \frac{42}{5} & -6 \\
+\end{array}
+\right| =-\hat{\text{j}} \left((-2)\, (-6)-\frac{2\ 42}{5}\right)=\frac{-\hat{\text{j}} (-24)}{5}=\fbox{$\frac{24 \hat{\text{j}}}{5}$}: \\
+  \text{= }\frac{-48 \hat{\text{i}}}{5}+\fbox{$\frac{24 \hat{\text{j}}}{5}$}+\hat{\text{k}} \left| 
+\begin{array}{cc}
+ -2 & \frac{18}{5} \\
+ \frac{42}{5} & -6 \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \hat{\text{k}} \left| 
+\begin{array}{cc}
+ -2 & \frac{18}{5} \\
+ \frac{42}{5} & -6 \\
+\end{array}
+\right| =\hat{\text{k}} \left((-2)\, (-6)-\frac{18\ 42}{5\ 5}\right)=\frac{\hat{\text{k}} (-456)}{25}=\fbox{$-\frac{456 \hat{\text{k}}}{25}$}: \\
+  \text{= }\frac{-48 \hat{\text{i}}}{5}+\frac{24 \hat{\text{j}}}{5}+\fbox{$-\frac{456 \hat{\text{k}}}{25}$} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Order }\text{the }\text{terms }\text{in }\text{a }\text{more }\text{natural }\text{way}: \\
+  \text{= }-\frac{48 \hat{\text{i}}}{5}+\frac{24 \hat{\text{j}}}{5}-\frac{456 \hat{\text{k}}}{25} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ -\frac{48 \hat{\text{i}}}{5}+\frac{24 \hat{\text{j}}}{5}-\frac{456 \hat{\text{k}}}{25}=\, \left(-\frac{48}{5},\frac{24}{5},-\frac{456}{25}\right)\, : \\
+ \fbox{$
+\begin{array}{ll}
+ \text{Answer:} &  \\
+ \text{} & \, \left(-\frac{48}{5},\frac{24}{5},-\frac{456}{25}\right)\,  \\
+\end{array}
+$} \\
+\end{array}
+ \\
+\end{array}

pretraining/mathematica/linear_algebra/cross_product_w_steps/2308.txt

@@ -0,0 +1,169 @@
+Problem:
+Find the cross product of the following vectors:
+$\left(
+\begin{array}{c}
+ \frac{88}{9} \\
+ \frac{59}{9} \\
+ -\frac{1}{3} \\
+\end{array}
+\right)$ and
+$\left(
+\begin{array}{c}
+ -\frac{23}{3} \\
+ \frac{28}{9} \\
+ \frac{44}{9} \\
+\end{array}
+\right)$
+Answer:
+\begin{array}{l}
+ 
+\begin{array}{l}
+ \text{Compute the following cross product}: \\
+ \, \left(\frac{88}{9},\frac{59}{9},-\frac{1}{3}\right)\, \times \, \left(-\frac{23}{3},\frac{28}{9},\frac{44}{9}\right)\,  \\
+\end{array}
+ \\
+\hline
+ 
+\begin{array}{l}
+ \text{Construct }\text{a }\text{matrix }\text{where }\text{the }\text{first }\text{row }\text{contains }\text{unit }\text{vectors }\hat{\text{i}}, \hat{\text{j}}, \text{and }\hat{\text{k}}; \text{and }\text{the }\text{second }\text{and }\text{third }\text{rows }\text{are }\text{made }\text{of }\text{vectors }\, \left(\frac{88}{9},\frac{59}{9},-\frac{1}{3}\right)\,  \text{and }\, \left(-\frac{23}{3},\frac{28}{9},\frac{44}{9}\right)\, : \\
+ \left(
+\begin{array}{ccc}
+ \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\
+ \frac{88}{9} & \frac{59}{9} & -\frac{1}{3} \\
+ -\frac{23}{3} & \frac{28}{9} & \frac{44}{9} \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Take }\text{the }\text{determinant }\text{of }\text{this }\text{matrix}: \\
+ \left| 
+\begin{array}{ccc}
+ \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\
+ \frac{88}{9} & \frac{59}{9} & -\frac{1}{3} \\
+ -\frac{23}{3} & \frac{28}{9} & \frac{44}{9} \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Expand }\text{with }\text{respect }\text{to }\text{row }1: \\
+  \text{= }\left| 
+\begin{array}{ccc}
+ \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\
+ \frac{88}{9} & \frac{59}{9} & -\frac{1}{3} \\
+ -\frac{23}{3} & \frac{28}{9} & \frac{44}{9} \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{The }\text{determinant }\text{of }\text{the }\text{matrix }\left(
+\begin{array}{ccc}
+ \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\
+ \frac{88}{9} & \frac{59}{9} & -\frac{1}{3} \\
+ -\frac{23}{3} & \frac{28}{9} & \frac{44}{9} \\
+\end{array}
+\right) \text{is }\text{given }\text{by }\hat{\text{i}} \left| 
+\begin{array}{cc}
+ \frac{59}{9} & -\frac{1}{3} \\
+ \frac{28}{9} & \frac{44}{9} \\
+\end{array}
+\right| +\left(-\hat{\text{j}}\right) \left| 
+\begin{array}{cc}
+ \frac{88}{9} & -\frac{1}{3} \\
+ -\frac{23}{3} & \frac{44}{9} \\
+\end{array}
+\right| +\hat{\text{k}} \left| 
+\begin{array}{cc}
+ \frac{88}{9} & \frac{59}{9} \\
+ -\frac{23}{3} & \frac{28}{9} \\
+\end{array}
+\right| : \\
+  \text{= }\hat{\text{i}} \left| 
+\begin{array}{cc}
+ \frac{59}{9} & -\frac{1}{3} \\
+ \frac{28}{9} & \frac{44}{9} \\
+\end{array}
+\right| +\left(-\hat{\text{j}}\right) \left| 
+\begin{array}{cc}
+ \frac{88}{9} & -\frac{1}{3} \\
+ -\frac{23}{3} & \frac{44}{9} \\
+\end{array}
+\right| +\hat{\text{k}} \left| 
+\begin{array}{cc}
+ \frac{88}{9} & \frac{59}{9} \\
+ -\frac{23}{3} & \frac{28}{9} \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \hat{\text{i}} \left| 
+\begin{array}{cc}
+ \frac{59}{9} & -\frac{1}{3} \\
+ \frac{28}{9} & \frac{44}{9} \\
+\end{array}
+\right| =\hat{\text{i}} \left(\frac{59\ 44}{9\ 9}-\frac{1}{3} \left(-\frac{28}{9}\right)\right)=\frac{\hat{\text{i}} 2680}{81}=\fbox{$\frac{2680 \hat{\text{i}}}{81}$}: \\
+  \text{= }\fbox{$\frac{2680 \hat{\text{i}}}{81}$}+\left(-\hat{\text{j}}\right) \left| 
+\begin{array}{cc}
+ \frac{88}{9} & -\frac{1}{3} \\
+ -\frac{23}{3} & \frac{44}{9} \\
+\end{array}
+\right| +\hat{\text{k}} \left| 
+\begin{array}{cc}
+ \frac{88}{9} & \frac{59}{9} \\
+ -\frac{23}{3} & \frac{28}{9} \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ -\hat{\text{j}} \left| 
+\begin{array}{cc}
+ \frac{88}{9} & -\frac{1}{3} \\
+ -\frac{23}{3} & \frac{44}{9} \\
+\end{array}
+\right| =-\hat{\text{j}} \left(\frac{88\ 44}{9\ 9}-\frac{23}{3\ 3}\right)=\frac{-\hat{\text{j}} 3665}{81}=\fbox{$-\frac{3665 \hat{\text{j}}}{81}$}: \\
+  \text{= }\frac{2680 \hat{\text{i}}}{81}+\fbox{$-\frac{3665 \hat{\text{j}}}{81}$}+\hat{\text{k}} \left| 
+\begin{array}{cc}
+ \frac{88}{9} & \frac{59}{9} \\
+ -\frac{23}{3} & \frac{28}{9} \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \hat{\text{k}} \left| 
+\begin{array}{cc}
+ \frac{88}{9} & \frac{59}{9} \\
+ -\frac{23}{3} & \frac{28}{9} \\
+\end{array}
+\right| =\hat{\text{k}} \left(\frac{88\ 28}{9\ 9}-\frac{59 (-23)}{9\ 3}\right)=\frac{\hat{\text{k}} 6535}{81}=\fbox{$\frac{6535 \hat{\text{k}}}{81}$}: \\
+  \text{= }\frac{2680 \hat{\text{i}}}{81}-\frac{3665 \hat{\text{j}}}{81}+\fbox{$\frac{6535 \hat{\text{k}}}{81}$} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Order }\text{the }\text{terms }\text{in }\text{a }\text{more }\text{natural }\text{way}: \\
+  \text{= }\frac{2680 \hat{\text{i}}}{81}-\frac{3665 \hat{\text{j}}}{81}+\frac{6535 \hat{\text{k}}}{81} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \frac{2680 \hat{\text{i}}}{81}-\frac{3665 \hat{\text{j}}}{81}+\frac{6535 \hat{\text{k}}}{81}=\, \left(\frac{2680}{81},-\frac{3665}{81},\frac{6535}{81}\right)\, : \\
+ \fbox{$
+\begin{array}{ll}
+ \text{Answer:} &  \\
+ \text{} & \, \left(\frac{2680}{81},-\frac{3665}{81},\frac{6535}{81}\right)\,  \\
+\end{array}
+$} \\
+\end{array}
+ \\
+\end{array}

pretraining/mathematica/linear_algebra/cross_product_w_steps/2354.txt

@@ -0,0 +1,169 @@
+Problem:
+Find the cross product of the following vectors:
+$\left(
+\begin{array}{c}
+ 0 \\
+ 9 \\
+ 1 \\
+\end{array}
+\right)$ and
+$\left(
+\begin{array}{c}
+ 4 \\
+ 5 \\
+ 2 \\
+\end{array}
+\right)$
+Answer:
+\begin{array}{l}
+ 
+\begin{array}{l}
+ \text{Compute the following cross product}: \\
+ \, (0,9,1)\, \times \, (4,5,2)\,  \\
+\end{array}
+ \\
+\hline
+ 
+\begin{array}{l}
+ \text{Construct }\text{a }\text{matrix }\text{where }\text{the }\text{first }\text{row }\text{contains }\text{unit }\text{vectors }\hat{\text{i}}, \hat{\text{j}}, \text{and }\hat{\text{k}}; \text{and }\text{the }\text{second }\text{and }\text{third }\text{rows }\text{are }\text{made }\text{of }\text{vectors }\, (0,9,1)\,  \text{and }\, (4,5,2)\, : \\
+ \left(
+\begin{array}{ccc}
+ \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\
+ 0 & 9 & 1 \\
+ 4 & 5 & 2 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Take }\text{the }\text{determinant }\text{of }\text{this }\text{matrix}: \\
+ \left| 
+\begin{array}{ccc}
+ \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\
+ 0 & 9 & 1 \\
+ 4 & 5 & 2 \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Expand }\text{with }\text{respect }\text{to }\text{row }1: \\
+  \text{= }\left| 
+\begin{array}{ccc}
+ \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\
+ 0 & 9 & 1 \\
+ 4 & 5 & 2 \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{The }\text{determinant }\text{of }\text{the }\text{matrix }\left(
+\begin{array}{ccc}
+ \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\
+ 0 & 9 & 1 \\
+ 4 & 5 & 2 \\
+\end{array}
+\right) \text{is }\text{given }\text{by }\hat{\text{i}} \left| 
+\begin{array}{cc}
+ 9 & 1 \\
+ 5 & 2 \\
+\end{array}
+\right| +\left(-\hat{\text{j}}\right) \left| 
+\begin{array}{cc}
+ 0 & 1 \\
+ 4 & 2 \\
+\end{array}
+\right| +\hat{\text{k}} \left| 
+\begin{array}{cc}
+ 0 & 9 \\
+ 4 & 5 \\
+\end{array}
+\right| : \\
+  \text{= }\hat{\text{i}} \left| 
+\begin{array}{cc}
+ 9 & 1 \\
+ 5 & 2 \\
+\end{array}
+\right| +\left(-\hat{\text{j}}\right) \left| 
+\begin{array}{cc}
+ 0 & 1 \\
+ 4 & 2 \\
+\end{array}
+\right| +\hat{\text{k}} \left| 
+\begin{array}{cc}
+ 0 & 9 \\
+ 4 & 5 \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \hat{\text{i}} \left| 
+\begin{array}{cc}
+ 9 & 1 \\
+ 5 & 2 \\
+\end{array}
+\right| =\hat{\text{i}} (9\ 2-1\ 5)=\hat{\text{i}} 13=\fbox{$13 \hat{\text{i}}$}: \\
+  \text{= }\fbox{$13 \hat{\text{i}}$}+\left(-\hat{\text{j}}\right) \left| 
+\begin{array}{cc}
+ 0 & 1 \\
+ 4 & 2 \\
+\end{array}
+\right| +\hat{\text{k}} \left| 
+\begin{array}{cc}
+ 0 & 9 \\
+ 4 & 5 \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ -\hat{\text{j}} \left| 
+\begin{array}{cc}
+ 0 & 1 \\
+ 4 & 2 \\
+\end{array}
+\right| =-\hat{\text{j}} (0\ 2-1\ 4)=-\hat{\text{j}} (-4)=\fbox{$4 \hat{\text{j}}$}: \\
+  \text{= }13 \hat{\text{i}}+\fbox{$4 \hat{\text{j}}$}+\hat{\text{k}} \left| 
+\begin{array}{cc}
+ 0 & 9 \\
+ 4 & 5 \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \hat{\text{k}} \left| 
+\begin{array}{cc}
+ 0 & 9 \\
+ 4 & 5 \\
+\end{array}
+\right| =\hat{\text{k}} (0\ 5-9\ 4)=\hat{\text{k}} (-36)=\fbox{$-36 \hat{\text{k}}$}: \\
+  \text{= }13 \hat{\text{i}}+4 \hat{\text{j}}+\fbox{$-36 \hat{\text{k}}$} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Order }\text{the }\text{terms }\text{in }\text{a }\text{more }\text{natural }\text{way}: \\
+  \text{= }13 \hat{\text{i}}+4 \hat{\text{j}}-36 \hat{\text{k}} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ 13 \hat{\text{i}}+4 \hat{\text{j}}-36 \hat{\text{k}}=\, (13,4,-36)\, : \\
+ \fbox{$
+\begin{array}{ll}
+ \text{Answer:} &  \\
+ \text{} & \, (13,4,-36)\,  \\
+\end{array}
+$} \\
+\end{array}
+ \\
+\end{array}

pretraining/mathematica/linear_algebra/cross_product_w_steps/241.txt

@@ -0,0 +1,169 @@
+Problem:
+Find the cross product of the following vectors:
+$\left(
+\begin{array}{c}
+ 4 \\
+ -6 \\
+ -8 \\
+\end{array}
+\right)$ and
+$\left(
+\begin{array}{c}
+ 9 \\
+ 4 \\
+ 7 \\
+\end{array}
+\right)$
+Answer:
+\begin{array}{l}
+ 
+\begin{array}{l}
+ \text{Compute the following cross product}: \\
+ \, (4,-6,-8)\, \times \, (9,4,7)\,  \\
+\end{array}
+ \\
+\hline
+ 
+\begin{array}{l}
+ \text{Construct }\text{a }\text{matrix }\text{where }\text{the }\text{first }\text{row }\text{contains }\text{unit }\text{vectors }\hat{\text{i}}, \hat{\text{j}}, \text{and }\hat{\text{k}}; \text{and }\text{the }\text{second }\text{and }\text{third }\text{rows }\text{are }\text{made }\text{of }\text{vectors }\, (4,-6,-8)\,  \text{and }\, (9,4,7)\, : \\
+ \left(
+\begin{array}{ccc}
+ \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\
+ 4 & -6 & -8 \\
+ 9 & 4 & 7 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Take }\text{the }\text{determinant }\text{of }\text{this }\text{matrix}: \\
+ \left| 
+\begin{array}{ccc}
+ \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\
+ 4 & -6 & -8 \\
+ 9 & 4 & 7 \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Expand }\text{with }\text{respect }\text{to }\text{row }1: \\
+  \text{= }\left| 
+\begin{array}{ccc}
+ \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\
+ 4 & -6 & -8 \\
+ 9 & 4 & 7 \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{The }\text{determinant }\text{of }\text{the }\text{matrix }\left(
+\begin{array}{ccc}
+ \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\
+ 4 & -6 & -8 \\
+ 9 & 4 & 7 \\
+\end{array}
+\right) \text{is }\text{given }\text{by }\hat{\text{i}} \left| 
+\begin{array}{cc}
+ -6 & -8 \\
+ 4 & 7 \\
+\end{array}
+\right| +\left(-\hat{\text{j}}\right) \left| 
+\begin{array}{cc}
+ 4 & -8 \\
+ 9 & 7 \\
+\end{array}
+\right| +\hat{\text{k}} \left| 
+\begin{array}{cc}
+ 4 & -6 \\
+ 9 & 4 \\
+\end{array}
+\right| : \\
+  \text{= }\hat{\text{i}} \left| 
+\begin{array}{cc}
+ -6 & -8 \\
+ 4 & 7 \\
+\end{array}
+\right| +\left(-\hat{\text{j}}\right) \left| 
+\begin{array}{cc}
+ 4 & -8 \\
+ 9 & 7 \\
+\end{array}
+\right| +\hat{\text{k}} \left| 
+\begin{array}{cc}
+ 4 & -6 \\
+ 9 & 4 \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \hat{\text{i}} \left| 
+\begin{array}{cc}
+ -6 & -8 \\
+ 4 & 7 \\
+\end{array}
+\right| =\hat{\text{i}} ((-6)\, \times \, 7-(-8)\, \times \, 4)=\hat{\text{i}} (-10)=\fbox{$-10 \hat{\text{i}}$}: \\
+  \text{= }\fbox{$-10 \hat{\text{i}}$}+\left(-\hat{\text{j}}\right) \left| 
+\begin{array}{cc}
+ 4 & -8 \\
+ 9 & 7 \\
+\end{array}
+\right| +\hat{\text{k}} \left| 
+\begin{array}{cc}
+ 4 & -6 \\
+ 9 & 4 \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ -\hat{\text{j}} \left| 
+\begin{array}{cc}
+ 4 & -8 \\
+ 9 & 7 \\
+\end{array}
+\right| =-\hat{\text{j}} (4\ 7-(-8)\, \times \, 9)=-\hat{\text{j}} 100=\fbox{$-100 \hat{\text{j}}$}: \\
+  \text{= }-10 \hat{\text{i}}+\fbox{$-100 \hat{\text{j}}$}+\hat{\text{k}} \left| 
+\begin{array}{cc}
+ 4 & -6 \\
+ 9 & 4 \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \hat{\text{k}} \left| 
+\begin{array}{cc}
+ 4 & -6 \\
+ 9 & 4 \\
+\end{array}
+\right| =\hat{\text{k}} (4\ 4-(-6)\, \times \, 9)=\hat{\text{k}} 70=\fbox{$70 \hat{\text{k}}$}: \\
+  \text{= }-10 \hat{\text{i}}-100 \hat{\text{j}}+\fbox{$70 \hat{\text{k}}$} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Order }\text{the }\text{terms }\text{in }\text{a }\text{more }\text{natural }\text{way}: \\
+  \text{= }-10 \hat{\text{i}}-100 \hat{\text{j}}+70 \hat{\text{k}} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ -10 \hat{\text{i}}-100 \hat{\text{j}}+70 \hat{\text{k}}=\, (-10,-100,70)\, : \\
+ \fbox{$
+\begin{array}{ll}
+ \text{Answer:} &  \\
+ \text{} & \, (-10,-100,70)\,  \\
+\end{array}
+$} \\
+\end{array}
+ \\
+\end{array}

pretraining/mathematica/linear_algebra/cross_product_w_steps/2450.txt

@@ -0,0 +1,169 @@
+Problem:
+Find the cross product of the following vectors:
+$\left(
+\begin{array}{c}
+ \frac{14}{3} \\
+ 6 \\
+ \frac{2}{3} \\
+\end{array}
+\right)$ and
+$\left(
+\begin{array}{c}
+ 3 \\
+ -\frac{8}{3} \\
+ -\frac{8}{3} \\
+\end{array}
+\right)$
+Answer:
+\begin{array}{l}
+ 
+\begin{array}{l}
+ \text{Compute the following cross product}: \\
+ \, \left(\frac{14}{3},6,\frac{2}{3}\right)\, \times \, \left(3,-\frac{8}{3},-\frac{8}{3}\right)\,  \\
+\end{array}
+ \\
+\hline
+ 
+\begin{array}{l}
+ \text{Construct }\text{a }\text{matrix }\text{where }\text{the }\text{first }\text{row }\text{contains }\text{unit }\text{vectors }\hat{\text{i}}, \hat{\text{j}}, \text{and }\hat{\text{k}}; \text{and }\text{the }\text{second }\text{and }\text{third }\text{rows }\text{are }\text{made }\text{of }\text{vectors }\, \left(\frac{14}{3},6,\frac{2}{3}\right)\,  \text{and }\, \left(3,-\frac{8}{3},-\frac{8}{3}\right)\, : \\
+ \left(
+\begin{array}{ccc}
+ \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\
+ \frac{14}{3} & 6 & \frac{2}{3} \\
+ 3 & -\frac{8}{3} & -\frac{8}{3} \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Take }\text{the }\text{determinant }\text{of }\text{this }\text{matrix}: \\
+ \left| 
+\begin{array}{ccc}
+ \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\
+ \frac{14}{3} & 6 & \frac{2}{3} \\
+ 3 & -\frac{8}{3} & -\frac{8}{3} \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Expand }\text{with }\text{respect }\text{to }\text{row }1: \\
+  \text{= }\left| 
+\begin{array}{ccc}
+ \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\
+ \frac{14}{3} & 6 & \frac{2}{3} \\
+ 3 & -\frac{8}{3} & -\frac{8}{3} \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{The }\text{determinant }\text{of }\text{the }\text{matrix }\left(
+\begin{array}{ccc}
+ \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\
+ \frac{14}{3} & 6 & \frac{2}{3} \\
+ 3 & -\frac{8}{3} & -\frac{8}{3} \\
+\end{array}
+\right) \text{is }\text{given }\text{by }\hat{\text{i}} \left| 
+\begin{array}{cc}
+ 6 & \frac{2}{3} \\
+ -\frac{8}{3} & -\frac{8}{3} \\
+\end{array}
+\right| +\left(-\hat{\text{j}}\right) \left| 
+\begin{array}{cc}
+ \frac{14}{3} & \frac{2}{3} \\
+ 3 & -\frac{8}{3} \\
+\end{array}
+\right| +\hat{\text{k}} \left| 
+\begin{array}{cc}
+ \frac{14}{3} & 6 \\
+ 3 & -\frac{8}{3} \\
+\end{array}
+\right| : \\
+  \text{= }\hat{\text{i}} \left| 
+\begin{array}{cc}
+ 6 & \frac{2}{3} \\
+ -\frac{8}{3} & -\frac{8}{3} \\
+\end{array}
+\right| +\left(-\hat{\text{j}}\right) \left| 
+\begin{array}{cc}
+ \frac{14}{3} & \frac{2}{3} \\
+ 3 & -\frac{8}{3} \\
+\end{array}
+\right| +\hat{\text{k}} \left| 
+\begin{array}{cc}
+ \frac{14}{3} & 6 \\
+ 3 & -\frac{8}{3} \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \hat{\text{i}} \left| 
+\begin{array}{cc}
+ 6 & \frac{2}{3} \\
+ -\frac{8}{3} & -\frac{8}{3} \\
+\end{array}
+\right| =\hat{\text{i}} \left(\frac{6 (-8)}{3}-\frac{2 (-8)}{3\ 3}\right)=\frac{\hat{\text{i}} (-128)}{9}=\fbox{$-\frac{128 \hat{\text{i}}}{9}$}: \\
+  \text{= }\fbox{$-\frac{128 \hat{\text{i}}}{9}$}+\left(-\hat{\text{j}}\right) \left| 
+\begin{array}{cc}
+ \frac{14}{3} & \frac{2}{3} \\
+ 3 & -\frac{8}{3} \\
+\end{array}
+\right| +\hat{\text{k}} \left| 
+\begin{array}{cc}
+ \frac{14}{3} & 6 \\
+ 3 & -\frac{8}{3} \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ -\hat{\text{j}} \left| 
+\begin{array}{cc}
+ \frac{14}{3} & \frac{2}{3} \\
+ 3 & -\frac{8}{3} \\
+\end{array}
+\right| =-\hat{\text{j}} \left(\frac{14 (-8)}{3\ 3}-\frac{2\ 3}{3}\right)=\frac{-\hat{\text{j}} (-130)}{9}=\fbox{$\frac{130 \hat{\text{j}}}{9}$}: \\
+  \text{= }\frac{-128 \hat{\text{i}}}{9}+\fbox{$\frac{130 \hat{\text{j}}}{9}$}+\hat{\text{k}} \left| 
+\begin{array}{cc}
+ \frac{14}{3} & 6 \\
+ 3 & -\frac{8}{3} \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \hat{\text{k}} \left| 
+\begin{array}{cc}
+ \frac{14}{3} & 6 \\
+ 3 & -\frac{8}{3} \\
+\end{array}
+\right| =\hat{\text{k}} \left(\frac{14 (-8)}{3\ 3}-6\ 3\right)=\frac{\hat{\text{k}} (-274)}{9}=\fbox{$-\frac{274 \hat{\text{k}}}{9}$}: \\
+  \text{= }\frac{-128 \hat{\text{i}}}{9}+\frac{130 \hat{\text{j}}}{9}+\fbox{$-\frac{274 \hat{\text{k}}}{9}$} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Order }\text{the }\text{terms }\text{in }\text{a }\text{more }\text{natural }\text{way}: \\
+  \text{= }-\frac{128 \hat{\text{i}}}{9}+\frac{130 \hat{\text{j}}}{9}-\frac{274 \hat{\text{k}}}{9} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ -\frac{128 \hat{\text{i}}}{9}+\frac{130 \hat{\text{j}}}{9}-\frac{274 \hat{\text{k}}}{9}=\, \left(-\frac{128}{9},\frac{130}{9},-\frac{274}{9}\right)\, : \\
+ \fbox{$
+\begin{array}{ll}
+ \text{Answer:} &  \\
+ \text{} & \, \left(-\frac{128}{9},\frac{130}{9},-\frac{274}{9}\right)\,  \\
+\end{array}
+$} \\
+\end{array}
+ \\
+\end{array}

pretraining/mathematica/linear_algebra/cross_product_w_steps/2549.txt

@@ -0,0 +1,169 @@
+Problem:
+Find the cross product of the following vectors:
+$\left(
+\begin{array}{c}
+ \frac{125}{16} \\
+ \frac{61}{8} \\
+ -4 \\
+\end{array}
+\right)$ and
+$\left(
+\begin{array}{c}
+ \frac{61}{8} \\
+ -\frac{41}{16} \\
+ -\frac{73}{16} \\
+\end{array}
+\right)$
+Answer:
+\begin{array}{l}
+ 
+\begin{array}{l}
+ \text{Compute the following cross product}: \\
+ \, \left(\frac{125}{16},\frac{61}{8},-4\right)\, \times \, \left(\frac{61}{8},-\frac{41}{16},-\frac{73}{16}\right)\,  \\
+\end{array}
+ \\
+\hline
+ 
+\begin{array}{l}
+ \text{Construct }\text{a }\text{matrix }\text{where }\text{the }\text{first }\text{row }\text{contains }\text{unit }\text{vectors }\hat{\text{i}}, \hat{\text{j}}, \text{and }\hat{\text{k}}; \text{and }\text{the }\text{second }\text{and }\text{third }\text{rows }\text{are }\text{made }\text{of }\text{vectors }\, \left(\frac{125}{16},\frac{61}{8},-4\right)\,  \text{and }\, \left(\frac{61}{8},-\frac{41}{16},-\frac{73}{16}\right)\, : \\
+ \left(
+\begin{array}{ccc}
+ \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\
+ \frac{125}{16} & \frac{61}{8} & -4 \\
+ \frac{61}{8} & -\frac{41}{16} & -\frac{73}{16} \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Take }\text{the }\text{determinant }\text{of }\text{this }\text{matrix}: \\
+ \left| 
+\begin{array}{ccc}
+ \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\
+ \frac{125}{16} & \frac{61}{8} & -4 \\
+ \frac{61}{8} & -\frac{41}{16} & -\frac{73}{16} \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Expand }\text{with }\text{respect }\text{to }\text{row }1: \\
+  \text{= }\left| 
+\begin{array}{ccc}
+ \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\
+ \frac{125}{16} & \frac{61}{8} & -4 \\
+ \frac{61}{8} & -\frac{41}{16} & -\frac{73}{16} \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{The }\text{determinant }\text{of }\text{the }\text{matrix }\left(
+\begin{array}{ccc}
+ \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\
+ \frac{125}{16} & \frac{61}{8} & -4 \\
+ \frac{61}{8} & -\frac{41}{16} & -\frac{73}{16} \\
+\end{array}
+\right) \text{is }\text{given }\text{by }\hat{\text{i}} \left| 
+\begin{array}{cc}
+ \frac{61}{8} & -4 \\
+ -\frac{41}{16} & -\frac{73}{16} \\
+\end{array}
+\right| +\left(-\hat{\text{j}}\right) \left| 
+\begin{array}{cc}
+ \frac{125}{16} & -4 \\
+ \frac{61}{8} & -\frac{73}{16} \\
+\end{array}
+\right| +\hat{\text{k}} \left| 
+\begin{array}{cc}
+ \frac{125}{16} & \frac{61}{8} \\
+ \frac{61}{8} & -\frac{41}{16} \\
+\end{array}
+\right| : \\
+  \text{= }\hat{\text{i}} \left| 
+\begin{array}{cc}
+ \frac{61}{8} & -4 \\
+ -\frac{41}{16} & -\frac{73}{16} \\
+\end{array}
+\right| +\left(-\hat{\text{j}}\right) \left| 
+\begin{array}{cc}
+ \frac{125}{16} & -4 \\
+ \frac{61}{8} & -\frac{73}{16} \\
+\end{array}
+\right| +\hat{\text{k}} \left| 
+\begin{array}{cc}
+ \frac{125}{16} & \frac{61}{8} \\
+ \frac{61}{8} & -\frac{41}{16} \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \hat{\text{i}} \left| 
+\begin{array}{cc}
+ \frac{61}{8} & -4 \\
+ -\frac{41}{16} & -\frac{73}{16} \\
+\end{array}
+\right| =\hat{\text{i}} \left(\frac{61 (-73)}{8\ 16}-(-4)\, \left(-\frac{41}{16}\right)\right)=\frac{\hat{\text{i}} (-5765)}{128}=\fbox{$-\frac{5765 \hat{\text{i}}}{128}$}: \\
+  \text{= }\fbox{$-\frac{5765 \hat{\text{i}}}{128}$}+\left(-\hat{\text{j}}\right) \left| 
+\begin{array}{cc}
+ \frac{125}{16} & -4 \\
+ \frac{61}{8} & -\frac{73}{16} \\
+\end{array}
+\right| +\hat{\text{k}} \left| 
+\begin{array}{cc}
+ \frac{125}{16} & \frac{61}{8} \\
+ \frac{61}{8} & -\frac{41}{16} \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ -\hat{\text{j}} \left| 
+\begin{array}{cc}
+ \frac{125}{16} & -4 \\
+ \frac{61}{8} & -\frac{73}{16} \\
+\end{array}
+\right| =-\hat{\text{j}} \left(\frac{125 (-73)}{16\ 16}-(-4)\, \times \, \frac{61}{8}\right)=\frac{-\hat{\text{j}} (-1317)}{256}=\fbox{$\frac{1317 \hat{\text{j}}}{256}$}: \\
+  \text{= }\frac{-5765 \hat{\text{i}}}{128}+\fbox{$\frac{1317 \hat{\text{j}}}{256}$}+\hat{\text{k}} \left| 
+\begin{array}{cc}
+ \frac{125}{16} & \frac{61}{8} \\
+ \frac{61}{8} & -\frac{41}{16} \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \hat{\text{k}} \left| 
+\begin{array}{cc}
+ \frac{125}{16} & \frac{61}{8} \\
+ \frac{61}{8} & -\frac{41}{16} \\
+\end{array}
+\right| =\hat{\text{k}} \left(\frac{125 (-41)}{16\ 16}-\frac{61\ 61}{8\ 8}\right)=\frac{\hat{\text{k}} (-20009)}{256}=\fbox{$-\frac{20009 \hat{\text{k}}}{256}$}: \\
+  \text{= }\frac{-5765 \hat{\text{i}}}{128}+\frac{1317 \hat{\text{j}}}{256}+\fbox{$-\frac{20009 \hat{\text{k}}}{256}$} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Order }\text{the }\text{terms }\text{in }\text{a }\text{more }\text{natural }\text{way}: \\
+  \text{= }-\frac{5765 \hat{\text{i}}}{128}+\frac{1317 \hat{\text{j}}}{256}-\frac{20009 \hat{\text{k}}}{256} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ -\frac{5765 \hat{\text{i}}}{128}+\frac{1317 \hat{\text{j}}}{256}-\frac{20009 \hat{\text{k}}}{256}=\, \left(-\frac{5765}{128},\frac{1317}{256},-\frac{20009}{256}\right)\, : \\
+ \fbox{$
+\begin{array}{ll}
+ \text{Answer:} &  \\
+ \text{} & \, \left(-\frac{5765}{128},\frac{1317}{256},-\frac{20009}{256}\right)\,  \\
+\end{array}
+$} \\
+\end{array}
+ \\
+\end{array}

pretraining/mathematica/linear_algebra/cross_product_w_steps/256.txt

@@ -0,0 +1,169 @@
+Problem:
+Find the cross product of the following vectors:
+$\left(
+\begin{array}{c}
+ \frac{2}{3} \\
+ -2 \\
+ -9 \\
+\end{array}
+\right)$ and
+$\left(
+\begin{array}{c}
+ -5 \\
+ 1 \\
+ -2 \\
+\end{array}
+\right)$
+Answer:
+\begin{array}{l}
+ 
+\begin{array}{l}
+ \text{Compute the following cross product}: \\
+ \, \left(\frac{2}{3},-2,-9\right)\, \times \, (-5,1,-2)\,  \\
+\end{array}
+ \\
+\hline
+ 
+\begin{array}{l}
+ \text{Construct }\text{a }\text{matrix }\text{where }\text{the }\text{first }\text{row }\text{contains }\text{unit }\text{vectors }\hat{\text{i}}, \hat{\text{j}}, \text{and }\hat{\text{k}}; \text{and }\text{the }\text{second }\text{and }\text{third }\text{rows }\text{are }\text{made }\text{of }\text{vectors }\, \left(\frac{2}{3},-2,-9\right)\,  \text{and }\, (-5,1,-2)\, : \\
+ \left(
+\begin{array}{ccc}
+ \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\
+ \frac{2}{3} & -2 & -9 \\
+ -5 & 1 & -2 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Take }\text{the }\text{determinant }\text{of }\text{this }\text{matrix}: \\
+ \left| 
+\begin{array}{ccc}
+ \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\
+ \frac{2}{3} & -2 & -9 \\
+ -5 & 1 & -2 \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Expand }\text{with }\text{respect }\text{to }\text{row }1: \\
+  \text{= }\left| 
+\begin{array}{ccc}
+ \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\
+ \frac{2}{3} & -2 & -9 \\
+ -5 & 1 & -2 \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{The }\text{determinant }\text{of }\text{the }\text{matrix }\left(
+\begin{array}{ccc}
+ \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\
+ \frac{2}{3} & -2 & -9 \\
+ -5 & 1 & -2 \\
+\end{array}
+\right) \text{is }\text{given }\text{by }\hat{\text{i}} \left| 
+\begin{array}{cc}
+ -2 & -9 \\
+ 1 & -2 \\
+\end{array}
+\right| +\left(-\hat{\text{j}}\right) \left| 
+\begin{array}{cc}
+ \frac{2}{3} & -9 \\
+ -5 & -2 \\
+\end{array}
+\right| +\hat{\text{k}} \left| 
+\begin{array}{cc}
+ \frac{2}{3} & -2 \\
+ -5 & 1 \\
+\end{array}
+\right| : \\
+  \text{= }\hat{\text{i}} \left| 
+\begin{array}{cc}
+ -2 & -9 \\
+ 1 & -2 \\
+\end{array}
+\right| +\left(-\hat{\text{j}}\right) \left| 
+\begin{array}{cc}
+ \frac{2}{3} & -9 \\
+ -5 & -2 \\
+\end{array}
+\right| +\hat{\text{k}} \left| 
+\begin{array}{cc}
+ \frac{2}{3} & -2 \\
+ -5 & 1 \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \hat{\text{i}} \left| 
+\begin{array}{cc}
+ -2 & -9 \\
+ 1 & -2 \\
+\end{array}
+\right| =\hat{\text{i}} ((-2)\, (-2)-(-9)\, \times \, 1)=\hat{\text{i}} 13=\fbox{$13 \hat{\text{i}}$}: \\
+  \text{= }\fbox{$13 \hat{\text{i}}$}+\left(-\hat{\text{j}}\right) \left| 
+\begin{array}{cc}
+ \frac{2}{3} & -9 \\
+ -5 & -2 \\
+\end{array}
+\right| +\hat{\text{k}} \left| 
+\begin{array}{cc}
+ \frac{2}{3} & -2 \\
+ -5 & 1 \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ -\hat{\text{j}} \left| 
+\begin{array}{cc}
+ \frac{2}{3} & -9 \\
+ -5 & -2 \\
+\end{array}
+\right| =-\hat{\text{j}} \left(\frac{2 (-2)}{3}-(-9)\, (-5)\right)=\frac{-\hat{\text{j}} (-139)}{3}=\fbox{$\frac{139 \hat{\text{j}}}{3}$}: \\
+  \text{= }13 \hat{\text{i}}+\fbox{$\frac{139 \hat{\text{j}}}{3}$}+\hat{\text{k}} \left| 
+\begin{array}{cc}
+ \frac{2}{3} & -2 \\
+ -5 & 1 \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \hat{\text{k}} \left| 
+\begin{array}{cc}
+ \frac{2}{3} & -2 \\
+ -5 & 1 \\
+\end{array}
+\right| =\hat{\text{k}} \left(\frac{2}{3}-(-2)\, (-5)\right)=\frac{\hat{\text{k}} (-28)}{3}=\fbox{$-\frac{28 \hat{\text{k}}}{3}$}: \\
+  \text{= }13 \hat{\text{i}}+\frac{139 \hat{\text{j}}}{3}+\fbox{$-\frac{28 \hat{\text{k}}}{3}$} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Order }\text{the }\text{terms }\text{in }\text{a }\text{more }\text{natural }\text{way}: \\
+  \text{= }13 \hat{\text{i}}+\frac{139 \hat{\text{j}}}{3}-\frac{28 \hat{\text{k}}}{3} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ 13 \hat{\text{i}}+\frac{139 \hat{\text{j}}}{3}-\frac{28 \hat{\text{k}}}{3}=\, \left(13,\frac{139}{3},-\frac{28}{3}\right)\, : \\
+ \fbox{$
+\begin{array}{ll}
+ \text{Answer:} &  \\
+ \text{} & \, \left(13,\frac{139}{3},-\frac{28}{3}\right)\,  \\
+\end{array}
+$} \\
+\end{array}
+ \\
+\end{array}

pretraining/mathematica/linear_algebra/cross_product_w_steps/2601.txt

@@ -0,0 +1,169 @@
+Problem:
+Find the cross product of the following vectors:
+$\left(
+\begin{array}{c}
+ -4 \\
+ -7 \\
+ -7 \\
+\end{array}
+\right)$ and
+$\left(
+\begin{array}{c}
+ -5 \\
+ 6 \\
+ 6 \\
+\end{array}
+\right)$
+Answer:
+\begin{array}{l}
+ 
+\begin{array}{l}
+ \text{Compute the following cross product}: \\
+ \, (-4,-7,-7)\, \times \, (-5,6,6)\,  \\
+\end{array}
+ \\
+\hline
+ 
+\begin{array}{l}
+ \text{Construct }\text{a }\text{matrix }\text{where }\text{the }\text{first }\text{row }\text{contains }\text{unit }\text{vectors }\hat{\text{i}}, \hat{\text{j}}, \text{and }\hat{\text{k}}; \text{and }\text{the }\text{second }\text{and }\text{third }\text{rows }\text{are }\text{made }\text{of }\text{vectors }\, (-4,-7,-7)\,  \text{and }\, (-5,6,6)\, : \\
+ \left(
+\begin{array}{ccc}
+ \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\
+ -4 & -7 & -7 \\
+ -5 & 6 & 6 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Take }\text{the }\text{determinant }\text{of }\text{this }\text{matrix}: \\
+ \left| 
+\begin{array}{ccc}
+ \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\
+ -4 & -7 & -7 \\
+ -5 & 6 & 6 \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Expand }\text{with }\text{respect }\text{to }\text{row }1: \\
+  \text{= }\left| 
+\begin{array}{ccc}
+ \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\
+ -4 & -7 & -7 \\
+ -5 & 6 & 6 \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{The }\text{determinant }\text{of }\text{the }\text{matrix }\left(
+\begin{array}{ccc}
+ \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\
+ -4 & -7 & -7 \\
+ -5 & 6 & 6 \\
+\end{array}
+\right) \text{is }\text{given }\text{by }\hat{\text{i}} \left| 
+\begin{array}{cc}
+ -7 & -7 \\
+ 6 & 6 \\
+\end{array}
+\right| +\left(-\hat{\text{j}}\right) \left| 
+\begin{array}{cc}
+ -4 & -7 \\
+ -5 & 6 \\
+\end{array}
+\right| +\hat{\text{k}} \left| 
+\begin{array}{cc}
+ -4 & -7 \\
+ -5 & 6 \\
+\end{array}
+\right| : \\
+  \text{= }\hat{\text{i}} \left| 
+\begin{array}{cc}
+ -7 & -7 \\
+ 6 & 6 \\
+\end{array}
+\right| +\left(-\hat{\text{j}}\right) \left| 
+\begin{array}{cc}
+ -4 & -7 \\
+ -5 & 6 \\
+\end{array}
+\right| +\hat{\text{k}} \left| 
+\begin{array}{cc}
+ -4 & -7 \\
+ -5 & 6 \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \hat{\text{i}} \left| 
+\begin{array}{cc}
+ -7 & -7 \\
+ 6 & 6 \\
+\end{array}
+\right| =\hat{\text{i}} ((-7)\, \times \, 6-(-7)\, \times \, 6)=\hat{\text{i}} 0=\fbox{$0$}: \\
+  \text{= }\fbox{$0$}+\left(-\hat{\text{j}}\right) \left| 
+\begin{array}{cc}
+ -4 & -7 \\
+ -5 & 6 \\
+\end{array}
+\right| +\hat{\text{k}} \left| 
+\begin{array}{cc}
+ -4 & -7 \\
+ -5 & 6 \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ -\hat{\text{j}} \left| 
+\begin{array}{cc}
+ -4 & -7 \\
+ -5 & 6 \\
+\end{array}
+\right| =-\hat{\text{j}} ((-4)\, \times \, 6-(-7)\, (-5))=-\hat{\text{j}} (-59)=\fbox{$59 \hat{\text{j}}$}: \\
+  \text{= }\fbox{$59 \hat{\text{j}}$}+\hat{\text{k}} \left| 
+\begin{array}{cc}
+ -4 & -7 \\
+ -5 & 6 \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \hat{\text{k}} \left| 
+\begin{array}{cc}
+ -4 & -7 \\
+ -5 & 6 \\
+\end{array}
+\right| =\hat{\text{k}} ((-4)\, \times \, 6-(-7)\, (-5))=\hat{\text{k}} (-59)=\fbox{$-59 \hat{\text{k}}$}: \\
+  \text{= }59 \hat{\text{j}}+\fbox{$-59 \hat{\text{k}}$} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Order }\text{the }\text{terms }\text{in }\text{a }\text{more }\text{natural }\text{way}: \\
+  \text{= }59 \hat{\text{j}}-59 \hat{\text{k}} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ 59 \hat{\text{j}}-59 \hat{\text{k}}=\, (0,59,-59)\, : \\
+ \fbox{$
+\begin{array}{ll}
+ \text{Answer:} &  \\
+ \text{} & \, (0,59,-59)\,  \\
+\end{array}
+$} \\
+\end{array}
+ \\
+\end{array}

pretraining/mathematica/linear_algebra/cross_product_w_steps/2744.txt

@@ -0,0 +1,169 @@
+Problem:
+Find the cross product of the following vectors:
+$\left(
+\begin{array}{c}
+ -\frac{33}{5} \\
+ -7 \\
+ -\frac{41}{5} \\
+\end{array}
+\right)$ and
+$\left(
+\begin{array}{c}
+ -\frac{28}{5} \\
+ \frac{6}{5} \\
+ -3 \\
+\end{array}
+\right)$
+Answer:
+\begin{array}{l}
+ 
+\begin{array}{l}
+ \text{Compute the following cross product}: \\
+ \, \left(-\frac{33}{5},-7,-\frac{41}{5}\right)\, \times \, \left(-\frac{28}{5},\frac{6}{5},-3\right)\,  \\
+\end{array}
+ \\
+\hline
+ 
+\begin{array}{l}
+ \text{Construct }\text{a }\text{matrix }\text{where }\text{the }\text{first }\text{row }\text{contains }\text{unit }\text{vectors }\hat{\text{i}}, \hat{\text{j}}, \text{and }\hat{\text{k}}; \text{and }\text{the }\text{second }\text{and }\text{third }\text{rows }\text{are }\text{made }\text{of }\text{vectors }\, \left(-\frac{33}{5},-7,-\frac{41}{5}\right)\,  \text{and }\, \left(-\frac{28}{5},\frac{6}{5},-3\right)\, : \\
+ \left(
+\begin{array}{ccc}
+ \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\
+ -\frac{33}{5} & -7 & -\frac{41}{5} \\
+ -\frac{28}{5} & \frac{6}{5} & -3 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Take }\text{the }\text{determinant }\text{of }\text{this }\text{matrix}: \\
+ \left| 
+\begin{array}{ccc}
+ \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\
+ -\frac{33}{5} & -7 & -\frac{41}{5} \\
+ -\frac{28}{5} & \frac{6}{5} & -3 \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Expand }\text{with }\text{respect }\text{to }\text{row }1: \\
+  \text{= }\left| 
+\begin{array}{ccc}
+ \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\
+ -\frac{33}{5} & -7 & -\frac{41}{5} \\
+ -\frac{28}{5} & \frac{6}{5} & -3 \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{The }\text{determinant }\text{of }\text{the }\text{matrix }\left(
+\begin{array}{ccc}
+ \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\
+ -\frac{33}{5} & -7 & -\frac{41}{5} \\
+ -\frac{28}{5} & \frac{6}{5} & -3 \\
+\end{array}
+\right) \text{is }\text{given }\text{by }\hat{\text{i}} \left| 
+\begin{array}{cc}
+ -7 & -\frac{41}{5} \\
+ \frac{6}{5} & -3 \\
+\end{array}
+\right| +\left(-\hat{\text{j}}\right) \left| 
+\begin{array}{cc}
+ -\frac{33}{5} & -\frac{41}{5} \\
+ -\frac{28}{5} & -3 \\
+\end{array}
+\right| +\hat{\text{k}} \left| 
+\begin{array}{cc}
+ -\frac{33}{5} & -7 \\
+ -\frac{28}{5} & \frac{6}{5} \\
+\end{array}
+\right| : \\
+  \text{= }\hat{\text{i}} \left| 
+\begin{array}{cc}
+ -7 & -\frac{41}{5} \\
+ \frac{6}{5} & -3 \\
+\end{array}
+\right| +\left(-\hat{\text{j}}\right) \left| 
+\begin{array}{cc}
+ -\frac{33}{5} & -\frac{41}{5} \\
+ -\frac{28}{5} & -3 \\
+\end{array}
+\right| +\hat{\text{k}} \left| 
+\begin{array}{cc}
+ -\frac{33}{5} & -7 \\
+ -\frac{28}{5} & \frac{6}{5} \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \hat{\text{i}} \left| 
+\begin{array}{cc}
+ -7 & -\frac{41}{5} \\
+ \frac{6}{5} & -3 \\
+\end{array}
+\right| =\hat{\text{i}} \left((-7)\, (-3)-\left(-\frac{41}{5}\right)\, \times \, \frac{6}{5}\right)=\frac{\hat{\text{i}} 771}{25}=\fbox{$\frac{771 \hat{\text{i}}}{25}$}: \\
+  \text{= }\fbox{$\frac{771 \hat{\text{i}}}{25}$}+\left(-\hat{\text{j}}\right) \left| 
+\begin{array}{cc}
+ -\frac{33}{5} & -\frac{41}{5} \\
+ -\frac{28}{5} & -3 \\
+\end{array}
+\right| +\hat{\text{k}} \left| 
+\begin{array}{cc}
+ -\frac{33}{5} & -7 \\
+ -\frac{28}{5} & \frac{6}{5} \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ -\hat{\text{j}} \left| 
+\begin{array}{cc}
+ -\frac{33}{5} & -\frac{41}{5} \\
+ -\frac{28}{5} & -3 \\
+\end{array}
+\right| =-\hat{\text{j}} \left(\left(-\frac{33}{5}\right)\, (-3)-\left(-\frac{41}{5}\right)\, \left(-\frac{28}{5}\right)\right)=\frac{-\hat{\text{j}} (-653)}{25}=\fbox{$\frac{653 \hat{\text{j}}}{25}$}: \\
+  \text{= }\frac{771 \hat{\text{i}}}{25}+\fbox{$\frac{653 \hat{\text{j}}}{25}$}+\hat{\text{k}} \left| 
+\begin{array}{cc}
+ -\frac{33}{5} & -7 \\
+ -\frac{28}{5} & \frac{6}{5} \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \hat{\text{k}} \left| 
+\begin{array}{cc}
+ -\frac{33}{5} & -7 \\
+ -\frac{28}{5} & \frac{6}{5} \\
+\end{array}
+\right| =\hat{\text{k}} \left(\left(-\frac{33}{5}\right)\, \times \, \frac{6}{5}-(-7)\, \left(-\frac{28}{5}\right)\right)=\frac{\hat{\text{k}} (-1178)}{25}=\fbox{$-\frac{1178 \hat{\text{k}}}{25}$}: \\
+  \text{= }\frac{771 \hat{\text{i}}}{25}+\frac{653 \hat{\text{j}}}{25}+\fbox{$-\frac{1178 \hat{\text{k}}}{25}$} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Order }\text{the }\text{terms }\text{in }\text{a }\text{more }\text{natural }\text{way}: \\
+  \text{= }\frac{771 \hat{\text{i}}}{25}+\frac{653 \hat{\text{j}}}{25}-\frac{1178 \hat{\text{k}}}{25} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \frac{771 \hat{\text{i}}}{25}+\frac{653 \hat{\text{j}}}{25}-\frac{1178 \hat{\text{k}}}{25}=\, \left(\frac{771}{25},\frac{653}{25},-\frac{1178}{25}\right)\, : \\
+ \fbox{$
+\begin{array}{ll}
+ \text{Answer:} &  \\
+ \text{} & \, \left(\frac{771}{25},\frac{653}{25},-\frac{1178}{25}\right)\,  \\
+\end{array}
+$} \\
+\end{array}
+ \\
+\end{array}

pretraining/mathematica/linear_algebra/cross_product_w_steps/280.txt

@@ -0,0 +1,169 @@
+Problem:
+Find the cross product of the following vectors:
+$\left(
+\begin{array}{c}
+ 1 \\
+ -6 \\
+ -8 \\
+\end{array}
+\right)$ and
+$\left(
+\begin{array}{c}
+ -7 \\
+ 7 \\
+ 6 \\
+\end{array}
+\right)$
+Answer:
+\begin{array}{l}
+ 
+\begin{array}{l}
+ \text{Compute the following cross product}: \\
+ \, (1,-6,-8)\, \times \, (-7,7,6)\,  \\
+\end{array}
+ \\
+\hline
+ 
+\begin{array}{l}
+ \text{Construct }\text{a }\text{matrix }\text{where }\text{the }\text{first }\text{row }\text{contains }\text{unit }\text{vectors }\hat{\text{i}}, \hat{\text{j}}, \text{and }\hat{\text{k}}; \text{and }\text{the }\text{second }\text{and }\text{third }\text{rows }\text{are }\text{made }\text{of }\text{vectors }\, (1,-6,-8)\,  \text{and }\, (-7,7,6)\, : \\
+ \left(
+\begin{array}{ccc}
+ \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\
+ 1 & -6 & -8 \\
+ -7 & 7 & 6 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Take }\text{the }\text{determinant }\text{of }\text{this }\text{matrix}: \\
+ \left| 
+\begin{array}{ccc}
+ \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\
+ 1 & -6 & -8 \\
+ -7 & 7 & 6 \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Expand }\text{with }\text{respect }\text{to }\text{row }1: \\
+  \text{= }\left| 
+\begin{array}{ccc}
+ \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\
+ 1 & -6 & -8 \\
+ -7 & 7 & 6 \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{The }\text{determinant }\text{of }\text{the }\text{matrix }\left(
+\begin{array}{ccc}
+ \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\
+ 1 & -6 & -8 \\
+ -7 & 7 & 6 \\
+\end{array}
+\right) \text{is }\text{given }\text{by }\hat{\text{i}} \left| 
+\begin{array}{cc}
+ -6 & -8 \\
+ 7 & 6 \\
+\end{array}
+\right| +\left(-\hat{\text{j}}\right) \left| 
+\begin{array}{cc}
+ 1 & -8 \\
+ -7 & 6 \\
+\end{array}
+\right| +\hat{\text{k}} \left| 
+\begin{array}{cc}
+ 1 & -6 \\
+ -7 & 7 \\
+\end{array}
+\right| : \\
+  \text{= }\hat{\text{i}} \left| 
+\begin{array}{cc}
+ -6 & -8 \\
+ 7 & 6 \\
+\end{array}
+\right| +\left(-\hat{\text{j}}\right) \left| 
+\begin{array}{cc}
+ 1 & -8 \\
+ -7 & 6 \\
+\end{array}
+\right| +\hat{\text{k}} \left| 
+\begin{array}{cc}
+ 1 & -6 \\
+ -7 & 7 \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \hat{\text{i}} \left| 
+\begin{array}{cc}
+ -6 & -8 \\
+ 7 & 6 \\
+\end{array}
+\right| =\hat{\text{i}} ((-6)\, \times \, 6-(-8)\, \times \, 7)=\hat{\text{i}} 20=\fbox{$20 \hat{\text{i}}$}: \\
+  \text{= }\fbox{$20 \hat{\text{i}}$}+\left(-\hat{\text{j}}\right) \left| 
+\begin{array}{cc}
+ 1 & -8 \\
+ -7 & 6 \\
+\end{array}
+\right| +\hat{\text{k}} \left| 
+\begin{array}{cc}
+ 1 & -6 \\
+ -7 & 7 \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ -\hat{\text{j}} \left| 
+\begin{array}{cc}
+ 1 & -8 \\
+ -7 & 6 \\
+\end{array}
+\right| =-\hat{\text{j}} (1\ 6-(-8)\, (-7))=-\hat{\text{j}} (-50)=\fbox{$50 \hat{\text{j}}$}: \\
+  \text{= }20 \hat{\text{i}}+\fbox{$50 \hat{\text{j}}$}+\hat{\text{k}} \left| 
+\begin{array}{cc}
+ 1 & -6 \\
+ -7 & 7 \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \hat{\text{k}} \left| 
+\begin{array}{cc}
+ 1 & -6 \\
+ -7 & 7 \\
+\end{array}
+\right| =\hat{\text{k}} (1\ 7-(-6)\, (-7))=\hat{\text{k}} (-35)=\fbox{$-35 \hat{\text{k}}$}: \\
+  \text{= }20 \hat{\text{i}}+50 \hat{\text{j}}+\fbox{$-35 \hat{\text{k}}$} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Order }\text{the }\text{terms }\text{in }\text{a }\text{more }\text{natural }\text{way}: \\
+  \text{= }20 \hat{\text{i}}+50 \hat{\text{j}}-35 \hat{\text{k}} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ 20 \hat{\text{i}}+50 \hat{\text{j}}-35 \hat{\text{k}}=\, (20,50,-35)\, : \\
+ \fbox{$
+\begin{array}{ll}
+ \text{Answer:} &  \\
+ \text{} & \, (20,50,-35)\,  \\
+\end{array}
+$} \\
+\end{array}
+ \\
+\end{array}

pretraining/mathematica/linear_algebra/cross_product_w_steps/2948.txt

@@ -0,0 +1,169 @@
+Problem:
+Find the cross product of the following vectors:
+$\left(
+\begin{array}{c}
+ -\frac{7}{2} \\
+ \frac{81}{10} \\
+ -9 \\
+\end{array}
+\right)$ and
+$\left(
+\begin{array}{c}
+ -\frac{87}{10} \\
+ -\frac{43}{10} \\
+ \frac{39}{10} \\
+\end{array}
+\right)$
+Answer:
+\begin{array}{l}
+ 
+\begin{array}{l}
+ \text{Compute the following cross product}: \\
+ \, \left(-\frac{7}{2},\frac{81}{10},-9\right)\, \times \, \left(-\frac{87}{10},-\frac{43}{10},\frac{39}{10}\right)\,  \\
+\end{array}
+ \\
+\hline
+ 
+\begin{array}{l}
+ \text{Construct }\text{a }\text{matrix }\text{where }\text{the }\text{first }\text{row }\text{contains }\text{unit }\text{vectors }\hat{\text{i}}, \hat{\text{j}}, \text{and }\hat{\text{k}}; \text{and }\text{the }\text{second }\text{and }\text{third }\text{rows }\text{are }\text{made }\text{of }\text{vectors }\, \left(-\frac{7}{2},\frac{81}{10},-9\right)\,  \text{and }\, \left(-\frac{87}{10},-\frac{43}{10},\frac{39}{10}\right)\, : \\
+ \left(
+\begin{array}{ccc}
+ \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\
+ -\frac{7}{2} & \frac{81}{10} & -9 \\
+ -\frac{87}{10} & -\frac{43}{10} & \frac{39}{10} \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Take }\text{the }\text{determinant }\text{of }\text{this }\text{matrix}: \\
+ \left| 
+\begin{array}{ccc}
+ \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\
+ -\frac{7}{2} & \frac{81}{10} & -9 \\
+ -\frac{87}{10} & -\frac{43}{10} & \frac{39}{10} \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Expand }\text{with }\text{respect }\text{to }\text{row }1: \\
+  \text{= }\left| 
+\begin{array}{ccc}
+ \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\
+ -\frac{7}{2} & \frac{81}{10} & -9 \\
+ -\frac{87}{10} & -\frac{43}{10} & \frac{39}{10} \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{The }\text{determinant }\text{of }\text{the }\text{matrix }\left(
+\begin{array}{ccc}
+ \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\
+ -\frac{7}{2} & \frac{81}{10} & -9 \\
+ -\frac{87}{10} & -\frac{43}{10} & \frac{39}{10} \\
+\end{array}
+\right) \text{is }\text{given }\text{by }\hat{\text{i}} \left| 
+\begin{array}{cc}
+ \frac{81}{10} & -9 \\
+ -\frac{43}{10} & \frac{39}{10} \\
+\end{array}
+\right| +\left(-\hat{\text{j}}\right) \left| 
+\begin{array}{cc}
+ -\frac{7}{2} & -9 \\
+ -\frac{87}{10} & \frac{39}{10} \\
+\end{array}
+\right| +\hat{\text{k}} \left| 
+\begin{array}{cc}
+ -\frac{7}{2} & \frac{81}{10} \\
+ -\frac{87}{10} & -\frac{43}{10} \\
+\end{array}
+\right| : \\
+  \text{= }\hat{\text{i}} \left| 
+\begin{array}{cc}
+ \frac{81}{10} & -9 \\
+ -\frac{43}{10} & \frac{39}{10} \\
+\end{array}
+\right| +\left(-\hat{\text{j}}\right) \left| 
+\begin{array}{cc}
+ -\frac{7}{2} & -9 \\
+ -\frac{87}{10} & \frac{39}{10} \\
+\end{array}
+\right| +\hat{\text{k}} \left| 
+\begin{array}{cc}
+ -\frac{7}{2} & \frac{81}{10} \\
+ -\frac{87}{10} & -\frac{43}{10} \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \hat{\text{i}} \left| 
+\begin{array}{cc}
+ \frac{81}{10} & -9 \\
+ -\frac{43}{10} & \frac{39}{10} \\
+\end{array}
+\right| =\hat{\text{i}} \left(\frac{81\ 39}{10\ 10}-(-9)\, \left(-\frac{43}{10}\right)\right)=\frac{\hat{\text{i}} (-711)}{100}=\fbox{$-\frac{711 \hat{\text{i}}}{100}$}: \\
+  \text{= }\fbox{$-\frac{711 \hat{\text{i}}}{100}$}+\left(-\hat{\text{j}}\right) \left| 
+\begin{array}{cc}
+ -\frac{7}{2} & -9 \\
+ -\frac{87}{10} & \frac{39}{10} \\
+\end{array}
+\right| +\hat{\text{k}} \left| 
+\begin{array}{cc}
+ -\frac{7}{2} & \frac{81}{10} \\
+ -\frac{87}{10} & -\frac{43}{10} \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ -\hat{\text{j}} \left| 
+\begin{array}{cc}
+ -\frac{7}{2} & -9 \\
+ -\frac{87}{10} & \frac{39}{10} \\
+\end{array}
+\right| =-\hat{\text{j}} \left(\left(-\frac{7}{2}\right)\, \times \, \frac{39}{10}-(-9)\, \left(-\frac{87}{10}\right)\right)=\frac{-\hat{\text{j}} (-1839)}{20}=\fbox{$\frac{1839 \hat{\text{j}}}{20}$}: \\
+  \text{= }\frac{-711 \hat{\text{i}}}{100}+\fbox{$\frac{1839 \hat{\text{j}}}{20}$}+\hat{\text{k}} \left| 
+\begin{array}{cc}
+ -\frac{7}{2} & \frac{81}{10} \\
+ -\frac{87}{10} & -\frac{43}{10} \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \hat{\text{k}} \left| 
+\begin{array}{cc}
+ -\frac{7}{2} & \frac{81}{10} \\
+ -\frac{87}{10} & -\frac{43}{10} \\
+\end{array}
+\right| =\hat{\text{k}} \left(\left(-\frac{7}{2}\right)\, \left(-\frac{43}{10}\right)-\frac{81 (-87)}{10\ 10}\right)=\frac{\hat{\text{k}} 2138}{25}=\fbox{$\frac{2138 \hat{\text{k}}}{25}$}: \\
+  \text{= }\frac{-711 \hat{\text{i}}}{100}+\frac{1839 \hat{\text{j}}}{20}+\fbox{$\frac{2138 \hat{\text{k}}}{25}$} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Order }\text{the }\text{terms }\text{in }\text{a }\text{more }\text{natural }\text{way}: \\
+  \text{= }-\frac{711 \hat{\text{i}}}{100}+\frac{1839 \hat{\text{j}}}{20}+\frac{2138 \hat{\text{k}}}{25} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ -\frac{711 \hat{\text{i}}}{100}+\frac{1839 \hat{\text{j}}}{20}+\frac{2138 \hat{\text{k}}}{25}=\, \left(-\frac{711}{100},\frac{1839}{20},\frac{2138}{25}\right)\, : \\
+ \fbox{$
+\begin{array}{ll}
+ \text{Answer:} &  \\
+ \text{} & \, \left(-\frac{711}{100},\frac{1839}{20},\frac{2138}{25}\right)\,  \\
+\end{array}
+$} \\
+\end{array}
+ \\
+\end{array}

pretraining/mathematica/linear_algebra/cross_product_w_steps/3013.txt

@@ -0,0 +1,169 @@
+Problem:
+Find the cross product of the following vectors:
+$\left(
+\begin{array}{c}
+ -8 \\
+ \frac{7}{2} \\
+ \frac{7}{2} \\
+\end{array}
+\right)$ and
+$\left(
+\begin{array}{c}
+ -2 \\
+ -\frac{15}{2} \\
+ \frac{19}{2} \\
+\end{array}
+\right)$
+Answer:
+\begin{array}{l}
+ 
+\begin{array}{l}
+ \text{Compute the following cross product}: \\
+ \, \left(-8,\frac{7}{2},\frac{7}{2}\right)\, \times \, \left(-2,-\frac{15}{2},\frac{19}{2}\right)\,  \\
+\end{array}
+ \\
+\hline
+ 
+\begin{array}{l}
+ \text{Construct }\text{a }\text{matrix }\text{where }\text{the }\text{first }\text{row }\text{contains }\text{unit }\text{vectors }\hat{\text{i}}, \hat{\text{j}}, \text{and }\hat{\text{k}}; \text{and }\text{the }\text{second }\text{and }\text{third }\text{rows }\text{are }\text{made }\text{of }\text{vectors }\, \left(-8,\frac{7}{2},\frac{7}{2}\right)\,  \text{and }\, \left(-2,-\frac{15}{2},\frac{19}{2}\right)\, : \\
+ \left(
+\begin{array}{ccc}
+ \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\
+ -8 & \frac{7}{2} & \frac{7}{2} \\
+ -2 & -\frac{15}{2} & \frac{19}{2} \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Take }\text{the }\text{determinant }\text{of }\text{this }\text{matrix}: \\
+ \left| 
+\begin{array}{ccc}
+ \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\
+ -8 & \frac{7}{2} & \frac{7}{2} \\
+ -2 & -\frac{15}{2} & \frac{19}{2} \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Expand }\text{with }\text{respect }\text{to }\text{row }1: \\
+  \text{= }\left| 
+\begin{array}{ccc}
+ \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\
+ -8 & \frac{7}{2} & \frac{7}{2} \\
+ -2 & -\frac{15}{2} & \frac{19}{2} \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{The }\text{determinant }\text{of }\text{the }\text{matrix }\left(
+\begin{array}{ccc}
+ \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\
+ -8 & \frac{7}{2} & \frac{7}{2} \\
+ -2 & -\frac{15}{2} & \frac{19}{2} \\
+\end{array}
+\right) \text{is }\text{given }\text{by }\hat{\text{i}} \left| 
+\begin{array}{cc}
+ \frac{7}{2} & \frac{7}{2} \\
+ -\frac{15}{2} & \frac{19}{2} \\
+\end{array}
+\right| +\left(-\hat{\text{j}}\right) \left| 
+\begin{array}{cc}
+ -8 & \frac{7}{2} \\
+ -2 & \frac{19}{2} \\
+\end{array}
+\right| +\hat{\text{k}} \left| 
+\begin{array}{cc}
+ -8 & \frac{7}{2} \\
+ -2 & -\frac{15}{2} \\
+\end{array}
+\right| : \\
+  \text{= }\hat{\text{i}} \left| 
+\begin{array}{cc}
+ \frac{7}{2} & \frac{7}{2} \\
+ -\frac{15}{2} & \frac{19}{2} \\
+\end{array}
+\right| +\left(-\hat{\text{j}}\right) \left| 
+\begin{array}{cc}
+ -8 & \frac{7}{2} \\
+ -2 & \frac{19}{2} \\
+\end{array}
+\right| +\hat{\text{k}} \left| 
+\begin{array}{cc}
+ -8 & \frac{7}{2} \\
+ -2 & -\frac{15}{2} \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \hat{\text{i}} \left| 
+\begin{array}{cc}
+ \frac{7}{2} & \frac{7}{2} \\
+ -\frac{15}{2} & \frac{19}{2} \\
+\end{array}
+\right| =\hat{\text{i}} \left(\frac{7\ 19}{2\ 2}-\frac{7 (-15)}{2\ 2}\right)=\frac{\hat{\text{i}} 119}{2}=\fbox{$\frac{119 \hat{\text{i}}}{2}$}: \\
+  \text{= }\fbox{$\frac{119 \hat{\text{i}}}{2}$}+\left(-\hat{\text{j}}\right) \left| 
+\begin{array}{cc}
+ -8 & \frac{7}{2} \\
+ -2 & \frac{19}{2} \\
+\end{array}
+\right| +\hat{\text{k}} \left| 
+\begin{array}{cc}
+ -8 & \frac{7}{2} \\
+ -2 & -\frac{15}{2} \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ -\hat{\text{j}} \left| 
+\begin{array}{cc}
+ -8 & \frac{7}{2} \\
+ -2 & \frac{19}{2} \\
+\end{array}
+\right| =-\hat{\text{j}} \left((-8)\, \times \, \frac{19}{2}-\frac{7 (-2)}{2}\right)=-\hat{\text{j}} (-69)=\fbox{$69 \hat{\text{j}}$}: \\
+  \text{= }\frac{119 \hat{\text{i}}}{2}+\fbox{$69 \hat{\text{j}}$}+\hat{\text{k}} \left| 
+\begin{array}{cc}
+ -8 & \frac{7}{2} \\
+ -2 & -\frac{15}{2} \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \hat{\text{k}} \left| 
+\begin{array}{cc}
+ -8 & \frac{7}{2} \\
+ -2 & -\frac{15}{2} \\
+\end{array}
+\right| =\hat{\text{k}} \left((-8)\, \left(-\frac{15}{2}\right)-\frac{7 (-2)}{2}\right)=\hat{\text{k}} 67=\fbox{$67 \hat{\text{k}}$}: \\
+  \text{= }\frac{119 \hat{\text{i}}}{2}+69 \hat{\text{j}}+\fbox{$67 \hat{\text{k}}$} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Order }\text{the }\text{terms }\text{in }\text{a }\text{more }\text{natural }\text{way}: \\
+  \text{= }\frac{119 \hat{\text{i}}}{2}+69 \hat{\text{j}}+67 \hat{\text{k}} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \frac{119 \hat{\text{i}}}{2}+69 \hat{\text{j}}+67 \hat{\text{k}}=\, \left(\frac{119}{2},69,67\right)\, : \\
+ \fbox{$
+\begin{array}{ll}
+ \text{Answer:} &  \\
+ \text{} & \, \left(\frac{119}{2},69,67\right)\,  \\
+\end{array}
+$} \\
+\end{array}
+ \\
+\end{array}

pretraining/mathematica/linear_algebra/cross_product_w_steps/3038.txt

@@ -0,0 +1,169 @@
+Problem:
+Find the cross product of the following vectors:
+$\left(
+\begin{array}{c}
+ \frac{48}{5} \\
+ -9 \\
+ -\frac{16}{5} \\
+\end{array}
+\right)$ and
+$\left(
+\begin{array}{c}
+ \frac{18}{5} \\
+ -\frac{36}{5} \\
+ -\frac{24}{5} \\
+\end{array}
+\right)$
+Answer:
+\begin{array}{l}
+ 
+\begin{array}{l}
+ \text{Compute the following cross product}: \\
+ \, \left(\frac{48}{5},-9,-\frac{16}{5}\right)\, \times \, \left(\frac{18}{5},-\frac{36}{5},-\frac{24}{5}\right)\,  \\
+\end{array}
+ \\
+\hline
+ 
+\begin{array}{l}
+ \text{Construct }\text{a }\text{matrix }\text{where }\text{the }\text{first }\text{row }\text{contains }\text{unit }\text{vectors }\hat{\text{i}}, \hat{\text{j}}, \text{and }\hat{\text{k}}; \text{and }\text{the }\text{second }\text{and }\text{third }\text{rows }\text{are }\text{made }\text{of }\text{vectors }\, \left(\frac{48}{5},-9,-\frac{16}{5}\right)\,  \text{and }\, \left(\frac{18}{5},-\frac{36}{5},-\frac{24}{5}\right)\, : \\
+ \left(
+\begin{array}{ccc}
+ \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\
+ \frac{48}{5} & -9 & -\frac{16}{5} \\
+ \frac{18}{5} & -\frac{36}{5} & -\frac{24}{5} \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Take }\text{the }\text{determinant }\text{of }\text{this }\text{matrix}: \\
+ \left| 
+\begin{array}{ccc}
+ \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\
+ \frac{48}{5} & -9 & -\frac{16}{5} \\
+ \frac{18}{5} & -\frac{36}{5} & -\frac{24}{5} \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Expand }\text{with }\text{respect }\text{to }\text{row }1: \\
+  \text{= }\left| 
+\begin{array}{ccc}
+ \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\
+ \frac{48}{5} & -9 & -\frac{16}{5} \\
+ \frac{18}{5} & -\frac{36}{5} & -\frac{24}{5} \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{The }\text{determinant }\text{of }\text{the }\text{matrix }\left(
+\begin{array}{ccc}
+ \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\
+ \frac{48}{5} & -9 & -\frac{16}{5} \\
+ \frac{18}{5} & -\frac{36}{5} & -\frac{24}{5} \\
+\end{array}
+\right) \text{is }\text{given }\text{by }\hat{\text{i}} \left| 
+\begin{array}{cc}
+ -9 & -\frac{16}{5} \\
+ -\frac{36}{5} & -\frac{24}{5} \\
+\end{array}
+\right| +\left(-\hat{\text{j}}\right) \left| 
+\begin{array}{cc}
+ \frac{48}{5} & -\frac{16}{5} \\
+ \frac{18}{5} & -\frac{24}{5} \\
+\end{array}
+\right| +\hat{\text{k}} \left| 
+\begin{array}{cc}
+ \frac{48}{5} & -9 \\
+ \frac{18}{5} & -\frac{36}{5} \\
+\end{array}
+\right| : \\
+  \text{= }\hat{\text{i}} \left| 
+\begin{array}{cc}
+ -9 & -\frac{16}{5} \\
+ -\frac{36}{5} & -\frac{24}{5} \\
+\end{array}
+\right| +\left(-\hat{\text{j}}\right) \left| 
+\begin{array}{cc}
+ \frac{48}{5} & -\frac{16}{5} \\
+ \frac{18}{5} & -\frac{24}{5} \\
+\end{array}
+\right| +\hat{\text{k}} \left| 
+\begin{array}{cc}
+ \frac{48}{5} & -9 \\
+ \frac{18}{5} & -\frac{36}{5} \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \hat{\text{i}} \left| 
+\begin{array}{cc}
+ -9 & -\frac{16}{5} \\
+ -\frac{36}{5} & -\frac{24}{5} \\
+\end{array}
+\right| =\hat{\text{i}} \left((-9)\, \left(-\frac{24}{5}\right)-\left(-\frac{16}{5}\right)\, \left(-\frac{36}{5}\right)\right)=\frac{\hat{\text{i}} 504}{25}=\fbox{$\frac{504 \hat{\text{i}}}{25}$}: \\
+  \text{= }\fbox{$\frac{504 \hat{\text{i}}}{25}$}+\left(-\hat{\text{j}}\right) \left| 
+\begin{array}{cc}
+ \frac{48}{5} & -\frac{16}{5} \\
+ \frac{18}{5} & -\frac{24}{5} \\
+\end{array}
+\right| +\hat{\text{k}} \left| 
+\begin{array}{cc}
+ \frac{48}{5} & -9 \\
+ \frac{18}{5} & -\frac{36}{5} \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ -\hat{\text{j}} \left| 
+\begin{array}{cc}
+ \frac{48}{5} & -\frac{16}{5} \\
+ \frac{18}{5} & -\frac{24}{5} \\
+\end{array}
+\right| =-\hat{\text{j}} \left(\frac{48 (-24)}{5\ 5}-\left(-\frac{16}{5}\right)\, \times \, \frac{18}{5}\right)=\frac{-\hat{\text{j}} (-864)}{25}=\fbox{$\frac{864 \hat{\text{j}}}{25}$}: \\
+  \text{= }\frac{504 \hat{\text{i}}}{25}+\fbox{$\frac{864 \hat{\text{j}}}{25}$}+\hat{\text{k}} \left| 
+\begin{array}{cc}
+ \frac{48}{5} & -9 \\
+ \frac{18}{5} & -\frac{36}{5} \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \hat{\text{k}} \left| 
+\begin{array}{cc}
+ \frac{48}{5} & -9 \\
+ \frac{18}{5} & -\frac{36}{5} \\
+\end{array}
+\right| =\hat{\text{k}} \left(\frac{48 (-36)}{5\ 5}-(-9)\, \times \, \frac{18}{5}\right)=\frac{\hat{\text{k}} (-918)}{25}=\fbox{$-\frac{918 \hat{\text{k}}}{25}$}: \\
+  \text{= }\frac{504 \hat{\text{i}}}{25}+\frac{864 \hat{\text{j}}}{25}+\fbox{$-\frac{918 \hat{\text{k}}}{25}$} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Order }\text{the }\text{terms }\text{in }\text{a }\text{more }\text{natural }\text{way}: \\
+  \text{= }\frac{504 \hat{\text{i}}}{25}+\frac{864 \hat{\text{j}}}{25}-\frac{918 \hat{\text{k}}}{25} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \frac{504 \hat{\text{i}}}{25}+\frac{864 \hat{\text{j}}}{25}-\frac{918 \hat{\text{k}}}{25}=\, \left(\frac{504}{25},\frac{864}{25},-\frac{918}{25}\right)\, : \\
+ \fbox{$
+\begin{array}{ll}
+ \text{Answer:} &  \\
+ \text{} & \, \left(\frac{504}{25},\frac{864}{25},-\frac{918}{25}\right)\,  \\
+\end{array}
+$} \\
+\end{array}
+ \\
+\end{array}

pretraining/mathematica/linear_algebra/cross_product_w_steps/3069.txt

@@ -0,0 +1,169 @@
+Problem:
+Find the cross product of the following vectors:
+$\left(
+\begin{array}{c}
+ -2 \\
+ 0 \\
+ 6 \\
+\end{array}
+\right)$ and
+$\left(
+\begin{array}{c}
+ -5 \\
+ 8 \\
+ 9 \\
+\end{array}
+\right)$
+Answer:
+\begin{array}{l}
+ 
+\begin{array}{l}
+ \text{Compute the following cross product}: \\
+ \, (-2,0,6)\, \times \, (-5,8,9)\,  \\
+\end{array}
+ \\
+\hline
+ 
+\begin{array}{l}
+ \text{Construct }\text{a }\text{matrix }\text{where }\text{the }\text{first }\text{row }\text{contains }\text{unit }\text{vectors }\hat{\text{i}}, \hat{\text{j}}, \text{and }\hat{\text{k}}; \text{and }\text{the }\text{second }\text{and }\text{third }\text{rows }\text{are }\text{made }\text{of }\text{vectors }\, (-2,0,6)\,  \text{and }\, (-5,8,9)\, : \\
+ \left(
+\begin{array}{ccc}
+ \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\
+ -2 & 0 & 6 \\
+ -5 & 8 & 9 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Take }\text{the }\text{determinant }\text{of }\text{this }\text{matrix}: \\
+ \left| 
+\begin{array}{ccc}
+ \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\
+ -2 & 0 & 6 \\
+ -5 & 8 & 9 \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Expand }\text{with }\text{respect }\text{to }\text{row }1: \\
+  \text{= }\left| 
+\begin{array}{ccc}
+ \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\
+ -2 & 0 & 6 \\
+ -5 & 8 & 9 \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{The }\text{determinant }\text{of }\text{the }\text{matrix }\left(
+\begin{array}{ccc}
+ \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\
+ -2 & 0 & 6 \\
+ -5 & 8 & 9 \\
+\end{array}
+\right) \text{is }\text{given }\text{by }\hat{\text{i}} \left| 
+\begin{array}{cc}
+ 0 & 6 \\
+ 8 & 9 \\
+\end{array}
+\right| +\left(-\hat{\text{j}}\right) \left| 
+\begin{array}{cc}
+ -2 & 6 \\
+ -5 & 9 \\
+\end{array}
+\right| +\hat{\text{k}} \left| 
+\begin{array}{cc}
+ -2 & 0 \\
+ -5 & 8 \\
+\end{array}
+\right| : \\
+  \text{= }\hat{\text{i}} \left| 
+\begin{array}{cc}
+ 0 & 6 \\
+ 8 & 9 \\
+\end{array}
+\right| +\left(-\hat{\text{j}}\right) \left| 
+\begin{array}{cc}
+ -2 & 6 \\
+ -5 & 9 \\
+\end{array}
+\right| +\hat{\text{k}} \left| 
+\begin{array}{cc}
+ -2 & 0 \\
+ -5 & 8 \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \hat{\text{i}} \left| 
+\begin{array}{cc}
+ 0 & 6 \\
+ 8 & 9 \\
+\end{array}
+\right| =\hat{\text{i}} (0\ 9-6\ 8)=\hat{\text{i}} (-48)=\fbox{$-48 \hat{\text{i}}$}: \\
+  \text{= }\fbox{$-48 \hat{\text{i}}$}+\left(-\hat{\text{j}}\right) \left| 
+\begin{array}{cc}
+ -2 & 6 \\
+ -5 & 9 \\
+\end{array}
+\right| +\hat{\text{k}} \left| 
+\begin{array}{cc}
+ -2 & 0 \\
+ -5 & 8 \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ -\hat{\text{j}} \left| 
+\begin{array}{cc}
+ -2 & 6 \\
+ -5 & 9 \\
+\end{array}
+\right| =-\hat{\text{j}} ((-2)\, \times \, 9-6 (-5))=-\hat{\text{j}} 12=\fbox{$-12 \hat{\text{j}}$}: \\
+  \text{= }-48 \hat{\text{i}}+\fbox{$-12 \hat{\text{j}}$}+\hat{\text{k}} \left| 
+\begin{array}{cc}
+ -2 & 0 \\
+ -5 & 8 \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \hat{\text{k}} \left| 
+\begin{array}{cc}
+ -2 & 0 \\
+ -5 & 8 \\
+\end{array}
+\right| =\hat{\text{k}} ((-2)\, \times \, 8-0 (-5))=\hat{\text{k}} (-16)=\fbox{$-16 \hat{\text{k}}$}: \\
+  \text{= }-48 \hat{\text{i}}-12 \hat{\text{j}}+\fbox{$-16 \hat{\text{k}}$} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Order }\text{the }\text{terms }\text{in }\text{a }\text{more }\text{natural }\text{way}: \\
+  \text{= }-48 \hat{\text{i}}-12 \hat{\text{j}}-16 \hat{\text{k}} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ -48 \hat{\text{i}}-12 \hat{\text{j}}-16 \hat{\text{k}}=\, (-48,-12,-16)\, : \\
+ \fbox{$
+\begin{array}{ll}
+ \text{Answer:} &  \\
+ \text{} & \, (-48,-12,-16)\,  \\
+\end{array}
+$} \\
+\end{array}
+ \\
+\end{array}

pretraining/mathematica/linear_algebra/cross_product_w_steps/3077.txt

@@ -0,0 +1,169 @@
+Problem:
+Find the cross product of the following vectors:
+$\left(
+\begin{array}{c}
+ -7 \\
+ 8 \\
+ 4 \\
+\end{array}
+\right)$ and
+$\left(
+\begin{array}{c}
+ 1 \\
+ 2 \\
+ 6 \\
+\end{array}
+\right)$
+Answer:
+\begin{array}{l}
+ 
+\begin{array}{l}
+ \text{Compute the following cross product}: \\
+ \, (-7,8,4)\, \times \, (1,2,6)\,  \\
+\end{array}
+ \\
+\hline
+ 
+\begin{array}{l}
+ \text{Construct }\text{a }\text{matrix }\text{where }\text{the }\text{first }\text{row }\text{contains }\text{unit }\text{vectors }\hat{\text{i}}, \hat{\text{j}}, \text{and }\hat{\text{k}}; \text{and }\text{the }\text{second }\text{and }\text{third }\text{rows }\text{are }\text{made }\text{of }\text{vectors }\, (-7,8,4)\,  \text{and }\, (1,2,6)\, : \\
+ \left(
+\begin{array}{ccc}
+ \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\
+ -7 & 8 & 4 \\
+ 1 & 2 & 6 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Take }\text{the }\text{determinant }\text{of }\text{this }\text{matrix}: \\
+ \left| 
+\begin{array}{ccc}
+ \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\
+ -7 & 8 & 4 \\
+ 1 & 2 & 6 \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Expand }\text{with }\text{respect }\text{to }\text{row }1: \\
+  \text{= }\left| 
+\begin{array}{ccc}
+ \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\
+ -7 & 8 & 4 \\
+ 1 & 2 & 6 \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{The }\text{determinant }\text{of }\text{the }\text{matrix }\left(
+\begin{array}{ccc}
+ \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\
+ -7 & 8 & 4 \\
+ 1 & 2 & 6 \\
+\end{array}
+\right) \text{is }\text{given }\text{by }\hat{\text{i}} \left| 
+\begin{array}{cc}
+ 8 & 4 \\
+ 2 & 6 \\
+\end{array}
+\right| +\left(-\hat{\text{j}}\right) \left| 
+\begin{array}{cc}
+ -7 & 4 \\
+ 1 & 6 \\
+\end{array}
+\right| +\hat{\text{k}} \left| 
+\begin{array}{cc}
+ -7 & 8 \\
+ 1 & 2 \\
+\end{array}
+\right| : \\
+  \text{= }\hat{\text{i}} \left| 
+\begin{array}{cc}
+ 8 & 4 \\
+ 2 & 6 \\
+\end{array}
+\right| +\left(-\hat{\text{j}}\right) \left| 
+\begin{array}{cc}
+ -7 & 4 \\
+ 1 & 6 \\
+\end{array}
+\right| +\hat{\text{k}} \left| 
+\begin{array}{cc}
+ -7 & 8 \\
+ 1 & 2 \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \hat{\text{i}} \left| 
+\begin{array}{cc}
+ 8 & 4 \\
+ 2 & 6 \\
+\end{array}
+\right| =\hat{\text{i}} (8\ 6-4\ 2)=\hat{\text{i}} 40=\fbox{$40 \hat{\text{i}}$}: \\
+  \text{= }\fbox{$40 \hat{\text{i}}$}+\left(-\hat{\text{j}}\right) \left| 
+\begin{array}{cc}
+ -7 & 4 \\
+ 1 & 6 \\
+\end{array}
+\right| +\hat{\text{k}} \left| 
+\begin{array}{cc}
+ -7 & 8 \\
+ 1 & 2 \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ -\hat{\text{j}} \left| 
+\begin{array}{cc}
+ -7 & 4 \\
+ 1 & 6 \\
+\end{array}
+\right| =-\hat{\text{j}} ((-7)\, \times \, 6-4\ 1)=-\hat{\text{j}} (-46)=\fbox{$46 \hat{\text{j}}$}: \\
+  \text{= }40 \hat{\text{i}}+\fbox{$46 \hat{\text{j}}$}+\hat{\text{k}} \left| 
+\begin{array}{cc}
+ -7 & 8 \\
+ 1 & 2 \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \hat{\text{k}} \left| 
+\begin{array}{cc}
+ -7 & 8 \\
+ 1 & 2 \\
+\end{array}
+\right| =\hat{\text{k}} ((-7)\, \times \, 2-8\ 1)=\hat{\text{k}} (-22)=\fbox{$-22 \hat{\text{k}}$}: \\
+  \text{= }40 \hat{\text{i}}+46 \hat{\text{j}}+\fbox{$-22 \hat{\text{k}}$} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Order }\text{the }\text{terms }\text{in }\text{a }\text{more }\text{natural }\text{way}: \\
+  \text{= }40 \hat{\text{i}}+46 \hat{\text{j}}-22 \hat{\text{k}} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ 40 \hat{\text{i}}+46 \hat{\text{j}}-22 \hat{\text{k}}=\, (40,46,-22)\, : \\
+ \fbox{$
+\begin{array}{ll}
+ \text{Answer:} &  \\
+ \text{} & \, (40,46,-22)\,  \\
+\end{array}
+$} \\
+\end{array}
+ \\
+\end{array}

pretraining/mathematica/linear_algebra/cross_product_w_steps/3108.txt

@@ -0,0 +1,169 @@
+Problem:
+Find the cross product of the following vectors:
+$\left(
+\begin{array}{c}
+ -5 \\
+ 9 \\
+ -6 \\
+\end{array}
+\right)$ and
+$\left(
+\begin{array}{c}
+ -5 \\
+ -7 \\
+ -5 \\
+\end{array}
+\right)$
+Answer:
+\begin{array}{l}
+ 
+\begin{array}{l}
+ \text{Compute the following cross product}: \\
+ \, (-5,9,-6)\, \times \, (-5,-7,-5)\,  \\
+\end{array}
+ \\
+\hline
+ 
+\begin{array}{l}
+ \text{Construct }\text{a }\text{matrix }\text{where }\text{the }\text{first }\text{row }\text{contains }\text{unit }\text{vectors }\hat{\text{i}}, \hat{\text{j}}, \text{and }\hat{\text{k}}; \text{and }\text{the }\text{second }\text{and }\text{third }\text{rows }\text{are }\text{made }\text{of }\text{vectors }\, (-5,9,-6)\,  \text{and }\, (-5,-7,-5)\, : \\
+ \left(
+\begin{array}{ccc}
+ \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\
+ -5 & 9 & -6 \\
+ -5 & -7 & -5 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Take }\text{the }\text{determinant }\text{of }\text{this }\text{matrix}: \\
+ \left| 
+\begin{array}{ccc}
+ \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\
+ -5 & 9 & -6 \\
+ -5 & -7 & -5 \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Expand }\text{with }\text{respect }\text{to }\text{row }1: \\
+  \text{= }\left| 
+\begin{array}{ccc}
+ \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\
+ -5 & 9 & -6 \\
+ -5 & -7 & -5 \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{The }\text{determinant }\text{of }\text{the }\text{matrix }\left(
+\begin{array}{ccc}
+ \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\
+ -5 & 9 & -6 \\
+ -5 & -7 & -5 \\
+\end{array}
+\right) \text{is }\text{given }\text{by }\hat{\text{i}} \left| 
+\begin{array}{cc}
+ 9 & -6 \\
+ -7 & -5 \\
+\end{array}
+\right| +\left(-\hat{\text{j}}\right) \left| 
+\begin{array}{cc}
+ -5 & -6 \\
+ -5 & -5 \\
+\end{array}
+\right| +\hat{\text{k}} \left| 
+\begin{array}{cc}
+ -5 & 9 \\
+ -5 & -7 \\
+\end{array}
+\right| : \\
+  \text{= }\hat{\text{i}} \left| 
+\begin{array}{cc}
+ 9 & -6 \\
+ -7 & -5 \\
+\end{array}
+\right| +\left(-\hat{\text{j}}\right) \left| 
+\begin{array}{cc}
+ -5 & -6 \\
+ -5 & -5 \\
+\end{array}
+\right| +\hat{\text{k}} \left| 
+\begin{array}{cc}
+ -5 & 9 \\
+ -5 & -7 \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \hat{\text{i}} \left| 
+\begin{array}{cc}
+ 9 & -6 \\
+ -7 & -5 \\
+\end{array}
+\right| =\hat{\text{i}} (9 (-5)-(-6)\, (-7))=\hat{\text{i}} (-87)=\fbox{$-87 \hat{\text{i}}$}: \\
+  \text{= }\fbox{$-87 \hat{\text{i}}$}+\left(-\hat{\text{j}}\right) \left| 
+\begin{array}{cc}
+ -5 & -6 \\
+ -5 & -5 \\
+\end{array}
+\right| +\hat{\text{k}} \left| 
+\begin{array}{cc}
+ -5 & 9 \\
+ -5 & -7 \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ -\hat{\text{j}} \left| 
+\begin{array}{cc}
+ -5 & -6 \\
+ -5 & -5 \\
+\end{array}
+\right| =-\hat{\text{j}} ((-5)\, (-5)-(-6)\, (-5))=-\hat{\text{j}} (-5)=\fbox{$5 \hat{\text{j}}$}: \\
+  \text{= }-87 \hat{\text{i}}+\fbox{$5 \hat{\text{j}}$}+\hat{\text{k}} \left| 
+\begin{array}{cc}
+ -5 & 9 \\
+ -5 & -7 \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \hat{\text{k}} \left| 
+\begin{array}{cc}
+ -5 & 9 \\
+ -5 & -7 \\
+\end{array}
+\right| =\hat{\text{k}} ((-5)\, (-7)-9 (-5))=\hat{\text{k}} 80=\fbox{$80 \hat{\text{k}}$}: \\
+  \text{= }-87 \hat{\text{i}}+5 \hat{\text{j}}+\fbox{$80 \hat{\text{k}}$} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Order }\text{the }\text{terms }\text{in }\text{a }\text{more }\text{natural }\text{way}: \\
+  \text{= }-87 \hat{\text{i}}+5 \hat{\text{j}}+80 \hat{\text{k}} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ -87 \hat{\text{i}}+5 \hat{\text{j}}+80 \hat{\text{k}}=\, (-87,5,80)\, : \\
+ \fbox{$
+\begin{array}{ll}
+ \text{Answer:} &  \\
+ \text{} & \, (-87,5,80)\,  \\
+\end{array}
+$} \\
+\end{array}
+ \\
+\end{array}

pretraining/mathematica/linear_algebra/cross_product_w_steps/3253.txt

@@ -0,0 +1,169 @@
+Problem:
+Find the cross product of the following vectors:
+$\left(
+\begin{array}{c}
+ 0 \\
+ 9 \\
+ 2 \\
+\end{array}
+\right)$ and
+$\left(
+\begin{array}{c}
+ -10 \\
+ 1 \\
+ -3 \\
+\end{array}
+\right)$
+Answer:
+\begin{array}{l}
+ 
+\begin{array}{l}
+ \text{Compute the following cross product}: \\
+ \, (0,9,2)\, \times \, (-10,1,-3)\,  \\
+\end{array}
+ \\
+\hline
+ 
+\begin{array}{l}
+ \text{Construct }\text{a }\text{matrix }\text{where }\text{the }\text{first }\text{row }\text{contains }\text{unit }\text{vectors }\hat{\text{i}}, \hat{\text{j}}, \text{and }\hat{\text{k}}; \text{and }\text{the }\text{second }\text{and }\text{third }\text{rows }\text{are }\text{made }\text{of }\text{vectors }\, (0,9,2)\,  \text{and }\, (-10,1,-3)\, : \\
+ \left(
+\begin{array}{ccc}
+ \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\
+ 0 & 9 & 2 \\
+ -10 & 1 & -3 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Take }\text{the }\text{determinant }\text{of }\text{this }\text{matrix}: \\
+ \left| 
+\begin{array}{ccc}
+ \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\
+ 0 & 9 & 2 \\
+ -10 & 1 & -3 \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Expand }\text{with }\text{respect }\text{to }\text{row }1: \\
+  \text{= }\left| 
+\begin{array}{ccc}
+ \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\
+ 0 & 9 & 2 \\
+ -10 & 1 & -3 \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{The }\text{determinant }\text{of }\text{the }\text{matrix }\left(
+\begin{array}{ccc}
+ \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\
+ 0 & 9 & 2 \\
+ -10 & 1 & -3 \\
+\end{array}
+\right) \text{is }\text{given }\text{by }\hat{\text{i}} \left| 
+\begin{array}{cc}
+ 9 & 2 \\
+ 1 & -3 \\
+\end{array}
+\right| +\left(-\hat{\text{j}}\right) \left| 
+\begin{array}{cc}
+ 0 & 2 \\
+ -10 & -3 \\
+\end{array}
+\right| +\hat{\text{k}} \left| 
+\begin{array}{cc}
+ 0 & 9 \\
+ -10 & 1 \\
+\end{array}
+\right| : \\
+  \text{= }\hat{\text{i}} \left| 
+\begin{array}{cc}
+ 9 & 2 \\
+ 1 & -3 \\
+\end{array}
+\right| +\left(-\hat{\text{j}}\right) \left| 
+\begin{array}{cc}
+ 0 & 2 \\
+ -10 & -3 \\
+\end{array}
+\right| +\hat{\text{k}} \left| 
+\begin{array}{cc}
+ 0 & 9 \\
+ -10 & 1 \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \hat{\text{i}} \left| 
+\begin{array}{cc}
+ 9 & 2 \\
+ 1 & -3 \\
+\end{array}
+\right| =\hat{\text{i}} (9 (-3)-2\ 1)=\hat{\text{i}} (-29)=\fbox{$-29 \hat{\text{i}}$}: \\
+  \text{= }\fbox{$-29 \hat{\text{i}}$}+\left(-\hat{\text{j}}\right) \left| 
+\begin{array}{cc}
+ 0 & 2 \\
+ -10 & -3 \\
+\end{array}
+\right| +\hat{\text{k}} \left| 
+\begin{array}{cc}
+ 0 & 9 \\
+ -10 & 1 \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ -\hat{\text{j}} \left| 
+\begin{array}{cc}
+ 0 & 2 \\
+ -10 & -3 \\
+\end{array}
+\right| =-\hat{\text{j}} (0 (-3)-2 (-10))=-\hat{\text{j}} 20=\fbox{$-20 \hat{\text{j}}$}: \\
+  \text{= }-29 \hat{\text{i}}+\fbox{$-20 \hat{\text{j}}$}+\hat{\text{k}} \left| 
+\begin{array}{cc}
+ 0 & 9 \\
+ -10 & 1 \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \hat{\text{k}} \left| 
+\begin{array}{cc}
+ 0 & 9 \\
+ -10 & 1 \\
+\end{array}
+\right| =\hat{\text{k}} (0\ 1-9 (-10))=\hat{\text{k}} 90=\fbox{$90 \hat{\text{k}}$}: \\
+  \text{= }-29 \hat{\text{i}}-20 \hat{\text{j}}+\fbox{$90 \hat{\text{k}}$} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Order }\text{the }\text{terms }\text{in }\text{a }\text{more }\text{natural }\text{way}: \\
+  \text{= }-29 \hat{\text{i}}-20 \hat{\text{j}}+90 \hat{\text{k}} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ -29 \hat{\text{i}}-20 \hat{\text{j}}+90 \hat{\text{k}}=\, (-29,-20,90)\, : \\
+ \fbox{$
+\begin{array}{ll}
+ \text{Answer:} &  \\
+ \text{} & \, (-29,-20,90)\,  \\
+\end{array}
+$} \\
+\end{array}
+ \\
+\end{array}

pretraining/mathematica/linear_algebra/cross_product_w_steps/3263.txt

@@ -0,0 +1,169 @@
+Problem:
+Find the cross product of the following vectors:
+$\left(
+\begin{array}{c}
+ \frac{17}{2} \\
+ -3 \\
+ \frac{1}{2} \\
+\end{array}
+\right)$ and
+$\left(
+\begin{array}{c}
+ 9 \\
+ \frac{19}{2} \\
+ \frac{7}{2} \\
+\end{array}
+\right)$
+Answer:
+\begin{array}{l}
+ 
+\begin{array}{l}
+ \text{Compute the following cross product}: \\
+ \, \left(\frac{17}{2},-3,\frac{1}{2}\right)\, \times \, \left(9,\frac{19}{2},\frac{7}{2}\right)\,  \\
+\end{array}
+ \\
+\hline
+ 
+\begin{array}{l}
+ \text{Construct }\text{a }\text{matrix }\text{where }\text{the }\text{first }\text{row }\text{contains }\text{unit }\text{vectors }\hat{\text{i}}, \hat{\text{j}}, \text{and }\hat{\text{k}}; \text{and }\text{the }\text{second }\text{and }\text{third }\text{rows }\text{are }\text{made }\text{of }\text{vectors }\, \left(\frac{17}{2},-3,\frac{1}{2}\right)\,  \text{and }\, \left(9,\frac{19}{2},\frac{7}{2}\right)\, : \\
+ \left(
+\begin{array}{ccc}
+ \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\
+ \frac{17}{2} & -3 & \frac{1}{2} \\
+ 9 & \frac{19}{2} & \frac{7}{2} \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Take }\text{the }\text{determinant }\text{of }\text{this }\text{matrix}: \\
+ \left| 
+\begin{array}{ccc}
+ \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\
+ \frac{17}{2} & -3 & \frac{1}{2} \\
+ 9 & \frac{19}{2} & \frac{7}{2} \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Expand }\text{with }\text{respect }\text{to }\text{row }1: \\
+  \text{= }\left| 
+\begin{array}{ccc}
+ \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\
+ \frac{17}{2} & -3 & \frac{1}{2} \\
+ 9 & \frac{19}{2} & \frac{7}{2} \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{The }\text{determinant }\text{of }\text{the }\text{matrix }\left(
+\begin{array}{ccc}
+ \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\
+ \frac{17}{2} & -3 & \frac{1}{2} \\
+ 9 & \frac{19}{2} & \frac{7}{2} \\
+\end{array}
+\right) \text{is }\text{given }\text{by }\hat{\text{i}} \left| 
+\begin{array}{cc}
+ -3 & \frac{1}{2} \\
+ \frac{19}{2} & \frac{7}{2} \\
+\end{array}
+\right| +\left(-\hat{\text{j}}\right) \left| 
+\begin{array}{cc}
+ \frac{17}{2} & \frac{1}{2} \\
+ 9 & \frac{7}{2} \\
+\end{array}
+\right| +\hat{\text{k}} \left| 
+\begin{array}{cc}
+ \frac{17}{2} & -3 \\
+ 9 & \frac{19}{2} \\
+\end{array}
+\right| : \\
+  \text{= }\hat{\text{i}} \left| 
+\begin{array}{cc}
+ -3 & \frac{1}{2} \\
+ \frac{19}{2} & \frac{7}{2} \\
+\end{array}
+\right| +\left(-\hat{\text{j}}\right) \left| 
+\begin{array}{cc}
+ \frac{17}{2} & \frac{1}{2} \\
+ 9 & \frac{7}{2} \\
+\end{array}
+\right| +\hat{\text{k}} \left| 
+\begin{array}{cc}
+ \frac{17}{2} & -3 \\
+ 9 & \frac{19}{2} \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \hat{\text{i}} \left| 
+\begin{array}{cc}
+ -3 & \frac{1}{2} \\
+ \frac{19}{2} & \frac{7}{2} \\
+\end{array}
+\right| =\hat{\text{i}} \left((-3)\, \times \, \frac{7}{2}-\frac{19}{2\ 2}\right)=\frac{\hat{\text{i}} (-61)}{4}=\fbox{$-\frac{61 \hat{\text{i}}}{4}$}: \\
+  \text{= }\fbox{$-\frac{61 \hat{\text{i}}}{4}$}+\left(-\hat{\text{j}}\right) \left| 
+\begin{array}{cc}
+ \frac{17}{2} & \frac{1}{2} \\
+ 9 & \frac{7}{2} \\
+\end{array}
+\right| +\hat{\text{k}} \left| 
+\begin{array}{cc}
+ \frac{17}{2} & -3 \\
+ 9 & \frac{19}{2} \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ -\hat{\text{j}} \left| 
+\begin{array}{cc}
+ \frac{17}{2} & \frac{1}{2} \\
+ 9 & \frac{7}{2} \\
+\end{array}
+\right| =-\hat{\text{j}} \left(\frac{17\ 7}{2\ 2}-\frac{9}{2}\right)=\frac{-\hat{\text{j}} 101}{4}=\fbox{$-\frac{101 \hat{\text{j}}}{4}$}: \\
+  \text{= }\frac{-61 \hat{\text{i}}}{4}+\fbox{$-\frac{101 \hat{\text{j}}}{4}$}+\hat{\text{k}} \left| 
+\begin{array}{cc}
+ \frac{17}{2} & -3 \\
+ 9 & \frac{19}{2} \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \hat{\text{k}} \left| 
+\begin{array}{cc}
+ \frac{17}{2} & -3 \\
+ 9 & \frac{19}{2} \\
+\end{array}
+\right| =\hat{\text{k}} \left(\frac{17\ 19}{2\ 2}-(-3)\, \times \, 9\right)=\frac{\hat{\text{k}} 431}{4}=\fbox{$\frac{431 \hat{\text{k}}}{4}$}: \\
+  \text{= }\frac{-61 \hat{\text{i}}}{4}-\frac{101 \hat{\text{j}}}{4}+\fbox{$\frac{431 \hat{\text{k}}}{4}$} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Order }\text{the }\text{terms }\text{in }\text{a }\text{more }\text{natural }\text{way}: \\
+  \text{= }-\frac{61 \hat{\text{i}}}{4}-\frac{101 \hat{\text{j}}}{4}+\frac{431 \hat{\text{k}}}{4} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ -\frac{61 \hat{\text{i}}}{4}-\frac{101 \hat{\text{j}}}{4}+\frac{431 \hat{\text{k}}}{4}=\, \left(-\frac{61}{4},-\frac{101}{4},\frac{431}{4}\right)\, : \\
+ \fbox{$
+\begin{array}{ll}
+ \text{Answer:} &  \\
+ \text{} & \, \left(-\frac{61}{4},-\frac{101}{4},\frac{431}{4}\right)\,  \\
+\end{array}
+$} \\
+\end{array}
+ \\
+\end{array}

pretraining/mathematica/linear_algebra/cross_product_w_steps/3309.txt

@@ -0,0 +1,169 @@
+Problem:
+Find the cross product of the following vectors:
+$\left(
+\begin{array}{c}
+ 6 \\
+ -\frac{21}{5} \\
+ \frac{49}{5} \\
+\end{array}
+\right)$ and
+$\left(
+\begin{array}{c}
+ \frac{13}{5} \\
+ -\frac{41}{5} \\
+ \frac{23}{5} \\
+\end{array}
+\right)$
+Answer:
+\begin{array}{l}
+ 
+\begin{array}{l}
+ \text{Compute the following cross product}: \\
+ \, \left(6,-\frac{21}{5},\frac{49}{5}\right)\, \times \, \left(\frac{13}{5},-\frac{41}{5},\frac{23}{5}\right)\,  \\
+\end{array}
+ \\
+\hline
+ 
+\begin{array}{l}
+ \text{Construct }\text{a }\text{matrix }\text{where }\text{the }\text{first }\text{row }\text{contains }\text{unit }\text{vectors }\hat{\text{i}}, \hat{\text{j}}, \text{and }\hat{\text{k}}; \text{and }\text{the }\text{second }\text{and }\text{third }\text{rows }\text{are }\text{made }\text{of }\text{vectors }\, \left(6,-\frac{21}{5},\frac{49}{5}\right)\,  \text{and }\, \left(\frac{13}{5},-\frac{41}{5},\frac{23}{5}\right)\, : \\
+ \left(
+\begin{array}{ccc}
+ \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\
+ 6 & -\frac{21}{5} & \frac{49}{5} \\
+ \frac{13}{5} & -\frac{41}{5} & \frac{23}{5} \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Take }\text{the }\text{determinant }\text{of }\text{this }\text{matrix}: \\
+ \left| 
+\begin{array}{ccc}
+ \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\
+ 6 & -\frac{21}{5} & \frac{49}{5} \\
+ \frac{13}{5} & -\frac{41}{5} & \frac{23}{5} \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Expand }\text{with }\text{respect }\text{to }\text{row }1: \\
+  \text{= }\left| 
+\begin{array}{ccc}
+ \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\
+ 6 & -\frac{21}{5} & \frac{49}{5} \\
+ \frac{13}{5} & -\frac{41}{5} & \frac{23}{5} \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{The }\text{determinant }\text{of }\text{the }\text{matrix }\left(
+\begin{array}{ccc}
+ \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\
+ 6 & -\frac{21}{5} & \frac{49}{5} \\
+ \frac{13}{5} & -\frac{41}{5} & \frac{23}{5} \\
+\end{array}
+\right) \text{is }\text{given }\text{by }\hat{\text{i}} \left| 
+\begin{array}{cc}
+ -\frac{21}{5} & \frac{49}{5} \\
+ -\frac{41}{5} & \frac{23}{5} \\
+\end{array}
+\right| +\left(-\hat{\text{j}}\right) \left| 
+\begin{array}{cc}
+ 6 & \frac{49}{5} \\
+ \frac{13}{5} & \frac{23}{5} \\
+\end{array}
+\right| +\hat{\text{k}} \left| 
+\begin{array}{cc}
+ 6 & -\frac{21}{5} \\
+ \frac{13}{5} & -\frac{41}{5} \\
+\end{array}
+\right| : \\
+  \text{= }\hat{\text{i}} \left| 
+\begin{array}{cc}
+ -\frac{21}{5} & \frac{49}{5} \\
+ -\frac{41}{5} & \frac{23}{5} \\
+\end{array}
+\right| +\left(-\hat{\text{j}}\right) \left| 
+\begin{array}{cc}
+ 6 & \frac{49}{5} \\
+ \frac{13}{5} & \frac{23}{5} \\
+\end{array}
+\right| +\hat{\text{k}} \left| 
+\begin{array}{cc}
+ 6 & -\frac{21}{5} \\
+ \frac{13}{5} & -\frac{41}{5} \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \hat{\text{i}} \left| 
+\begin{array}{cc}
+ -\frac{21}{5} & \frac{49}{5} \\
+ -\frac{41}{5} & \frac{23}{5} \\
+\end{array}
+\right| =\hat{\text{i}} \left(\left(-\frac{21}{5}\right)\, \times \, \frac{23}{5}-\frac{49 (-41)}{5\ 5}\right)=\frac{\hat{\text{i}} 1526}{25}=\fbox{$\frac{1526 \hat{\text{i}}}{25}$}: \\
+  \text{= }\fbox{$\frac{1526 \hat{\text{i}}}{25}$}+\left(-\hat{\text{j}}\right) \left| 
+\begin{array}{cc}
+ 6 & \frac{49}{5} \\
+ \frac{13}{5} & \frac{23}{5} \\
+\end{array}
+\right| +\hat{\text{k}} \left| 
+\begin{array}{cc}
+ 6 & -\frac{21}{5} \\
+ \frac{13}{5} & -\frac{41}{5} \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ -\hat{\text{j}} \left| 
+\begin{array}{cc}
+ 6 & \frac{49}{5} \\
+ \frac{13}{5} & \frac{23}{5} \\
+\end{array}
+\right| =-\hat{\text{j}} \left(\frac{6\ 23}{5}-\frac{49\ 13}{5\ 5}\right)=\frac{-\hat{\text{j}} 53}{25}=\fbox{$-\frac{53 \hat{\text{j}}}{25}$}: \\
+  \text{= }\frac{1526 \hat{\text{i}}}{25}+\fbox{$-\frac{53 \hat{\text{j}}}{25}$}+\hat{\text{k}} \left| 
+\begin{array}{cc}
+ 6 & -\frac{21}{5} \\
+ \frac{13}{5} & -\frac{41}{5} \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \hat{\text{k}} \left| 
+\begin{array}{cc}
+ 6 & -\frac{21}{5} \\
+ \frac{13}{5} & -\frac{41}{5} \\
+\end{array}
+\right| =\hat{\text{k}} \left(\frac{6 (-41)}{5}-\left(-\frac{21}{5}\right)\, \times \, \frac{13}{5}\right)=\frac{\hat{\text{k}} (-957)}{25}=\fbox{$-\frac{957 \hat{\text{k}}}{25}$}: \\
+  \text{= }\frac{1526 \hat{\text{i}}}{25}-\frac{53 \hat{\text{j}}}{25}+\fbox{$-\frac{957 \hat{\text{k}}}{25}$} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Order }\text{the }\text{terms }\text{in }\text{a }\text{more }\text{natural }\text{way}: \\
+  \text{= }\frac{1526 \hat{\text{i}}}{25}-\frac{53 \hat{\text{j}}}{25}-\frac{957 \hat{\text{k}}}{25} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \frac{1526 \hat{\text{i}}}{25}-\frac{53 \hat{\text{j}}}{25}-\frac{957 \hat{\text{k}}}{25}=\, \left(\frac{1526}{25},-\frac{53}{25},-\frac{957}{25}\right)\, : \\
+ \fbox{$
+\begin{array}{ll}
+ \text{Answer:} &  \\
+ \text{} & \, \left(\frac{1526}{25},-\frac{53}{25},-\frac{957}{25}\right)\,  \\
+\end{array}
+$} \\
+\end{array}
+ \\
+\end{array}

pretraining/mathematica/linear_algebra/cross_product_w_steps/3339.txt

@@ -0,0 +1,169 @@
+Problem:
+Find the cross product of the following vectors:
+$\left(
+\begin{array}{c}
+ \frac{3}{4} \\
+ -6 \\
+ 6 \\
+\end{array}
+\right)$ and
+$\left(
+\begin{array}{c}
+ \frac{23}{4} \\
+ \frac{5}{2} \\
+ \frac{13}{2} \\
+\end{array}
+\right)$
+Answer:
+\begin{array}{l}
+ 
+\begin{array}{l}
+ \text{Compute the following cross product}: \\
+ \, \left(\frac{3}{4},-6,6\right)\, \times \, \left(\frac{23}{4},\frac{5}{2},\frac{13}{2}\right)\,  \\
+\end{array}
+ \\
+\hline
+ 
+\begin{array}{l}
+ \text{Construct }\text{a }\text{matrix }\text{where }\text{the }\text{first }\text{row }\text{contains }\text{unit }\text{vectors }\hat{\text{i}}, \hat{\text{j}}, \text{and }\hat{\text{k}}; \text{and }\text{the }\text{second }\text{and }\text{third }\text{rows }\text{are }\text{made }\text{of }\text{vectors }\, \left(\frac{3}{4},-6,6\right)\,  \text{and }\, \left(\frac{23}{4},\frac{5}{2},\frac{13}{2}\right)\, : \\
+ \left(
+\begin{array}{ccc}
+ \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\
+ \frac{3}{4} & -6 & 6 \\
+ \frac{23}{4} & \frac{5}{2} & \frac{13}{2} \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Take }\text{the }\text{determinant }\text{of }\text{this }\text{matrix}: \\
+ \left| 
+\begin{array}{ccc}
+ \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\
+ \frac{3}{4} & -6 & 6 \\
+ \frac{23}{4} & \frac{5}{2} & \frac{13}{2} \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Expand }\text{with }\text{respect }\text{to }\text{row }1: \\
+  \text{= }\left| 
+\begin{array}{ccc}
+ \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\
+ \frac{3}{4} & -6 & 6 \\
+ \frac{23}{4} & \frac{5}{2} & \frac{13}{2} \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{The }\text{determinant }\text{of }\text{the }\text{matrix }\left(
+\begin{array}{ccc}
+ \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\
+ \frac{3}{4} & -6 & 6 \\
+ \frac{23}{4} & \frac{5}{2} & \frac{13}{2} \\
+\end{array}
+\right) \text{is }\text{given }\text{by }\hat{\text{i}} \left| 
+\begin{array}{cc}
+ -6 & 6 \\
+ \frac{5}{2} & \frac{13}{2} \\
+\end{array}
+\right| +\left(-\hat{\text{j}}\right) \left| 
+\begin{array}{cc}
+ \frac{3}{4} & 6 \\
+ \frac{23}{4} & \frac{13}{2} \\
+\end{array}
+\right| +\hat{\text{k}} \left| 
+\begin{array}{cc}
+ \frac{3}{4} & -6 \\
+ \frac{23}{4} & \frac{5}{2} \\
+\end{array}
+\right| : \\
+  \text{= }\hat{\text{i}} \left| 
+\begin{array}{cc}
+ -6 & 6 \\
+ \frac{5}{2} & \frac{13}{2} \\
+\end{array}
+\right| +\left(-\hat{\text{j}}\right) \left| 
+\begin{array}{cc}
+ \frac{3}{4} & 6 \\
+ \frac{23}{4} & \frac{13}{2} \\
+\end{array}
+\right| +\hat{\text{k}} \left| 
+\begin{array}{cc}
+ \frac{3}{4} & -6 \\
+ \frac{23}{4} & \frac{5}{2} \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \hat{\text{i}} \left| 
+\begin{array}{cc}
+ -6 & 6 \\
+ \frac{5}{2} & \frac{13}{2} \\
+\end{array}
+\right| =\hat{\text{i}} \left((-6)\, \times \, \frac{13}{2}-\frac{6\ 5}{2}\right)=\hat{\text{i}} (-54)=\fbox{$-54 \hat{\text{i}}$}: \\
+  \text{= }\fbox{$-54 \hat{\text{i}}$}+\left(-\hat{\text{j}}\right) \left| 
+\begin{array}{cc}
+ \frac{3}{4} & 6 \\
+ \frac{23}{4} & \frac{13}{2} \\
+\end{array}
+\right| +\hat{\text{k}} \left| 
+\begin{array}{cc}
+ \frac{3}{4} & -6 \\
+ \frac{23}{4} & \frac{5}{2} \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ -\hat{\text{j}} \left| 
+\begin{array}{cc}
+ \frac{3}{4} & 6 \\
+ \frac{23}{4} & \frac{13}{2} \\
+\end{array}
+\right| =-\hat{\text{j}} \left(\frac{3\ 13}{4\ 2}-\frac{6\ 23}{4}\right)=\frac{-\hat{\text{j}} (-237)}{8}=\fbox{$\frac{237 \hat{\text{j}}}{8}$}: \\
+  \text{= }-54 \hat{\text{i}}+\fbox{$\frac{237 \hat{\text{j}}}{8}$}+\hat{\text{k}} \left| 
+\begin{array}{cc}
+ \frac{3}{4} & -6 \\
+ \frac{23}{4} & \frac{5}{2} \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \hat{\text{k}} \left| 
+\begin{array}{cc}
+ \frac{3}{4} & -6 \\
+ \frac{23}{4} & \frac{5}{2} \\
+\end{array}
+\right| =\hat{\text{k}} \left(\frac{3\ 5}{4\ 2}-(-6)\, \times \, \frac{23}{4}\right)=\frac{\hat{\text{k}} 291}{8}=\fbox{$\frac{291 \hat{\text{k}}}{8}$}: \\
+  \text{= }-54 \hat{\text{i}}+\frac{237 \hat{\text{j}}}{8}+\fbox{$\frac{291 \hat{\text{k}}}{8}$} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Order }\text{the }\text{terms }\text{in }\text{a }\text{more }\text{natural }\text{way}: \\
+  \text{= }-54 \hat{\text{i}}+\frac{237 \hat{\text{j}}}{8}+\frac{291 \hat{\text{k}}}{8} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ -54 \hat{\text{i}}+\frac{237 \hat{\text{j}}}{8}+\frac{291 \hat{\text{k}}}{8}=\, \left(-54,\frac{237}{8},\frac{291}{8}\right)\, : \\
+ \fbox{$
+\begin{array}{ll}
+ \text{Answer:} &  \\
+ \text{} & \, \left(-54,\frac{237}{8},\frac{291}{8}\right)\,  \\
+\end{array}
+$} \\
+\end{array}
+ \\
+\end{array}

pretraining/mathematica/linear_algebra/cross_product_w_steps/3371.txt

@@ -0,0 +1,169 @@
+Problem:
+Find the cross product of the following vectors:
+$\left(
+\begin{array}{c}
+ -5 \\
+ -5 \\
+ 8 \\
+\end{array}
+\right)$ and
+$\left(
+\begin{array}{c}
+ 3 \\
+ 1 \\
+ 8 \\
+\end{array}
+\right)$
+Answer:
+\begin{array}{l}
+ 
+\begin{array}{l}
+ \text{Compute the following cross product}: \\
+ \, (-5,-5,8)\, \times \, (3,1,8)\,  \\
+\end{array}
+ \\
+\hline
+ 
+\begin{array}{l}
+ \text{Construct }\text{a }\text{matrix }\text{where }\text{the }\text{first }\text{row }\text{contains }\text{unit }\text{vectors }\hat{\text{i}}, \hat{\text{j}}, \text{and }\hat{\text{k}}; \text{and }\text{the }\text{second }\text{and }\text{third }\text{rows }\text{are }\text{made }\text{of }\text{vectors }\, (-5,-5,8)\,  \text{and }\, (3,1,8)\, : \\
+ \left(
+\begin{array}{ccc}
+ \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\
+ -5 & -5 & 8 \\
+ 3 & 1 & 8 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Take }\text{the }\text{determinant }\text{of }\text{this }\text{matrix}: \\
+ \left| 
+\begin{array}{ccc}
+ \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\
+ -5 & -5 & 8 \\
+ 3 & 1 & 8 \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Expand }\text{with }\text{respect }\text{to }\text{row }1: \\
+  \text{= }\left| 
+\begin{array}{ccc}
+ \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\
+ -5 & -5 & 8 \\
+ 3 & 1 & 8 \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{The }\text{determinant }\text{of }\text{the }\text{matrix }\left(
+\begin{array}{ccc}
+ \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\
+ -5 & -5 & 8 \\
+ 3 & 1 & 8 \\
+\end{array}
+\right) \text{is }\text{given }\text{by }\hat{\text{i}} \left| 
+\begin{array}{cc}
+ -5 & 8 \\
+ 1 & 8 \\
+\end{array}
+\right| +\left(-\hat{\text{j}}\right) \left| 
+\begin{array}{cc}
+ -5 & 8 \\
+ 3 & 8 \\
+\end{array}
+\right| +\hat{\text{k}} \left| 
+\begin{array}{cc}
+ -5 & -5 \\
+ 3 & 1 \\
+\end{array}
+\right| : \\
+  \text{= }\hat{\text{i}} \left| 
+\begin{array}{cc}
+ -5 & 8 \\
+ 1 & 8 \\
+\end{array}
+\right| +\left(-\hat{\text{j}}\right) \left| 
+\begin{array}{cc}
+ -5 & 8 \\
+ 3 & 8 \\
+\end{array}
+\right| +\hat{\text{k}} \left| 
+\begin{array}{cc}
+ -5 & -5 \\
+ 3 & 1 \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \hat{\text{i}} \left| 
+\begin{array}{cc}
+ -5 & 8 \\
+ 1 & 8 \\
+\end{array}
+\right| =\hat{\text{i}} ((-5)\, \times \, 8-8\ 1)=\hat{\text{i}} (-48)=\fbox{$-48 \hat{\text{i}}$}: \\
+  \text{= }\fbox{$-48 \hat{\text{i}}$}+\left(-\hat{\text{j}}\right) \left| 
+\begin{array}{cc}
+ -5 & 8 \\
+ 3 & 8 \\
+\end{array}
+\right| +\hat{\text{k}} \left| 
+\begin{array}{cc}
+ -5 & -5 \\
+ 3 & 1 \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ -\hat{\text{j}} \left| 
+\begin{array}{cc}
+ -5 & 8 \\
+ 3 & 8 \\
+\end{array}
+\right| =-\hat{\text{j}} ((-5)\, \times \, 8-8\ 3)=-\hat{\text{j}} (-64)=\fbox{$64 \hat{\text{j}}$}: \\
+  \text{= }-48 \hat{\text{i}}+\fbox{$64 \hat{\text{j}}$}+\hat{\text{k}} \left| 
+\begin{array}{cc}
+ -5 & -5 \\
+ 3 & 1 \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \hat{\text{k}} \left| 
+\begin{array}{cc}
+ -5 & -5 \\
+ 3 & 1 \\
+\end{array}
+\right| =\hat{\text{k}} ((-5)\, \times \, 1-(-5)\, \times \, 3)=\hat{\text{k}} 10=\fbox{$10 \hat{\text{k}}$}: \\
+  \text{= }-48 \hat{\text{i}}+64 \hat{\text{j}}+\fbox{$10 \hat{\text{k}}$} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Order }\text{the }\text{terms }\text{in }\text{a }\text{more }\text{natural }\text{way}: \\
+  \text{= }-48 \hat{\text{i}}+64 \hat{\text{j}}+10 \hat{\text{k}} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ -48 \hat{\text{i}}+64 \hat{\text{j}}+10 \hat{\text{k}}=\, (-48,64,10)\, : \\
+ \fbox{$
+\begin{array}{ll}
+ \text{Answer:} &  \\
+ \text{} & \, (-48,64,10)\,  \\
+\end{array}
+$} \\
+\end{array}
+ \\
+\end{array}

pretraining/mathematica/linear_algebra/cross_product_w_steps/3406.txt

@@ -0,0 +1,169 @@
+Problem:
+Find the cross product of the following vectors:
+$\left(
+\begin{array}{c}
+ -8 \\
+ 10 \\
+ -2 \\
+\end{array}
+\right)$ and
+$\left(
+\begin{array}{c}
+ -6 \\
+ -4 \\
+ -10 \\
+\end{array}
+\right)$
+Answer:
+\begin{array}{l}
+ 
+\begin{array}{l}
+ \text{Compute the following cross product}: \\
+ \, (-8,10,-2)\, \times \, (-6,-4,-10)\,  \\
+\end{array}
+ \\
+\hline
+ 
+\begin{array}{l}
+ \text{Construct }\text{a }\text{matrix }\text{where }\text{the }\text{first }\text{row }\text{contains }\text{unit }\text{vectors }\hat{\text{i}}, \hat{\text{j}}, \text{and }\hat{\text{k}}; \text{and }\text{the }\text{second }\text{and }\text{third }\text{rows }\text{are }\text{made }\text{of }\text{vectors }\, (-8,10,-2)\,  \text{and }\, (-6,-4,-10)\, : \\
+ \left(
+\begin{array}{ccc}
+ \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\
+ -8 & 10 & -2 \\
+ -6 & -4 & -10 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Take }\text{the }\text{determinant }\text{of }\text{this }\text{matrix}: \\
+ \left| 
+\begin{array}{ccc}
+ \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\
+ -8 & 10 & -2 \\
+ -6 & -4 & -10 \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Expand }\text{with }\text{respect }\text{to }\text{row }1: \\
+  \text{= }\left| 
+\begin{array}{ccc}
+ \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\
+ -8 & 10 & -2 \\
+ -6 & -4 & -10 \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{The }\text{determinant }\text{of }\text{the }\text{matrix }\left(
+\begin{array}{ccc}
+ \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\
+ -8 & 10 & -2 \\
+ -6 & -4 & -10 \\
+\end{array}
+\right) \text{is }\text{given }\text{by }\hat{\text{i}} \left| 
+\begin{array}{cc}
+ 10 & -2 \\
+ -4 & -10 \\
+\end{array}
+\right| +\left(-\hat{\text{j}}\right) \left| 
+\begin{array}{cc}
+ -8 & -2 \\
+ -6 & -10 \\
+\end{array}
+\right| +\hat{\text{k}} \left| 
+\begin{array}{cc}
+ -8 & 10 \\
+ -6 & -4 \\
+\end{array}
+\right| : \\
+  \text{= }\hat{\text{i}} \left| 
+\begin{array}{cc}
+ 10 & -2 \\
+ -4 & -10 \\
+\end{array}
+\right| +\left(-\hat{\text{j}}\right) \left| 
+\begin{array}{cc}
+ -8 & -2 \\
+ -6 & -10 \\
+\end{array}
+\right| +\hat{\text{k}} \left| 
+\begin{array}{cc}
+ -8 & 10 \\
+ -6 & -4 \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \hat{\text{i}} \left| 
+\begin{array}{cc}
+ 10 & -2 \\
+ -4 & -10 \\
+\end{array}
+\right| =\hat{\text{i}} (10 (-10)-(-2)\, (-4))=\hat{\text{i}} (-108)=\fbox{$-108 \hat{\text{i}}$}: \\
+  \text{= }\fbox{$-108 \hat{\text{i}}$}+\left(-\hat{\text{j}}\right) \left| 
+\begin{array}{cc}
+ -8 & -2 \\
+ -6 & -10 \\
+\end{array}
+\right| +\hat{\text{k}} \left| 
+\begin{array}{cc}
+ -8 & 10 \\
+ -6 & -4 \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ -\hat{\text{j}} \left| 
+\begin{array}{cc}
+ -8 & -2 \\
+ -6 & -10 \\
+\end{array}
+\right| =-\hat{\text{j}} ((-8)\, (-10)-(-2)\, (-6))=-\hat{\text{j}} 68=\fbox{$-68 \hat{\text{j}}$}: \\
+  \text{= }-108 \hat{\text{i}}+\fbox{$-68 \hat{\text{j}}$}+\hat{\text{k}} \left| 
+\begin{array}{cc}
+ -8 & 10 \\
+ -6 & -4 \\
+\end{array}
+\right|  \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \hat{\text{k}} \left| 
+\begin{array}{cc}
+ -8 & 10 \\
+ -6 & -4 \\
+\end{array}
+\right| =\hat{\text{k}} ((-8)\, (-4)-10 (-6))=\hat{\text{k}} 92=\fbox{$92 \hat{\text{k}}$}: \\
+  \text{= }-108 \hat{\text{i}}-68 \hat{\text{j}}+\fbox{$92 \hat{\text{k}}$} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Order }\text{the }\text{terms }\text{in }\text{a }\text{more }\text{natural }\text{way}: \\
+  \text{= }-108 \hat{\text{i}}-68 \hat{\text{j}}+92 \hat{\text{k}} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ -108 \hat{\text{i}}-68 \hat{\text{j}}+92 \hat{\text{k}}=\, (-108,-68,92)\, : \\
+ \fbox{$
+\begin{array}{ll}
+ \text{Answer:} &  \\
+ \text{} & \, (-108,-68,92)\,  \\
+\end{array}
+$} \\
+\end{array}
+ \\
+\end{array}