question stringlengths 79 9.83k | answer stringlengths 33 9.39k |
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<p>Let S be the set of all a $$\in R$$ for which the angle between the vectors $$
\vec{u}=a\left(\log _{e} b\right) \hat{i}-6 \hat{j}+3 \hat{k}$$ and $$\vec{v}=\left(\log _{e} b\right) \hat{i}+2 \hat{j}+2 a\left(\log _{e} b\right) \hat{k}$$, $$(b>1)$$ is acute. Then S is equal to :</p>
Options:
[{"identifier": "A",... | ["B"]
Explanation:
<p>$$\overrightarrow u = a({\log _e}b)\widehat i - 6\widehat j + 3\widehat k$$</p>
<p>$$\overrightarrow v = ({\log _e}b)\widehat i + 2\widehat j + 2a({\log _e}b)\widehat k$$</p>
<p>For acute angle $$\overrightarrow u \,.\,\overrightarrow v > 0$$</p>
<p>$$ \Rightarrow a{({\log _e}b)^2} - 12 + 6a({... |
<p>Let $$\vec{a}=5 \hat{i}-\hat{j}-3 \hat{k}$$ and $$\vec{b}=\hat{i}+3 \hat{j}+5 \hat{k}$$ be two vectors. Then which one of the following statements is TRUE ?</p>
Options:
[{"identifier": "A", "content": "Projection of $$\\vec{a}$$ on $$\\vec{b}$$ is $$\\frac{-13}{\\sqrt{35}}$$ and the direction of the projection vec... | ["A"]
Explanation:
$\begin{aligned} & \text { Projection of }\vec{a} \text { on } \vec{b} =\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|} \\\\ & = \frac{(5 \hat{i}-\hat{j}-3 \hat{k}) \cdot(\hat{i}+3 \hat{j}+5 \hat{k})}{\sqrt{1^2+3^2+5^2}}=\frac{5-3-15}{\sqrt{35}} \\\\ & = \frac{-13}{\sqrt{35}}\end{aligned}$
<br/><br/>Negativ... |
<p>Let $$\vec{a}=2 \hat{i}-7 \hat{j}+5 \hat{k}, \vec{b}=\hat{i}+\hat{k}$$ and $$\vec{c}=\hat{i}+2 \hat{j}-3 \hat{k}$$ be three given vectors. If $$\overrightarrow{\mathrm{r}}$$ is a vector such that $$\vec{r} \times \vec{a}=\vec{c} \times \vec{a}$$ and $$\vec{r} \cdot \vec{b}=0$$, then $$|\vec{r}|$$ is equal to :</p>
... | ["D"]
Explanation:
$\begin{aligned} & \vec{r} \times \vec{a}=\vec{c} \times \vec{a} \\\\ & \Rightarrow(\vec{r}-\vec{c}) \times \vec{a}=0 \Rightarrow \vec{r}-\vec{c}=\lambda \vec{a}((\vec{r}-\vec{c} ) \text{and} \overrightarrow{a} \text { are parallel }) \\\\ & \Rightarrow \vec{r}=\vec{c}+\lambda \vec{a} \\\\ & \Right... |
<p>Let $$\vec{a}=2 \hat{i}+\hat{j}+\hat{k}$$, and $$\vec{b}$$ and $$\vec{c}$$ be two nonzero vectors such that $$|\vec{a}+\vec{b}+\vec{c}|=|\vec{a}+\vec{b}-\vec{c}|$$ and $$\vec{b} \cdot \vec{c}=0$$. Consider the following two statements:</p>
<p>(A) $$|\vec{a}+\lambda \vec{c}| \geq|\vec{a}|$$ for all $$\lambda \in \mat... | ["C"]
Explanation:
$|\vec{a}+\vec{b}+\vec{c}|=|\vec{a}+\vec{b}-\vec{c}|$
<br/><br/>$$ \Rightarrow $$ $|\vec{a}+\vec{b}+\vec{c}|^{2}=|\vec{a}+\vec{b}-\vec{c}|^{2}$
<br/><br/>$$
\begin{aligned}
& \Rightarrow |\vec{a}|^{2}+|\vec{b}|^{2}+|\vec{c}|^{2}+2(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \vec{a}) \\\\
& =... |
<p>If the vectors $$\overrightarrow a = \lambda \widehat i + \mu \widehat j + 4\widehat k$$, $$\overrightarrow b = - 2\widehat i + 4\widehat j - 2\widehat k$$ and $$\overrightarrow c = 2\widehat i + 3\widehat j + \widehat k$$ are coplanar and the projection of $$\overrightarrow a $$ on the vector $$\overrightarrow ... | ["A"]
Explanation:
$\vec{a}=\lambda \hat{i}+\mu \hat{j}+4 \hat{k}, \vec{b}=-2 \hat{i}+4 \hat{j}-2 \hat{k}, \vec{c}=2 \hat{i}+3 \hat{j}+\hat{k}$
<br/><br/>
Now, $\vec{a} \cdot \vec{b}=\sqrt{54} \Rightarrow \frac{-2 \lambda+4 \mu-8}{\sqrt{24}}=\sqrt{54}$
<br/><br/>
$\Rightarrow-2 \lambda+4 \mu-8=36$
<br/><br/>
$\Rightar... |
<p>The vector $$\overrightarrow a = - \widehat i + 2\widehat j + \widehat k$$ is rotated through a right angle, passing through the y-axis in its way and the resulting vector is $$\overrightarrow b $$. Then the projection of $$3\overrightarrow a + \sqrt 2 \overrightarrow b $$ on $$\overrightarrow c = 5\widehat i + ... | ["D"]
Explanation:
<p>First, we write $\overrightarrow{b}$ as a linear combination of $\overrightarrow{a}$ and $\overrightarrow{j}$ since $\overrightarrow{b}$ is a rotation of $\overrightarrow{a}$ about the y-axis.</p>
<p>$\vec{b}=\lambda \vec{a}+\mu \hat{j}=\lambda(-\hat{i}+2 \hat{j}+\hat{k})+\mu \hat{j}=-\lambda \ha... |
<p>Let $$\overrightarrow \alpha = 4\widehat i + 3\widehat j + 5\widehat k$$ and $$\overrightarrow \beta = \widehat i + 2\widehat j - 4\widehat k$$. Let $${\overrightarrow \beta _1}$$ be parallel to $$\overrightarrow \alpha $$ and $${\overrightarrow \beta _2}$$ be perpendicular to $$\overrightarrow \alpha $$. If... | ["B"]
Explanation:
Let $\vec{\beta}_1=\lambda \vec{\alpha}$<br/><br/>
Now $\vec{\beta}_2=\vec{\beta}-\vec{\beta}_1$<br/><br/>
$$
\begin{aligned}
& =(\hat{\mathrm{i}}+2 \hat{\mathrm{j}}-4 \hat{\mathrm{k}})-\lambda(4 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+5 \hat{\mathrm{k}}) \\\\
& =(1-4 \lambda) \hat{\mathrm{i}}+(2-3 \lam... |
The least positive integral value of $\alpha$, for which the angle between the vectors $\alpha \hat{i}-2 \hat{j}+2 \hat{k}$ and $\alpha \hat{i}+2 \alpha \hat{j}-2 \hat{k}$ is acute, is ___________.
Options:
[] | 5
Explanation:
<p>$$\begin{aligned}
& \cos \theta=\frac{(\alpha \hat{\mathrm{i}}-2 \hat{\mathrm{j}}+2 \hat{\mathrm{k}}) \cdot(\alpha \hat{\mathrm{i}}+2 \alpha \hat{\mathrm{j}}-2 \hat{\mathrm{k}})}{\sqrt{\alpha^2+4+4} \sqrt{\alpha^2+4 \alpha^2+4}} \\
& \cos \theta=\frac{\alpha^2-4 \alpha-4}{\sqrt{\alpha^2+8} \sqrt{5 \a... |
<p>Let a unit vector $$\hat{u}=x \hat{i}+y \hat{j}+z \hat{k}$$ make angles $$\frac{\pi}{2}, \frac{\pi}{3}$$ and $$\frac{2 \pi}{3}$$ with the vectors $$\frac{1}{\sqrt{2}} \hat{i}+\frac{1}{\sqrt{2}} \hat{k}, \frac{1}{\sqrt{2}} \hat{j}+\frac{1}{\sqrt{2}} \hat{k}$$ and $$\frac{1}{\sqrt{2}} \hat{i}+\frac{1}{\sqrt{2}} \hat{j... | ["B"]
Explanation:
<p>Unit vector $$\hat{\mathrm{u}}=\mathrm{x} \hat{\mathrm{i}}+\mathrm{y} \hat{\mathrm{j}}+\mathrm{z} \hat{\mathrm{k}}$$</p>
<p>$$\begin{aligned}
& \overrightarrow{\mathrm{p}}_1=\frac{1}{\sqrt{2}} \hat{\mathrm{i}}+\frac{1}{\sqrt{2}} \hat{\mathrm{k}}, \overrightarrow{\mathrm{p}}_2=\frac{1}{\sqrt{2}} \... |
<p>Let a unit vector which makes an angle of $$60^{\circ}$$ with $$2 \hat{i}+2 \hat{j}-\hat{k}$$ and an angle of $$45^{\circ}$$ with $$\hat{i}-\hat{k}$$ be $$\vec{C}$$. Then $$\vec{C}+\left(-\frac{1}{2} \hat{i}+\frac{1}{3 \sqrt{2}} \hat{j}-\frac{\sqrt{2}}{3} \hat{k}\right)$$ is:</p>
Options:
[{"identifier": "A", "cont... | ["C"]
Explanation:
<p>$$\begin{aligned}
& \text { Let } \vec{C}=a \hat{i}+b \hat{j}+c \hat{k} \\
& (a \hat{i}+b \hat{j}+c \hat{k}) \cdot(2 \hat{i}+2 \hat{j}-\hat{k})=1 \times 3 \times \frac{1}{2} \\
& 2 a+2 b-c=\frac{3}{2} \qquad \text{... (1)}\\
& (a \hat{i}+b \hat{j}+c \hat{k}) \cdot(\hat{i}-\hat{k})=1 \times \sqrt{... |
<p>For $$\lambda>0$$, let $$\theta$$ be the angle between the vectors $$\vec{a}=\hat{i}+\lambda \hat{j}-3 \hat{k}$$ and $$\vec{b}=3 \hat{i}-\hat{j}+2 \hat{k}$$. If the vectors $$\vec{a}+\vec{b}$$ and $$\vec{a}-\vec{b}$$ are mutually perpendicular, then the value of (14 cos $$\theta)^2$$ is equal to</p>
Options:
[{"... | ["A"]
Explanation:
<p>$$\begin{aligned}
& \text { Given } \vec{a}=\hat{i}+\lambda \hat{j}-3 \hat{k} \\
& \vec{b}=3 \hat{i}-\hat{j}+2 \hat{k} \\
& \vec{a}+\vec{b}=4 \hat{i}+(\lambda-1) \hat{j}-\hat{k} \\
& \vec{a}-\vec{b}=-2 \hat{i}+(\lambda+1) \hat{j}-5 \hat{k} \\
& (\vec{a}+\vec{b}) \cdot(\vec{a}-\vec{b})=0 \\
& -8+\... |
<p>Let $$\overrightarrow{\mathrm{a}}=\hat{i}+2 \hat{j}+3 \hat{k}, \overrightarrow{\mathrm{b}}=2 \hat{i}+3 \hat{j}-5 \hat{k}$$ and $$\overrightarrow{\mathrm{c}}=3 \hat{i}-\hat{j}+\lambda \hat{k}$$ be three vectors. Let $$\overrightarrow{\mathrm{r}}$$ be a unit vector along $$\vec{b}+\vec{c}$$. If $$\vec{r} \cdot \vec{a}... | ["B"]
Explanation:
<p>$$\begin{aligned}
& \vec{a}=\hat{i}+2 \hat{j}+3 \hat{k} \\
& \vec{b}=2 \hat{i}+3 \hat{j}-5 \hat{k} \\
& \vec{c}=3 \hat{i}-\hat{j}+\lambda \hat{k} \\
& \vec{b}+\vec{c}=5 \hat{i}+2 \hat{j}+(\lambda-5) \hat{k}
\end{aligned}$$</p>
<p>$$\vec{r}$$ is a unit vector along $$\vec{b}+\vec{c}$$</p>
<p>$$\th... |
<p>The set of all $$\alpha$$, for which the vectors $$\vec{a}=\alpha t \hat{i}+6 \hat{j}-3 \hat{k}$$ and $$\vec{b}=t \hat{i}-2 \hat{j}-2 \alpha t \hat{k}$$ are inclined at an obtuse angle for all $$t \in \mathbb{R}$$, is</p>
Options:
[{"identifier": "A", "content": "$$[0,1)$$\n"}, {"identifier": "B", "content": "$$\\l... | ["B"]
Explanation:
<p>Given $$\vec{a}=\alpha t \hat{i}+6 \hat{j}-3 \hat{k}$$</p>
<p>and $$\vec{b}=t \hat{i}-2 \hat{j}-2 \alpha t \hat{k}$$</p>
<p>angle between $$\vec{a}$$ and $$\vec{b}$$ is given by</p>
<p>$$\cos \theta=\frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|}$$</p>
<p>We have, $$\cos \theta < 0(\because$$ an... |
If the vectors $\overrightarrow{\mathbf{a}}, \overrightarrow{\mathbf{b}}$ and $\overrightarrow{\mathbf{c}}$ from the sides $B C, C A$ and $A B$ respectively of a triangle $A B C$, then :
Options:
[{"identifier": "A", "content": "$\\overrightarrow{\\mathbf{a}} \\cdot \\overrightarrow{\\mathbf{b}}=\\overrightarrow{\\mat... | ["B"]
Explanation:
If $\overrightarrow{\mathbf{a}}, \overrightarrow{\mathbf{b}}$ and $\overrightarrow{\mathbf{c}}$ are the sides of $\mathbf{a}$ triangle, then <br/><br/>$\overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{b}}+\overrightarrow{\mathbf{c}}=\overrightarrow{\mathbf{0}}$
<br/><br/>Since,
<br/><br/>$$
\begi... |
If $$\left| {\overrightarrow a } \right| = 4,\left| {\overrightarrow b } \right| = 2$$ and the angle between $${\overrightarrow a }$$ and $${\overrightarrow b }$$ is $$\pi /6$$ then $${\left( {\overrightarrow a \times \overrightarrow b } \right)^2}$$ is equal to :
Options:
[{"identifier": "A", "content": "$$48$$ "}, ... | ["B"]
Explanation:
$${\left( {\overrightarrow a \times \overrightarrow b } \right)^2} = {\left| {\overrightarrow a } \right|^2}{\left| {\overrightarrow b } \right|^2}\,\,{\sin ^2}{\pi \over 6}$$
<br><br>$$ = 16 \times 4 \times {1 \over 4} = 16$$ |
If the vectors $$\overrightarrow c ,\overrightarrow a = x\widehat i + y\widehat j + z\widehat k$$ and $$\widehat b = \widehat j$$ are such that $$\overrightarrow a ,\overrightarrow c $$ and $$\overrightarrow b $$ form a right handed system then $${\overrightarrow c }$$ is :
Options:
[{"identifier": "A", "content": "... | ["A"]
Explanation:
Since $$\overrightarrow a ,\overrightarrow c ,\overrightarrow b $$ form a right handed system,
<br><br>$$\therefore$$ $$\overrightarrow c = \overrightarrow b \times \overrightarrow a = \left| {\matrix{
{\widehat i} & {\widehat j} & {\widehat k} \cr
0 & 1 & 0 \cr
x &am... |
$$\overrightarrow a = 3\widehat i - 5\widehat j$$ and $$\overrightarrow b = 6\widehat i + 3\widehat j$$ are two vectors and $$\overrightarrow c $$ is a vector such that $$\overrightarrow c = \overrightarrow a \times \overrightarrow b $$ then $$\left| {\overrightarrow a } \right|:\left| {\overrightarrow b } \right|:... | ["B"]
Explanation:
<p>To solve this problem, let's take it step by step, beginning with calculating each of the vector magnitudes (or norms) and then finding the magnitude of the cross product vector $\overrightarrow{c}$.</p>
<p>Given vectors $\overrightarrow{a}$ and $\overrightarrow{b}$ are:</p>
<p>$\overrightarrow... |
If $$\overrightarrow a \times \overrightarrow b = \overrightarrow b \times \overrightarrow c = \overrightarrow c \times \overrightarrow a $$ then $$\overrightarrow a + \overrightarrow b + \overrightarrow c = $$
Options:
[{"identifier": "A", "content": "$$abc$$ "}, {"identifier": "B", "content": "$$-1$$"}, {"i... | ["C"]
Explanation:
Let $$\overrightarrow a + \overrightarrow b + \overrightarrow c = \overrightarrow r .$$ Then
<br><br>$$\overrightarrow a \times \left( {\overrightarrow a + \overrightarrow b + \overrightarrow c } \right) = \overrightarrow a \times \overrightarrow r $$
<br><br>$$ \Rightarrow 0 + \overrightarro... |
A tetrahedron has vertices at $$O(0,0,0), A(1,2,1) B(2,1,3)$$ and $$C(-1,1,2).$$ Then the angle between the faces $$OAB$$ and $$ABC$$ will be :
Options:
[{"identifier": "A", "content": "$${90^ \\circ }$$ "}, {"identifier": "B", "content": "$${\\cos ^{ - 1}}\\left( {{{19} \\over {35}}} \\right)$$ "}, {"identifier": "C"... | ["B"]
Explanation:
Vector perpendicular to the face $$OAB$$
<br><br>$$ = \overrightarrow {OA} \times \overrightarrow {OB} = \left| {\matrix{
{\widehat i} & {\widehat j} & {\widehat k} \cr
1 & 2 & 1 \cr
2 & 1 & 3 \cr
} } \right| = 5\widehat i - \widehat j - 3\widehat k$$
<br><... |
Let $$\overrightarrow u = \widehat i + \widehat j,\,\overrightarrow v = \widehat i - \widehat j$$ and $$\overrightarrow w = \widehat i + 2\widehat j + 3\widehat k\,\,.$$ If $$\widehat n$$ is a unit vector such that $$\overrightarrow u .\widehat n = 0$$ and $$\overrightarrow v .\widehat n = 0\,\,,$$ then $$\left| {\o... | ["A"]
Explanation:
Since, $\hat{\mathbf{n}} \perp \overrightarrow{\mathbf{u}}$ and $\hat{\mathbf{n}} \perp \overrightarrow{\mathbf{v}}$
<br/><br/>$$
\begin{aligned}
\Rightarrow \hat{\mathbf{n}} & =\frac{\overrightarrow{\mathbf{u}} \times \overrightarrow{\mathbf{v}}}{|\overrightarrow{\mathbf{u}} \times \overrightarrow{... |
Let $$\overrightarrow a ,\overrightarrow b $$ and $$\overrightarrow c $$ be non-zero vectors such that $$\left( {\overrightarrow a \times \overrightarrow b } \right) \times \overrightarrow c = {1 \over 3}\left| {\overrightarrow b } \right|\left| {\overrightarrow c } \right|\overrightarrow a \,\,.$$ If $$\theta $$ is ... | ["A"]
Explanation:
Given $$\left( {\overrightarrow a \times \overrightarrow b } \right) \times \overrightarrow c = {1 \over 3}\left| {\overrightarrow b } \right|\left| {\overrightarrow c } \right|\overrightarrow a $$
<br><br>Clearly $$\overrightarrow a $$ and $$\overrightarrow b $$ are noncollinear
<br><br>$$ \Right... |
For any vector $${\overrightarrow a }$$ , the value of $${\left( {\overrightarrow a \times \widehat i} \right)^2} + {\left( {\overrightarrow a \times \widehat j} \right)^2} + {\left( {\overrightarrow a \times \widehat k} \right)^2}$$ is equal to :
Options:
[{"identifier": "A", "content": "$$3{\\overrightarrow a ^2... | ["C"]
Explanation:
Let $$\overrightarrow a = x\overrightarrow i + y\overrightarrow j + z\overrightarrow k $$
<br><br>$$\overrightarrow a \times \overrightarrow i = z\overrightarrow j - y\overrightarrow k $$
<br><br>$$ \Rightarrow {\left( {\overrightarrow a \times \overrightarrow i } \right)^2} = {y^2} + {z^2}$$... |
If $$\widehat u$$ and $$\widehat v$$ are unit vectors and $$\theta $$ is the acute angle between them, then $$2\widehat u \times 3\widehat v$$ is a unit vector for :
Options:
[{"identifier": "A", "content": "no value of $$\\theta $$ "}, {"identifier": "B", "content": "exactly one value of $$\\theta $$ "}, {"identifier... | ["B"]
Explanation:
Given $$\left| {2\widehat u \times 3\widehat v} \right| = 1$$
<br><br>and $$\theta $$ is acute angle between $$\widehat u$$
<br><br>and $$\widehat v,\,\,\left| {\widehat u} \right| = 1,\,\,\left| {\widehat v} \right| = 1\,\,\,$$
<br><br>$$ \Rightarrow \,\,\,6\left| {\widehat u} \right|\left| {\wid... |
The vectors $$\overrightarrow a $$ and $$\overrightarrow b $$ are not perpendicular and $$\overrightarrow c $$ and $$\overrightarrow d $$ are two vectors satisfying $$\overrightarrow b \times \overrightarrow c = \overrightarrow b \times \overrightarrow d $$ and $$\overrightarrow a .\overrightarrow d = 0\,\,.$$ Then... | ["C"]
Explanation:
$$\overrightarrow a .\overrightarrow b \ne 0,\overrightarrow a .\overrightarrow d = 0$$
<br><br>Now, $$\overrightarrow b \times \overrightarrow c = \overrightarrow b \times \overrightarrow d $$
<br><br>$$ \Rightarrow \overrightarrow a \times \left( {\overrightarrow b \times \overrightarrow c ... |
Let $$\overrightarrow a = 2\widehat i + \widehat j -2 \widehat k$$ and $$\overrightarrow b = \widehat i + \widehat j$$.
<br/><br>Let $$\overrightarrow c $$ be a vector such that $$\left| {\overrightarrow c - \overrightarrow a } \right| = 3$$,
<br/><br>$$\left| {\left( {\overrightarrow a \times \overrightarrow b } \... | ["A"]
Explanation:
Given:
<br><br>$$\overrightarrow a = 2\widehat i + \widehat j - 2\widehat k,\,\,\overrightarrow b = \widehat i + \widehat j$$
<br><br>$$ \Rightarrow $$ $$\left| {\overrightarrow a } \right| = 3$$
<br><br>$$ \therefore $$ $$\overrightarrow a \times \overrightarrow b = 2\wide... |
The area (in sq. units) of the parallelogram whose diagonals are along the vectors $$8\widehat i - 6\widehat j$$ and $$3\widehat i + 4\widehat j - 12\widehat k,$$ is :
Options:
[{"identifier": "A", "content": "26"}, {"identifier": "B", "content": "65"}, {"identifier": "C", "content": "20"}, {"identifier": "D", "conte... | ["B"]
Explanation:
When diagonal $${\overrightarrow {{d_1}} }$$ and $${\overrightarrow {{d_2}} }$$ are given of a parallelogram then the area of parallelogram = $${1 \over 2}\left| {\overrightarrow {{d_1}} \times \overrightarrow {{d_2}} } \right|$$
<br><br>Given, $${\overrightarrow {{d_1}} }$$ = 8$$\widehat i$$ $$-$... |
If the vector $$\overrightarrow b = 3\widehat j + 4\widehat k$$ is written as the
sum of a vector $$\overrightarrow {{b_1}} ,$$ paralel to $$\overrightarrow a = \widehat i + \widehat j$$ and a vector $$\overrightarrow {{b_2}} ,$$ perpendicular to $$\overrightarrow a ,$$ then $$\overrightarrow {{b_1}} \times \over... | ["B"]
Explanation:
$$\overrightarrow {{b_1}} = {{\left( {\overrightarrow {{b_1}} .\overrightarrow a } \right)\widehat a} \over 1}$$
<br><br>= $$\left\{ {{{\left( {3\widehat j + 4\widehat k} \right).\left( {\widehat i + \widehat j} \right)} \over {\sqrt 2 }}} \right\}\left( {{{\widehat i + \widehat j}... |
If $$\overrightarrow a ,\,\,\overrightarrow b ,$$ and $$\overrightarrow C $$ are unit vectors such that $$\overrightarrow a + 2\overrightarrow b + 2\overrightarrow c = \overrightarrow 0 ,$$ then $$\left| {\overrightarrow a \times \overrightarrow c } \right|$$ is equal to :
Options:
[{"identifier": "A", "content":... | ["A"]
Explanation:
Given, <br><br>
$$\overrightarrow a + 2\overrightarrow b + 2\overrightarrow c = \overrightarrow 0 $$<br><br/>
$$ \Rightarrow $$ $$\overrightarrow a + 2\overrightarrow c = - 2\overrightarrow b $$<br>
<br>
Squaring both sides,<br><br/>
$${\left| {\overrightarrow a } \right|^2} + 4\overrightarro... |
Let $$\overrightarrow a = \widehat i + \widehat j + \widehat k,\overrightarrow c = \widehat j - \widehat k$$ and a vector $$\overrightarrow b $$ be such that $$\overrightarrow a \times \overrightarrow b = \overrightarrow c $$ and $$\overrightarrow a .\overrightarrow b = 3.$$ Then $$\left| {\overrightarrow b } \rig... | ["C"]
Explanation:
$$ \because $$ $$\overrightarrow a $$ $$=$$ $$\widehat i + \widehat j + \widehat k \Rightarrow \left| {\overrightarrow a } \right| = \sqrt 3 $$
<br><br>& $$\overrightarrow c = \widehat j - \widehat k \Rightarrow \left| {\overrightarrow c } \right|\sqrt 2 $$
<br><br>Now, $$\over... |
Let $$\mathop a\limits^ \to = 3\mathop i\limits^ \wedge + 2\mathop j\limits^ \wedge + x\mathop k\limits^ \wedge $$ and $$\mathop b\limits^ \to = \mathop i\limits^ \wedge - \mathop j\limits^ \wedge + \mathop k\limits^ \wedge $$
, for some real x. Then $$\left| {\mathop a\limits^ \to \times \mathop b\limi... | ["C"]
Explanation:
$$\overrightarrow a \times \overrightarrow b = \left| {\matrix{
{\widehat i} & {\widehat j} & {\widehat k} \cr
3 & 2 & x \cr
1 & { - 1} & 1 \cr
} } \right|$$
<br><br>= (2 + x)$${\widehat i}$$ + (3 - x)$${\widehat j}$$ - 5$${\widehat k}$$
<br><br>$$\left| {\... |
Let $$\overrightarrow a = 3\widehat i + 2\widehat j + 2\widehat k$$ and $$\overrightarrow b = \widehat i + 2\widehat j - 2\widehat k$$ be two vectors. If a vector perpendicular to both the vectors
$$\overrightarrow a + \overrightarrow b $$ and $$\overrightarrow a - \overrightarrow b $$ has the magnitude 12 then one... | ["A"]
Explanation:
Required vector is $\overrightarrow r$ = $$\lambda \left( {\left( {\overline a + \overline b } \right) \times \left( {\overline a - \overline b } \right)} \right)$$<br><br>
$$ \Rightarrow \left| {\matrix{
{\widehat i} & {\widehat j} & {\widehat k} \cr
4 & 4 & 0 \cr
2 &... |
The distance of the point having position vector $$ - \widehat i + 2\widehat j + 6\widehat k$$
from the straight line passing through the point
(2, 3, – 4) and parallel to the vector, $$6\widehat i + 3\widehat j - 4\widehat k$$ is :
Options:
[{"identifier": "A", "content": "6"}, {"identifier": "B", "content": "7"}, {... | ["B"]
Explanation:
<picture><source media="(max-width: 320px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264693/exam_images/n0zytcbrpcd3kgpqtlsb.webp"><source media="(max-width: 500px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734263784/exam_images/axuiomv56i9jmm3qd8t0.webp"><so... |
Let $$\overrightarrow \alpha = 3\widehat i + \widehat j$$ and $$\overrightarrow \beta = 2\widehat i - \widehat j + 3 \widehat k$$
. If $$\overrightarrow \beta = {\overrightarrow \beta _1} - \overrightarrow {{\beta _2}} $$,
where $${\overrightarrow \beta _1}$$
is parallel to $$\overrightarrow \alpha $$ and $$\o... | ["B"]
Explanation:
Given $$\overrightarrow \alpha = 3\widehat i + \widehat j$$<br><br>$$\overrightarrow \beta = 2\widehat i - \widehat j + 3 \widehat k$$
<br><br>$${\overrightarrow \beta _1}$$
is parallel to $$\overrightarrow \alpha $$
<br><br>$$ \therefore $$ $${\overrightarrow \beta _1}$$ = $$\lambda $$ $$\ov... |
Let $$\overrightarrow a = \widehat i + 2\widehat j + 4\widehat k,$$ $$\overrightarrow b = \widehat i + \lambda \widehat j + 4\widehat k$$ and $$\overrightarrow c = 2\widehat i + 4\widehat j + \left( {{\lambda ^2} - 1} \right)\widehat k$$ be coplanar vectors. Then the non-zero vector $$\overrightarrow a \times \ov... | ["B"]
Explanation:
$$\left[ {\overrightarrow a \,\,\overrightarrow b \,\,\overrightarrow c } \right] = 0$$
<br><br>$$ \Rightarrow \left| {\matrix{
1 & 2 & 4 \cr
1 & \lambda & 4 \cr
2 & 4 & {{\lambda ^2} - 1} \cr
} } \right| = 0$$
<br><br>$$ \Rightarrow {\lambda ^3} - 2{\lambda... |
Let $$\overrightarrow a $$
, $$\overrightarrow b $$
and $$\overrightarrow c $$
be three unit vectors such that
<br/>$$\overrightarrow a + \vec b + \overrightarrow c = \overrightarrow 0 $$. If $$\lambda = \overrightarrow a .\vec b + \vec b.\overrightarrow c + \overrightarrow c .\overrightarrow a $$ and
<br/>$$\ove... | ["C"]
Explanation:
$$\overrightarrow a + \vec b + \overrightarrow c = \overrightarrow 0 $$
<br><br>$$ \Rightarrow $$ $${\left| {\overrightarrow a + \overrightarrow b + \overrightarrow c } \right|^2}$$ = 0
<br><br>$$ \Rightarrow $$ $${{{\left| {\overrightarrow a } \right|}^2}}$$ + $${{{\left| {\overrightarrow b } \... |
Let $$\overrightarrow a = \widehat i - 2\widehat j + \widehat k$$ and $$\overrightarrow b = \widehat i - \widehat j + \widehat k$$ be two
vectors. If $$\overrightarrow c $$ is a vector such that $$\overrightarrow b \times \overrightarrow c = \overrightarrow b \times \overrightarrow a $$ and $$\overrightarrow c .\o... | ["A"]
Explanation:
$$\overrightarrow a = \widehat i - 2\widehat j + \widehat k$$
<br><br>$$\overrightarrow b = \widehat i - \widehat j + \widehat k$$
<br><br>$$\left| {\overrightarrow a } \right|$$ = $$\sqrt 6 $$, $$\left| {\overrightarrow b } \right|$$ = $$\sqrt 3 $$
<br><br>and $${\overrightarrow a .\overrightarro... |
Let $$\overrightarrow a $$, $$\overrightarrow b $$ and $$\overrightarrow c $$ be three vectors such that $$\left| {\overrightarrow a } \right| = \sqrt 3 $$,
$$\left| {\overrightarrow b } \right| = 5,\overrightarrow b .\overrightarrow c = 10$$ and the angle between $$\overrightarrow b $$ and $$\overrightarrow c $$
is ... | 30
Explanation:
Given $$\left| {\overrightarrow a } \right| = \sqrt 3 $$,
$$\left| {\overrightarrow b } \right| = 5$$
<br><br>Given $$\overrightarrow b .\overrightarrow c = 10$$
<br><br>And the angle between $$\overrightarrow b $$ and $$\overrightarrow c $$
is $${\pi \over 3}$$
<br><br>$$ \therefore $$ $$bc\cos {\p... |
Let the position vectors of points 'A' and 'B' be
<br/>$$\widehat i + \widehat j + \widehat k$$ and $$2\widehat i + \widehat j + 3\widehat k$$, respectively. A point
'P' divides the line segment AB internally in the
ratio
$$\lambda $$ : 1 (
$$\lambda $$ > 0). If O is the origin and
<br/>$$\overrightarrow {OB} .\ove... | 0.8
Explanation:
Let, $$\overrightarrow a $$ = $$\widehat i + \widehat j + \widehat k$$
<br><br>and $$\overrightarrow b $$ = $$2\widehat i + \widehat j + 3\widehat k$$
<picture><source media="(max-width: 320px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265164/exam_images/hoezbwyywsjaggdoi7yh.webp... |
If $$\overrightarrow a = 2\widehat i + \widehat j + 2\widehat k$$, then the value of<br/><br/>
$${\left| {\widehat i \times \left( {\overrightarrow a \times \widehat i} \right)} \right|^2} + {\left| {\widehat j \times \left( {\overrightarrow a \times \widehat j} \right)} \right|^2} + {\left| {\widehat k \times \lef... | 18
Explanation:
Let $$\overrightarrow a = x\widehat i + y\widehat j + z\widehat k$$<br><br>Now $$\widehat i \times \left( {\overrightarrow a \times \widehat i} \right) = \left( {\widehat i.\widehat i} \right)\overrightarrow a - \left( {\widehat i.\overrightarrow a } \right)\widehat i$$<br><br>= $$y\widehat j + z\wi... |
Let $$\overrightarrow a = \widehat i + 2\widehat j - \widehat k$$, $$\overrightarrow b = \widehat i - \widehat j$$ and $$\overrightarrow c = \widehat i - \widehat j - \widehat k$$ be three given vectors. If $$\overrightarrow r $$ is a vector such that $$\overrightarrow r \times \overrightarrow a = \overrightarrow ... | 12
Explanation:
<p>Given, $$\overrightarrow a = \widehat i + 2\widehat j - \widehat k$$,</p>
<p>$$\overrightarrow b = \widehat i - \widehat j$$,</p>
<p>$$\overrightarrow c = \widehat i - \widehat j - \widehat k$$</p>
<p>$$\overrightarrow r \times \overrightarrow a = \overrightarrow c \times \overrightarrow a $$<... |
Let $$\overrightarrow a = \widehat i + \alpha \widehat j + 3\widehat k$$ and $$\overrightarrow b = 3\widehat i - \alpha \widehat j + \widehat k$$. If the area of the parallelogram whose adjacent sides are represented by the vectors $$\overrightarrow a $$ and $$\overrightarrow b $$ is $$8\sqrt 3 $$ square units, then ... | 2
Explanation:
$$\overrightarrow a = \widehat i + \alpha \widehat j + 3\widehat k$$<br><br>$$\overrightarrow b = 3\widehat i - \alpha \widehat j + \widehat k$$<br><br>Area of parallelogram = $$\left| {\overrightarrow a \times \overrightarrow b } \right|$$<br><br>$$ = \left| {(\widehat i + \alpha \widehat j + 3\wide... |
Let $$\overrightarrow a $$ = $$\widehat i$$ + 2$$\widehat j$$ $$-$$ 3$$\widehat k$$ and $$\overrightarrow b = 2\widehat i$$ $$-$$ 3$$\widehat j$$ + 5$$\widehat k$$. If $$\overrightarrow r $$ $$\times$$ $$\overrightarrow a $$ = $$\overrightarrow b $$ $$\times$$ $$\overrightarrow r $$, <br/><br/>$$\overrightarrow r $$ .... | ["D"]
Explanation:
Given $$\overrightarrow r $$ $$\times$$ $$\overrightarrow a $$ = $$\overrightarrow b $$ $$\times$$ $$\overrightarrow r $$
<br><br>$$ \Rightarrow $$ $$\overrightarrow r \times \overrightarrow a = - \overrightarrow r \times \overrightarrow b $$<br><br>$$\overrightarrow r \times (\overrightarrow ... |
Let $$\overrightarrow a $$ and $$\overrightarrow b $$ be two non-zero vectors perpendicular to each other and $$|\overrightarrow a | = |\overrightarrow b |$$. If $$|\overrightarrow a \times \overrightarrow b | = |\overrightarrow a |$$, then the angle between the vectors $$\left( {\overrightarrow a + \overrightarrow b... | ["D"]
Explanation:
$$\overrightarrow a $$ is perpendicular to $$\overrightarrow b $$<br><br>$$ \therefore $$ $$\overrightarrow a $$ . $$\overrightarrow b $$ = 0<br><br>Given, | $$\overrightarrow a $$ $$\times$$ $$\overrightarrow b $$ | = | $$\overrightarrow a $$ |<br><br>and | $$\overrightarrow a $$ | = | $$\overright... |
If the shortest distance between the lines $$\overrightarrow {{r_1}} = \alpha \widehat i + 2\widehat j + 2\widehat k + \lambda (\widehat i - 2\widehat j + 2\widehat k)$$, $$\lambda$$ $$\in$$ R, $$\alpha$$ > 0 and $$\overrightarrow {{r_2}} = - 4\widehat i - \widehat k + \mu (3\widehat i - 2\widehat j - 2\widehat k... | 6
Explanation:
If $$\overrightarrow r = \overrightarrow a + \lambda \overrightarrow b $$ and $$\overrightarrow r = \overrightarrow c + \lambda \overrightarrow d $$ then shortest distance between two lines is <br><br>$$L = {{(\overrightarrow a - \overrightarrow c ).(\overrightarrow b \times \overrightarrow d )} \... |
Let $$\overrightarrow p = 2\widehat i + 3\widehat j + \widehat k$$ and $$\overrightarrow q = \widehat i + 2\widehat j + \widehat k$$ be two vectors. If a vector $$\overrightarrow r = (\alpha \widehat i + \beta \widehat j + \gamma \widehat k)$$ is perpendicular to each of the vectors ($$(\overrightarrow p + \overrig... | 3
Explanation:
$$\overrightarrow p = 2\widehat i + 3\widehat j + \widehat k$$ (Given )<br><br>$$\overrightarrow q = \widehat i + 2\widehat j + \widehat k$$<br><br>Now, $$(\overrightarrow p + \overrightarrow q ) \times (\overrightarrow p - \overrightarrow q ) = \left| {\matrix{
{\widehat i} & {\widehat j} &a... |
If $$\left| {\overrightarrow a } \right| = 2,\left| {\overrightarrow b } \right| = 5$$ and $$\left| {\overrightarrow a \times \overrightarrow b } \right|$$ = 8, then $$\left| {\overrightarrow a .\,\overrightarrow b } \right|$$ is equal to :
Options:
[{"identifier": "A", "content": "6"}, {"identifier": "B", "content":... | ["A"]
Explanation:
$$\left| {\overrightarrow a } \right| = 2,\left| {\overrightarrow b } \right| = 5$$<br><br>$$\left| {\overrightarrow a \times \overrightarrow b } \right| = \left| {\overrightarrow a } \right|\left| {\overrightarrow b } \right|\sin \theta = \pm 8$$<br><br>$$\sin \theta = \pm \,{4 \over 5}$$<br><... |
Let $$\overrightarrow a = \widehat i + \widehat j + 2\widehat k$$ and $$\overrightarrow b = - \widehat i + 2\widehat j + 3\widehat k$$. Then the vector product $$\left( {\overrightarrow a + \overrightarrow b } \right) \times \left( {\left( {\overrightarrow a \times \left( {\left( {\overrightarrow a - \overrightar... | ["B"]
Explanation:
$$\overrightarrow a = \widehat i + \widehat j + 2\widehat k$$<br><br>$$\overrightarrow b = - \widehat i + 2\widehat j + 3\widehat k$$<br><br>$$\overrightarrow a + \overrightarrow b = 3\widehat j + 5\widehat k;\overrightarrow a.\overrightarrow b = - 1 + 2 + 6 = 7$$<br><br>$$\left( {\left( {\ov... |
Let $$\overrightarrow a = \widehat i + \widehat j + \widehat k,\overrightarrow b $$ and $$\overrightarrow c = \widehat j - \widehat k$$ be three vectors such that $$\overrightarrow a \times \overrightarrow b = \overrightarrow c $$ and $$\overrightarrow a \,.\,\overrightarrow b = 1$$. If the length of projection ve... | 2
Explanation:
$$\overrightarrow a \times \overrightarrow b = \overrightarrow c $$<br><br>Take Dot with $$\overrightarrow c $$<br><br>$$\left( {\overrightarrow a \times \overrightarrow b } \right).\,\overrightarrow c = {\left| {\overrightarrow c } \right|^2} = 2$$<br><br>Projection of $$\overrightarrow b $$ or $$\... |
Let $$\overrightarrow a = \widehat i + 5\widehat j + \alpha \widehat k$$, $$\overrightarrow b = \widehat i + 3\widehat j + \beta \widehat k$$ and $$\overrightarrow c = - \widehat i + 2\widehat j - 3\widehat k$$ be three vectors such that, $$\left| {\overrightarrow b \times \overrightarrow c } \right| = 5\sqrt 3 $$... | 90
Explanation:
Since, $$\overrightarrow a .\,\overrightarrow b = 0$$<br><br>$$1 + 15 + \alpha \beta = 0 \Rightarrow \alpha \beta = - 16$$ .... (1)<br><br>Also, <br><br>$${\left| {\overrightarrow b \, \times \overrightarrow c } \right|^2} = 75 \Rightarrow (10 + {\beta ^2})14 - {(5 - 3\beta )^2} = 75$$<br><br>$$\Ri... |
Let $$\overrightarrow a = 2\widehat i - \widehat j + 2\widehat k$$ and $$\overrightarrow b = \widehat i + 2\widehat j - \widehat k$$. Let a vector $$\overrightarrow v $$ be in the plane containing $$\overrightarrow a $$ and $$\overrightarrow b $$. If $$\overrightarrow v $$ is perpendicular to the vector $$3\widehat i... | 1494
Explanation:
$$\overrightarrow a = 2\widehat i - \widehat j + 2\widehat k$$<br><br>$$\overrightarrow b = \widehat i + 2\widehat j - \widehat k$$<br><br>$$\overrightarrow c = 3\widehat i + 2\widehat j - \widehat k$$<br><br>$$\overrightarrow v = x\overrightarrow a + y\overrightarrow b $$<br><br>$$\overrightarr... |
<p>Let $$\overrightarrow a = \alpha \widehat i + 3\widehat j - \widehat k$$, $$\overrightarrow b = 3\widehat i - \beta \widehat j + 4\widehat k$$ and $$\overrightarrow c = \widehat i + 2\widehat j - 2\widehat k$$ where $$\alpha ,\,\beta \in R$$, be three vectors. If the projection of $$\overrightarrow a $$ on $$\ov... | ["A"]
Explanation:
<p>$$\overrightarrow a = \alpha \widehat i + 3\widehat j - \widehat k$$</p>
<p>$$\overrightarrow b = 3\widehat i - \beta \widehat j + 4\widehat k$$</p>
<p>$$\overrightarrow c = \widehat i + 2\widehat j - 2\widehat k$$</p>
<p>Projection of $$\overrightarrow a $$ on $$\overrightarrow c $$ is</p>
<p... |
<p>Let $$\overrightarrow a = \widehat i - 2\widehat j + 3\widehat k$$, $$\overrightarrow b = \widehat i + \widehat j + \widehat k$$ and $$\overrightarrow c $$ be a vector such that $$\overrightarrow a + \left( {\overrightarrow b \times \overrightarrow c } \right) = \overrightarrow 0 $$ and $$\overrig... | BONUS
Explanation:
$$
\vec{a} \cdot \vec{b}=(\hat{i}-2 \hat{j}+3 \hat{k}) \cdot(\hat{i}+\hat{j}+\hat{k})=2
$$ ........(i)
<br/><br/>Given: $\vec{a}+(\vec{b} \times \vec{c})=0$
<br/><br/>$$
\Rightarrow \vec{a} \cdot \vec{b}=0
$$ ........(ii)
<br/><br/>Equation (i) and equation (ii) are contradicting. |
<p>Let $$\overrightarrow a = \alpha \widehat i + 2\widehat j - \widehat k$$ and $$\overrightarrow b = - 2\widehat i + \alpha \widehat j + \widehat k$$, where $$\alpha \in R$$. If the area of the parallelogram whose adjacent sides are represented by the vectors $$\overrightarrow a $$ and $$\overrightarrow b $$ is $$... | ["D"]
Explanation:
<p>$$\overrightarrow a = \alpha \widehat i + 2\widehat j - \widehat k$$ and $$\overrightarrow b = - 2\widehat i + \alpha \widehat j + \widehat k$$</p>
<p>$$\therefore$$ $$\overrightarrow a \times \overrightarrow b = \left| {\matrix{
{\widehat i} & {\widehat j} & {\widehat k} \cr
\alpha ... |
<p>Let $$\overrightarrow a $$ be a vector which is perpendicular to the vector $$3\widehat i + {1 \over 2}\widehat j + 2\widehat k$$. If $$\overrightarrow a \times \left( {2\widehat i + \widehat k} \right) = 2\widehat i - 13\widehat j - 4\widehat k$$, then the projection of the vector $$\overrightarrow a $$ on the vec... | ["C"]
Explanation:
<p>Let $$\overrightarrow a = {a_1}\widehat i + {a_2}\widehat j + {a_3}\widehat k$$</p>
<p>and $$\overrightarrow a \,.\,\left( {3\widehat i - {1 \over 2}\widehat j + 2\widehat k} \right) = 0 \Rightarrow 3{a_1} + {{{a_2}} \over 2} + 2{a_3} = 0$$ ..... (i)</p>
<p>and $$\overrightarrow a \times (2\wid... |
<p>If $$\overrightarrow a = 2\widehat i + \widehat j + 3\widehat k$$, $$\overrightarrow b = 3\widehat i + 3\widehat j + \widehat k$$ and $$\overrightarrow c = {c_1}\widehat i + {c_2}\widehat j + {c_3}\widehat k$$ are coplanar vectors and $$\overrightarrow a \,.\,\overrightarrow c = 5$$, $$\overrightarrow b \bot \o... | 150
Explanation:
<p>$$2{C_1} + {C_2} + 3{C_3} = 5$$ ...... (i)</p>
<p>$$3{C_1} + 3{C_2} + {C_3} = 0$$ ...... (ii)</p>
<p>$$\left[ {\overrightarrow a \overrightarrow b \overrightarrow c } \right] = \left| {\matrix{
2 & 1 & 3 \cr
3 & 3 & 1 \cr
{{C_1}} & {{C_2}} & {{C_3}} \cr
} } \right|$$</p>
<p>$$ = 2(... |
<p>Let $$\overrightarrow a $$ and $$\overrightarrow b $$ be the vectors along the diagonals of a parallelogram having area $$2\sqrt 2 $$. Let the angle between $$\overrightarrow a $$ and $$\overrightarrow b $$ be acute, $$|\overrightarrow a | = 1$$, and $$|\overrightarrow a \,.\,\overrightarrow b | = |\overrightarrow a... | ["D"]
Explanation:
<p>$$\because$$ $$\overrightarrow a $$ and $$\overrightarrow b $$ be the vectors along the diagonals of a parallelogram having area 2$$\sqrt2$$.</p>
<p>$$\therefore$$ $${1 \over 2}|\overrightarrow a \times \overrightarrow b | = 2\sqrt 2 $$</p>
<p>$$|\overrightarrow a ||\overrightarrow b |\sin \thet... |
<p>Let $$\overrightarrow a = \widehat i + \widehat j - \widehat k$$ and $$\overrightarrow c = 2\widehat i - 3\widehat j + 2\widehat k$$. Then the number of vectors $$\overrightarrow b $$ such that $$\overrightarrow b \times \overrightarrow c = \overrightarrow a $$ and $$|\overrightarrow b | \in $$ {1, 2, ........, ... | ["A"]
Explanation:
<p>$$\overrightarrow a = \widehat i + \widehat j - \widehat k$$</p>
<p>$$\overrightarrow c = 2\widehat i - 3\widehat j + 2\widehat k$$</p>
<p>Now, $$\overrightarrow b \times \overrightarrow c = \overrightarrow a $$</p>
<p>$$\overrightarrow c \,.\,(\overrightarrow b \times \overrightarrow c ) = ... |
<p>Let $$\overrightarrow b = \widehat i + \widehat j + \lambda \widehat k$$, $$\lambda$$ $$\in$$ R. If $$\overrightarrow a $$ is a vector such that $$\overrightarrow a \times \overrightarrow b = 13\widehat i - \widehat j - 4\widehat k$$ and $$\overrightarrow a \,.\,\overrightarrow b + 21 = 0$$, then $$\left( {\over... | 14
Explanation:
<p>Let $$\overrightarrow a = x\widehat i = y\widehat j + z\widehat k$$</p>
<p>So, $$\left| {\matrix{
{\widehat i} & {\widehat j} & {\widehat k} \cr
x & y & z \cr
1 & 1 & \lambda \cr
} } \right| = \widehat i(\lambda y - z) + \widehat j(z - \lambda x) + \widehat k(x - y)$$</p>
<p>$$ \R... |
<p>Let $$\theta$$ be the angle between the vectors $$\overrightarrow a $$ and $$\overrightarrow b $$, where $$|\overrightarrow a | = 4,$$ $$|\overrightarrow b | = 3$$ and $$\theta \in \left( {{\pi \over 4},{\pi \over 3}} \right)$$. Then $${\left| {\left( {\overrightarrow a - \overrightarrow b } \right) \times \left... | 576
Explanation:
<p>$${\left| {\left( {\overrightarrow a - \overrightarrow b } \right) \times \left( {\overrightarrow a + \overrightarrow b } \right)} \right|^2} + 4{\left( {\overrightarrow a \,.\,\overrightarrow b } \right)^2}$$</p>
<p>$$ \Rightarrow {\left| {\overrightarrow a \times \overrightarrow a + \overrigh... |
<p>Let $$\widehat a$$ and $$\widehat b$$ be two unit vectors such that $$|(\widehat a + \widehat b) + 2(\widehat a \times \widehat b)| = 2$$. If $$\theta$$ $$\in$$ (0, $$\pi$$) is the angle between $$\widehat a$$ and $$\widehat b$$, then among the statements :</p>
<p>(S1) : $$2|\widehat a \times \widehat b| = |\widehat... | ["C"]
Explanation:
<p>$$\left| {\widehat a + \widehat b + 2(\widehat a \times \widehat b)} \right| = 2,\,\theta \in (0,\,\pi )$$</p>
<p>$$ \Rightarrow {\left| {\widehat a + \widehat b + 2(\widehat a \times \widehat b)} \right|^2} = 4$$</p>
<p>$$ \Rightarrow {\left| {\widehat a} \right|^2} + {\left| {\widehat b} \righ... |
<p>Let $$\widehat a$$, $$\widehat b$$ be unit vectors. If $$\overrightarrow c $$ be a vector such that the angle between $$\widehat a$$ and $$\overrightarrow c $$ is $${\pi \over {12}}$$, and $$\widehat b = \overrightarrow c + 2\left( {\overrightarrow c \times \widehat a} \right)$$, then $${\left| {6\overrightarrow ... | ["C"]
Explanation:
$\because \quad \hat{b}=\vec{c}+2(\vec{c} \times \hat{a})$
<br/><br/>
$$
\begin{aligned}
&\Rightarrow \hat{b} \cdot \vec{c}=|\vec{c}|^{2} \\\\
&\therefore \hat{b}-\vec{c}=2(\vec{c} \times \vec{a})
\end{aligned}
$$
<br/><br/>
$$
\begin{aligned}
&\Rightarrow|\hat{b}|^{2}+|\vec{c}|^{2}-2 \hat{b} \cdo... |
<p>Let $$\mathrm{ABC}$$ be a triangle such that $$\overrightarrow{\mathrm{BC}}=\overrightarrow{\mathrm{a}}, \overrightarrow{\mathrm{CA}}=\overrightarrow{\mathrm{b}}, \overrightarrow{\mathrm{AB}}=\overrightarrow{\mathrm{c}},|\overrightarrow{\mathrm{a}}|=6 \sqrt{2},|\overrightarrow{\mathrm{b}}|=2 \sqrt{3}$$ and $$\vec{b}... | ["C"]
Explanation:
<img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l97sapyl/779dd912-c32e-40a3-8f62-d0f26f1743e1/b6acf9d0-4b5b-11ed-bfde-e1cb3fafe700/file-1l97sapym.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l97sapyl/779dd912-c32e-40a3-8f62-d0f26f1743e1/b6acf9d0-4b5b-11ed-... |
<p>Let $$\vec{a}=\hat{i}-\hat{j}+2 \hat{k}$$ and let $$\vec{b}$$ be a vector such that $$\vec{a} \times \vec{b}=2 \hat{i}-\hat{k}$$ and $$\vec{a} \cdot \vec{b}=3$$. Then the projection of $$\vec{b}$$ on the vector $$\vec{a}-\vec{b}$$ is :</p>
Options:
[{"identifier": "A", "content": "$$\\frac{2}{\\sqrt{21}}$$"}, {"ide... | ["A"]
Explanation:
<p>$$\overrightarrow a = \widehat i - \widehat j + 2\widehat k$$</p>
<p>$$\overrightarrow a \times \overrightarrow b = 2\widehat i - \widehat k$$</p>
<p>$$\overrightarrow a \,.\,\overrightarrow b = 3$$</p>
<p>$$|\overrightarrow a \times \overrightarrow b {|^2} + |\overrightarrow a \,.\,\overrig... |
<p>Let $$\overrightarrow{\mathrm{a}}=\alpha \hat{i}+\hat{j}-\hat{k}$$ and $$\overrightarrow{\mathrm{b}}=2 \hat{i}+\hat{j}-\alpha \hat{k}, \alpha>0$$. If the projection of $$\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}$$ on the vector $$-\hat{i}+2 \hat{j}-2 \hat{k}$$ is 30, then $$\alpha$$ is equal ... | ["D"]
Explanation:
<p>Given : $$\overrightarrow a = (\alpha ,1, - 1)$$ and $$\overrightarrow b = (2,1, - \alpha )$$</p>
<p>$$\overrightarrow c = \overrightarrow a \times \overrightarrow b = \left| {\matrix{
{\widehat i} & {\widehat j} & {\widehat k} \cr
\alpha & 1 & { - 1} \cr
2 & 1 & { - \alpha } ... |
<p>Let $$\vec{a}=\alpha \hat{i}+\hat{j}+\beta \hat{k}$$ and $$\vec{b}=3 \hat{i}-5 \hat{j}+4 \hat{k}$$ be two vectors, such that $$\vec{a} \times \vec{b}=-\hat{i}+9 \hat{j}+12 \hat{k}$$. Then the projection of $$\vec{b}-2 \vec{a}$$ on $$\vec{b}+\vec{a}$$ is equal to :</p>
Options:
[{"identifier": "A", "content": "2"}, ... | ["D"]
Explanation:
<p>$$\overrightarrow a = \alpha \widehat i + \widehat j + \beta \widehat k$$, $$\overrightarrow b = 3\widehat i - 5\widehat j + 4\widehat k$$</p>
<p>$$\overrightarrow a \times \overrightarrow b = - \widehat i + 9\widehat j + 12\widehat k$$</p>
<p>$$\left| {\matrix{
{\widehat i} & {\widehat j... |
<p>Let $$\overrightarrow a $$, $$\overrightarrow b $$, $$\overrightarrow c $$ be three non-coplanar vectors such that $$\overrightarrow a $$ $$\times$$ $$\overrightarrow b $$ = 4$$\overrightarrow c $$, $$\overrightarrow b $$ $$\times$$ $$\overrightarrow c $$ = 9$$\overrightarrow a $$ and $$\overrightarrow c $$ $$\times... | 36
Explanation:
<p>Given,</p>
<p>$$\overrightarrow a \times \overrightarrow b = 4\,.\,\overrightarrow c $$ ..... (i)</p>
<p>$$\overrightarrow b \times \overrightarrow c = 9\,.\,\overrightarrow a $$ ..... (ii)</p>
<p>$$\overrightarrow c \times \overrightarrow a = \alpha \,.\,\overrightarrow b $$ .... (iii)</p>
<p... |
<p>Let a vector $$\vec{a}$$ has magnitude 9. Let a vector $$\vec{b}$$ be such that for every $$(x, y) \in \mathbf{R} \times \mathbf{R}-\{(0,0)\}$$, the vector $$(x \vec{a}+y \vec{b})$$ is perpendicular to the vector $$(6 y \vec{a}-18 x \vec{b})$$. Then the value of $$|\vec{a} \times \vec{b}|$$ is equal to :</p>
Option... | ["B"]
Explanation:
<p>$$\left( {x\overrightarrow a + y\overrightarrow b } \right).\left( {6y\overrightarrow a - 18x\overrightarrow b } \right) = 0$$</p>
<p>$$ \Rightarrow \left( {6xy|\overrightarrow a {|^2} - 18xy|\overrightarrow b {|^2}} \right) + \left( {6{y^2} - 18{x^2}} \right)\overrightarrow a .\overrightarrow ... |
<p>Let $$\hat{a}$$ and $$\hat{b}$$ be two unit vectors such that the angle between them is $$\frac{\pi}{4}$$. If $$\theta$$ is the angle between the vectors $$(\hat{a}+\hat{b})$$ and $$(\hat{a}+2 \hat{b}+2(\hat{a} \times \hat{b}))$$, then the value of $$164 \,\cos ^{2} \theta$$ is equal to :</p>
Options:
[{"identifier... | ["A"]
Explanation:
<p>$$\widehat a\,.\,\widehat b = {1 \over {\sqrt 2 }}$$ and $$|\widehat a \times \widehat b| = {1 \over {\sqrt 2 }}$$</p>
<p>$${{\left( {\widehat a + \widehat b} \right)\,.\,\left( {\widehat a + 2\widehat b + 2\left( {\widehat a \times \widehat b} \right)} \right)} \over {\left| {\widehat a + \wideh... |
<p>Let $$\vec{a}, \vec{b}, \vec{c}$$ be three coplanar concurrent vectors such that angles between any two of them is same. If the product of their magnitudes is 14 and
$$(\vec{a} \times \vec{b}) \cdot(\vec{b} \times \vec{c})+(\vec{b} \times \vec{c}) \cdot(\vec{c} \times \vec{a})+(\vec{c} \times \vec{a}) \cdot(\vec{a} ... | ["C"]
Explanation:
$|\vec{a}||\vec{b}||\vec{c}|=14$
<br/><br/>
$$
\begin{aligned}
& \vec{a} \wedge \vec{b}=\vec{b} \wedge \vec{c}=\vec{c} \wedge \vec{a}=\theta=\frac{2 \pi}{3} \\\\
& \vec{a} \cdot \vec{b}=-\frac{1}{2}|\vec{a}||\vec{b}| \\\\
& \vec{b} \cdot \vec{c}=-\frac{1}{2}|\vec{b}||\vec{c}| \\\\
& \vec{c} \cdot \v... |
<p>Let $$\vec{a}$$ and $$\vec{b}$$ be two vectors such that $$|\vec{a}+\vec{b}|^{2}=|\vec{a}|^{2}+2|\vec{b}|^{2}, \vec{a} \cdot \vec{b}=3$$ and $$|\vec{a} \times \vec{b}|^{2}=75$$. Then $$|\vec{a}|^{2}$$ is equal to __________.</p>
Options:
[] | 14
Explanation:
$\because|\vec{a}+\dot{b}|^{2}=|\vec{a}|^{2}+2|b|^{2}$
<br/><br/>or $|\vec{a}|^{2}+|\vec{b}|^{2}+2 \vec{a} \cdot \vec{b}=|\vec{a}|^{2}+2|\vec{b}|^{2}$
<br/><br/>$\therefore|\vec{b}|^{2}=6$
<br/><br/>Now $|\vec{a} \times \vec{b}|^{2}=|\vec{a}|^{2}|\vec{b}|^{2}-(\vec{a} \cdot \vec{b})^{2}$
<br/><br/>... |
Let $\vec{a}=\hat{i}+2 \hat{j}+3 \hat{k}, \vec{b}=\hat{i}-\hat{j}+2 \hat{k}$ and $\vec{c}=5 \hat{i}-3 \hat{j}+3 \hat{k}$ be three vectors. If $\vec{r}$ is a vector such
that, $\vec{r} \times \vec{b}=\vec{c} \times \vec{b}$ and $\vec{r} \cdot \vec{a}=0$, then $25|\vec{r}|^{2}$ is equal to :
Options:
[{"identifier": "... | ["C"]
Explanation:
$\overrightarrow{\mathrm{a}}=\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}$
<br/><br/>$\overrightarrow{\mathrm{b}}=\hat{\mathrm{i}}-\hat{\mathrm{j}}+2 \hat{\mathrm{k}}$
<br/><br/>$\overrightarrow{\mathrm{c}}=\hat{5 \mathrm{i}}-3 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}$
<br/><br/>$(\overright... |
Let $\vec{a}, \vec{b}, \vec{c}$ be three vectors such that
<br/><br/>$|\vec{a}|=\sqrt{31}, 4|\vec{b}|=|\vec{c}|=2$ and $2(\vec{a} \times \vec{b})=3(\vec{c} \times \vec{a})$.
<br/><br/>If the angle between $\vec{b}$ and $\vec{c}$ is $\frac{2 \pi}{3}$, then $\left(\frac{\vec{a} \times \vec{c}}{\vec{a} \cdot \vec{b}}\righ... | 3
Explanation:
$2(\vec{a} \times \vec{b})=3(\vec{c} \times \vec{a})$
<br/><br/>$\vec{a} \times(2 \vec{b}+3 \vec{c})=0$
<br/><br/>$$
\begin{aligned}
& \vec{a}=\lambda(2 \vec{b}+3 \vec{c}) \\\\
& |\vec{a}|^{2}=\lambda^{2}\left(4|b|^{2}+9|c|^{2}+12 \vec{b} \cdot \vec{c}\right) \\\\
& 31=31 \lambda^{2} \\\\
& \lambda=\pm... |
<p>$$A(2,6,2), B(-4,0, \lambda), C(2,3,-1)$$ and $$D(4,5,0),|\lambda| \leq 5$$ are the vertices of a quadrilateral $$A B C D$$. If its area is 18 square units, then $$5-6 \lambda$$ is equal to __________.</p>
Options:
[] | 11
Explanation:
$$
\begin{aligned}
& \mathrm{A}(2,6,2) \quad \mathrm{B}(-4,0, \lambda), \mathrm{C}(2,3,-1) \mathrm{D}(4,5,0) \\\\
& \text { Area }=\frac{1}{2}|\overrightarrow{B D} \times \overrightarrow{A C}|=18 \\\\
& \overrightarrow{A C} \times \overrightarrow{B D}=\left|\begin{array}{ccc}
\hat{i} & j & k \\\\
0 & -... |
<p>Let $$\vec{a}$$ and $$\vec{b}$$ be two vectors such that $$|\vec{a}|=\sqrt{14},|\vec{b}|=\sqrt{6}$$ and $$|\vec{a} \times \vec{b}|=\sqrt{48}$$. Then $$(\vec{a} \cdot \vec{b})^{2}$$ is equal to ___________.</p>
Options:
[] | 36
Explanation:
$|\vec{a}|=\sqrt{14},|\vec{b}|=\sqrt{6}$ and $|\vec{a} \times \vec{b}|=\sqrt{48}$
<br/><br/>$$
\begin{aligned}
& \Rightarrow |\vec{a} \times \vec{b}|^{2}+(\vec{a} \cdot \vec{b})^{2}=|\vec{a}|^{2}|\vec{b}|^{2} \\\\
& \Rightarrow 48+(\vec{a} \cdot \vec{b})^{2}=6 \times 14 \\\\
& \Rightarrow (\vec{a} \cd... |
Let $\vec{a}$ and $\vec{b}$ be two vectors, Let $|\vec{a}|=1,|\vec{b}|=4$ and $\vec{a} \cdot \vec{b}=2$. If $\vec{c}=(2 \vec{a} \times \vec{b})-3 \vec{b}$, then the value of $\vec{b} \cdot \vec{c}$ is :
Options:
[{"identifier": "A", "content": "$-48$"}, {"identifier": "B", "content": "$-60$"}, {"identifier": "C", "con... | ["A"]
Explanation:
<p>$$\overrightarrow b .\overrightarrow c = \overrightarrow b .(2\overline a \times \overrightarrow b ) - 3\overline b .\overrightarrow b $$</p>
<p>$$ = 0 - 3|\overline b {|^2} = - 48$$</p> |
<p>Let a unit vector $$\widehat{O P}$$ make angles $$\alpha, \beta, \gamma$$ with the positive directions of the co-ordinate axes $$\mathrm{OX}$$, $$\mathrm{OY}, \mathrm{OZ}$$ respectively, where $$\beta \in\left(0, \frac{\pi}{2}\right)$$. If $$\widehat{\mathrm{OP}}$$ is perpendicular to the plane through points $$(1,2... | ["A"]
Explanation:
<p>Let $$A \equiv (1,2,3),B \equiv (2,3,4),C \equiv (1,5,7)$$</p>
<p>$$\overrightarrow n = \overrightarrow {AB} \times \overrightarrow {AC} = \left| {\matrix{
i & j & k \cr
1 & 1 & 1 \cr
0 & 3 & 4 \cr
} } \right|$$</p>
<p>$$ = \widehat i - 4\widehat j + 3\widehat k$$</p>
<p>$$\wi... |
<p>If $$\overrightarrow a = \widehat i + 2\widehat k,\overrightarrow b = \widehat i + \widehat j + \widehat k,\overrightarrow c = 7\widehat i - 3\widehat j + 4\widehat k,\overrightarrow r \times \overrightarrow b + \overrightarrow b \times \overrightarrow c = \overrightarrow 0 $$ and $$\overrightarrow r \,.\,\ov... | ["C"]
Explanation:
<p>$$(\overrightarrow r - \overrightarrow c ) \times \overrightarrow b = 0$$</p>
<p>$$\overrightarrow r = \lambda \overrightarrow b + \overrightarrow c $$</p>
<p>$$ \Rightarrow \lambda \overrightarrow b \,.\,\overrightarrow a + \overrightarrow c \,.\,\overrightarrow a = 0$$</p>
<p>$$ \Rightarr... |
<p>Let $$\overrightarrow a = 4\widehat i + 3\widehat j$$ and $$\overrightarrow b = 3\widehat i - 4\widehat j + 5\widehat k$$. If $$\overrightarrow c $$ is a vector such that $$\overrightarrow c .\left( {\overrightarrow a \times \overrightarrow b } \right) + 25 = 0,\overrightarrow c \,.(\widehat i + \widehat j + \wid... | ["D"]
Explanation:
<p>$$[\matrix{
{\overrightarrow c } & {\overrightarrow a } & {\overrightarrow b } \cr
} ] = - 25$$</p>
<p>Let $$\overrightarrow c = l\widehat i + n\widehat j + n\widehat k$$</p>
<p>$$\left| {\matrix{
l & m & n \cr
4 & 3 & 0 \cr
3 & { - 4} & 5 \cr
} } \right| = - 25$$</p>
<... |
<p>Let $$\overrightarrow a = \widehat i + 2\widehat j + \lambda \widehat k,\overrightarrow b = 3\widehat i - 5\widehat j - \lambda \widehat k,\overrightarrow a \,.\,\overrightarrow c = 7,2\overrightarrow b \,.\,\overrightarrow c + 43 = 0,\overrightarrow a \times \overrightarrow c = \overrightarrow b \times \over... | 8
Explanation:
$$
\begin{aligned}
& \vec{a}=\hat{i}+2 \hat{j}+\lambda \hat{k}, \vec{b}=3 \hat{i}-5 \hat{j}-\lambda \hat{k}, \vec{a} \cdot \vec{c}=7 \\\\
& \vec{a} \times \vec{c}-\vec{b} \times \vec{c}=\overrightarrow{0} \\\\
& (\vec{a}-\vec{b}) \times \vec{c}=0 \Rightarrow(\vec{a}-\vec{b}) \text { is paralleled to } \... |
<p>Let $$|\vec{a}|=2,|\vec{b}|=3$$ and the angle between the vectors $$\vec{a}$$ and $$\vec{b}$$ be $$\frac{\pi}{4}$$. Then $$|(\vec{a}+2 \vec{b}) \times(2 \vec{a}-3 \vec{b})|^{2}$$ is equal to :</p>
Options:
[{"identifier": "A", "content": "441"}, {"identifier": "B", "content": "482"}, {"identifier": "C", "content": ... | ["D"]
Explanation:
$$
\begin{aligned}
& |\vec{a}|=2 \\\\
& |\vec{b}|=3 \\\\
& \vec{a} \cdot \vec{b}=\frac{\pi}{4}
\end{aligned}
$$
<br/><br/>$$
\begin{aligned}
& |(\vec{a}+2 \vec{b}) \times(2 \vec{a}-3 \vec{b})|^2 \\\\
& = |-3 \vec{a} \times \vec{b}+4 \vec{b} \times \vec{a}|^2 \\\\
& = |-3 \vec{a} \times \vec{b}-4 \ve... |
<p>Let for a triangle $$\mathrm{ABC}$$,</p>
<p>$$\overrightarrow{\mathrm{AB}}=-2 \hat{i}+\hat{j}+3 \hat{k}$$</p>
<p>$$\overrightarrow{\mathrm{CB}}=\alpha \hat{i}+\beta \hat{j}+\gamma \hat{k}$$</p>
<p>$$\overrightarrow{\mathrm{CA}}=4 \hat{i}+3 \hat{j}+\delta \hat{k}$$</p>
<p>If $$\delta > 0$$ and the area of the tria... | ["A"]
Explanation:
<img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lh1phdby/e74db975-74e1-4987-80c4-b8ae4380dc69/bb54dde0-e665-11ed-9cfb-9bcc868d407f/file-1lh1phdbz.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lh1phdby/e74db975-74e1-4987-80c4-b8ae4380dc69/bb54dde0-e665-11ed-... |
<p>Let $$\vec{a}=\hat{i}+4 \hat{j}+2 \hat{k}, \vec{b}=3 \hat{i}-2 \hat{j}+7 \hat{k}$$ and $$\vec{c}=2 \hat{i}-\hat{j}+4 \hat{k}$$. If a vector $$\vec{d}$$ satisfies $$\vec{d} \times \vec{b}=\vec{c} \times \vec{b}$$ and $$\vec{d} \cdot \vec{a}=24$$, then $$|\vec{d}|^{2}$$ is equal to :</p>
Options:
[{"identifier": "A",... | ["B"]
Explanation:
Given that $$\vec{d} \times \vec{b} = \vec{c} \times \vec{b}$$, we can rewrite this as:
<br/><br/>$$(\vec{d} - \vec{c}) \times \vec{b} = \vec{0}$$
<br/><br/>This implies that the vector $$\vec{d} - \vec{c}$$ is a scalar multiple of $$\vec{b}$$:
<br/><br/>$$\vec{d} = \vec{c} + \lambda \vec{b}$$
<... |
<p>Let $$\vec{a}=3 \hat{i}+\hat{j}-\hat{k}$$ and $$\vec{c}=2 \hat{i}-3 \hat{j}+3 \hat{k}$$. If $$\vec{b}$$ is a vector such that $$\vec{a}=\vec{b} \times \vec{c}$$ and $$|\vec{b}|^{2}=50$$, then $$|72-| \vec{b}+\left.\vec{c}\right|^{2} \mid$$ is equal to __________.</p>
Options:
[] | 66
Explanation:
<p>Given that $$\vec{a} = \vec{b} \times \vec{c}$$, we can find the magnitudes of $$\vec{a}$$ and $$\vec{c}$$:</p>
<p>$$|\vec{a}| = \sqrt{3^2 + 1^2 + (-1)^2} = \sqrt{11}$$
<br/><br/>$$|\vec{c}| = \sqrt{2^2 + (-3)^2 + 3^2} = \sqrt{22}$$</p>
<p>We know that the magnitude of the cross product of two vecto... |
<p>Let $$\vec{a}=\hat{i}+2 \hat{j}+3 \hat{k}$$ and $$\vec{b}=\hat{i}+\hat{j}-\hat{k}$$. If $$\vec{c}$$ is a vector such that $$\vec{a} \cdot \vec{c}=11,
\vec{b} \cdot(\vec{a} \times \vec{c})=27$$ and $$\vec{b} \cdot \vec{c}=-\sqrt{3}|\vec{b}|$$, then $$|\vec{a} \times \vec{c}|^{2}$$ is equal to _________.</p>
Options:... | 285
Explanation:
Given,
<br/><br/>$$
\begin{aligned}
& \vec{a}=\hat{i}+2 \hat{j}+3 \hat{k} \\\\
& \vec{b}=\hat{i}+\hat{j}-\hat{k} \\\\
& \vec{a} \cdot \vec{c}=11 \\\\
& \vec{b} \cdot(\vec{a} \times \vec{c})=27 \\\\
& \vec{b} \cdot \vec{c}=-\sqrt{3}|\vec{b}| \\\\
& (\vec{b} \times \vec{a}) \cdot \vec{c}=27
\end{aligne... |
<p>Let $$\vec{a}$$ be a non-zero vector parallel to the line of intersection of the two planes described by $$\hat{i}+\hat{j}, \hat{i}+\hat{k}$$ and $$\hat{i}-\hat{j}, \hat{j}-\hat{k}$$. If $$\theta$$ is the angle between the vector $$\vec{a}$$ and the vector $$\vec{b}=2 \hat{i}-2 \hat{j}+\hat{k}$$ and $$\vec{a} \cdot ... | ["D"]
Explanation:
We have, $$\vec{a}$$ is non-zero vector parallel to the line of intersection of the two planes described by $\hat{\mathbf{i}}+\hat{\mathbf{j}}, \hat{\mathbf{i}}+\hat{\mathbf{k}}$ and $\hat{\mathbf{i}}-\hat{\mathbf{j}}, \hat{\mathbf{j}}-\hat{\mathbf{k}}$.
<br/><br/>Let $\mathbf{n}_1$ and $\mathbf{n}_... |
<p>Let $$\vec{a}=2 \hat{i}+7 \hat{j}-\hat{k}, \vec{b}=3 \hat{i}+5 \hat{k}$$ and $$\vec{c}=\hat{i}-\hat{j}+2 \hat{k}$$. Let $$\vec{d}$$ be a vector which is perpendicular to both $$\vec{a}$$ and $$\vec{b}$$, and $$\vec{c} \cdot \vec{d}=12$$. Then $$(-\hat{i}+\hat{j}-\hat{k}) \cdot(\vec{c} \times \vec{d})$$ is equal to :... | ["C"]
Explanation:
If $\vec{d}$ is $\perp$ to both $\vec{a}$ and $\vec{b}$ then
<br/><br/>$$
\vec{d}=\lambda(\vec{a} \times \vec{b})=\lambda\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
2 & 7 & -1 \\
3 & 0 & 5
\end{array}\right|=(35 \hat{i}-13 \hat{j}-21 \hat{k}) \lambda
$$
<br/><br/>$$
\begin{aligned}
& \te... |
<p>Let O be the origin and the position vector of the point P be $$ - \widehat i - 2\widehat j + 3\widehat k$$. If the position vectors of the points A, B and C are $$ - 2\widehat i + \widehat j - 3\widehat k,2\widehat i + 4\widehat j - 2\widehat k$$ and $$ - 4\widehat i + 2\widehat j - \widehat k$$ respectively, then ... | ["B"]
Explanation:
Given, the position vector of point P is :
$ \overrightarrow{OP} = -\widehat{i} - 2\widehat{j} + 3\widehat{k} $
<br/><br/>Position vectors of points A, B, and C are :
<br/><br/>$ \overrightarrow{OA} = -2\widehat{i} + \widehat{j} - 3\widehat{k} $
<br/><br/>$ \overrightarrow{OB} = 2\widehat{i} + 4\wi... |
<p>The area of the quadrilateral $$\mathrm{ABCD}$$ with vertices $$\mathrm{A}(2,1,1), \mathrm{B}(1,2,5), \mathrm{C}(-2,-3,5)$$ and $$\mathrm{D}(1,-6,-7)$$ is equal to :</p>
Options:
[{"identifier": "A", "content": "48"}, {"identifier": "B", "content": "$$8 \\sqrt{38}$$"}, {"identifier": "C", "content": "54"}, {"identi... | ["B"]
Explanation:
$$
\begin{aligned}
& \text { Here } \overrightarrow{\mathrm{AC}}=(-2-2) \hat{i}+(-3-1) \hat{j}+(5-1) \hat{k} \\\\
& =-4 \hat{i}-4 \hat{j}+4 \hat{k} \\\\
& \overrightarrow{\mathrm{BD}}=(1-1) \hat{i}+(-6-2) \hat{j}+(-7-5) \hat{k} \\\\
& =-8 \hat{j}-12 \hat{k}
\end{aligned}
$$
<br/><br/>So, area of qua... |
<p>Let $$\vec{a}=6 \hat{i}+9 \hat{j}+12 \hat{k}, \vec{b}=\alpha \hat{i}+11 \hat{j}-2 \hat{k}$$ and $$\vec{c}$$ be vectors such that $$\vec{a} \times \vec{c}=\vec{a} \times \vec{b}$$. If <br/><br/>$$\vec{a} \cdot \vec{c}=-12, \vec{c} \cdot(\hat{i}-2 \hat{j}+\hat{k})=5$$, then $$\vec{c} \cdot(\hat{i}+\hat{j}+\hat{k})$$ i... | 11
Explanation:
Let $\vec{c}=c_1 \hat{i}+c_2 \hat{j}+c_3 \hat{k}$
<br/><br/>Now, $\vec{a} \cdot \vec{c}=-12$
<br/><br/>$$
\Rightarrow 6 c_1+9 c_2+12 c_3=-12
$$ ..............(i)
<br/><br/>Also, $\vec{c} \cdot(\hat{i}-2 \hat{j}+\hat{k})=5$
<br/><br/>$$
\Rightarrow c_1-2 c_2+c_3=5
$$ ................(ii)
<br/><br/>$$
\b... |
<p>Let $$\vec{a}=2 \hat{i}+3 \hat{j}+4 \hat{k}, \vec{b}=\hat{i}-2 \hat{j}-2 \hat{k}$$ and $$\vec{c}=-\hat{i}+4 \hat{j}+3 \hat{k}$$.
If $$\vec{d}$$ is a vector perpendicular to both $$\vec{b}$$ and $$\vec{c}$$, and $$\vec{a} \cdot \vec{d}=18$$, then $$|\vec{a} \times \vec{d}|^{2}$$ is equal to :</p>
Options:
[{"identi... | ["B"]
Explanation:
<p>Given vectors :
<br/><br/>$ \vec{a} = 2\hat{i} + 3\hat{j} + 4\hat{k} $
<br/><br/>$ \vec{b} = \hat{i} - 2\hat{j} - 2\hat{k} $
<br/><br/>$ \vec{c} = -\hat{i} + 4\hat{j} + 3\hat{k} $</p>
<p>Since $ \vec{d} $ is perpendicular to both $ \vec{b} $ and $ \vec{c} $, its direction is given by their cross ... |
Let $\overrightarrow{\mathrm{a}}=\hat{i}+\hat{j}+\hat{k}, \overrightarrow{\mathrm{b}}=-\hat{i}-8 \hat{j}+2 \hat{k}$ and $\overrightarrow{\mathrm{c}}=4 \hat{i}+\mathrm{c}_2 \hat{j}+\mathrm{c}_3 \hat{k}$ be three vectors such that $\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{a}}=\overrightarrow{\mathrm{c}}... | 38
Explanation:
$\begin{aligned} & \vec{a}=\hat{i}+\hat{j}+k \\\\ & \vec{b}=\hat{i}+8 \hat{j}+2 k \\\\ & \vec{c}=4 \hat{i}+c_2 \hat{j}+c_3 k \\\\ & \vec{b} \times \vec{a}=\vec{c} \times \vec{a} \\\\ & (\vec{b}-\vec{c}) \times \vec{a}=0 \\\\ & \vec{b}-\vec{c}=\lambda \vec{\alpha} \\\\ & \vec{b}=\vec{c}+\lambda \vec{\al... |
Let $\overrightarrow{\mathrm{a}}=-5 \hat{i}+\hat{j}-3 \hat{k}, \overrightarrow{\mathrm{b}}=\hat{i}+2 \hat{j}-4 \hat{k}$ and
<br/><br/>$\overrightarrow{\mathrm{c}}=(((\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}) \times \hat{i}) \times \hat{i}) \times \hat{i}$. Then $\vec{c} \cdot(-\hat{i}+\hat{j}+\ha... | ["A"]
Explanation:
$\begin{aligned} & \vec{a}=-5 \cdot \hat{i}+\hat{j}-3 \hat{k}, \vec{b}=\hat{i}+2 \hat{j}-4 \hat{k} \\\\ & \vec{c}=(((\vec{a} \times \vec{b}) \times \hat{i}) \times \hat{i}) \times \hat{i} \\\\ & =(((\vec{a} \cdot \hat{i}) \vec{b}-(\vec{b} \cdot \hat{i}) \vec{a}) \times \hat{i}) \times \hat{i} \\\\ &... |
<p>Let the position vectors of the vertices $$\mathrm{A}, \mathrm{B}$$ and $$\mathrm{C}$$ of a triangle be $$2 \hat{i}+2 \hat{j}+\hat{k}, \hat{i}+2 \hat{j}+2 \hat{k}$$ and $$2 \hat{i}+\hat{j}+2 \hat{k}$$ respectively. Let $$l_1, l_2$$ and $$l_3$$ be the lengths of perpendiculars drawn from the ortho center of the trian... | ["D"]
Explanation:
<p>$$\triangle \mathrm{ABC}$$ is equilateral</p>
<p>Orthocentre and centroid will be same</p>
<p>$$\mathrm{G}\left(\frac{5}{3}, \frac{5}{3}, \frac{5}{3}\right)$$</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lt1vv91f/efa66592-d248-4f32-8e31-2416eaa514c4/4caa5d30-d411-11ee-b... |
<p>Let $$\vec{a}=3 \hat{i}+2 \hat{j}+\hat{k}, \vec{b}=2 \hat{i}-\hat{j}+3 \hat{k}$$ and $$\vec{c}$$ be a vector such that $$(\vec{a}+\vec{b}) \times \vec{c}=2(\vec{a} \times \vec{b})+24 \hat{j}-6 \hat{k}$$ and $$(\vec{a}-\vec{b}+\hat{i}) \cdot \vec{c}=-3$$. Then $$|\vec{c}|^2$$ is equal to ________.</p>
Options:
[] | 38
Explanation:
<p>$$\begin{aligned}
& (\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}) \times \overrightarrow{\mathrm{c}}=2(\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}})+24 \hat{\mathrm{j}}-6 \hat{\mathrm{k}} \\
& (5 \hat{\mathrm{i}}+\hat{\mathrm{j}}+4 \hat{\mathrm{k}}) \times \overrightarr... |
<p>Let $$\vec{a}=3 \hat{i}+\hat{j}-2 \hat{k}, \vec{b}=4 \hat{i}+\hat{j}+7 \hat{k}$$ and $$\vec{c}=\hat{i}-3 \hat{j}+4 \hat{k}$$ be three vectors. If a vectors $$\vec{p}$$ satisfies $$\vec{p} \times \vec{b}=\vec{c} \times \vec{b}$$ and $$\vec{p} \cdot \vec{a}=0$$, then $$\vec{p} \cdot(\hat{i}-\hat{j}-\hat{k})$$ is equal... | ["B"]
Explanation:
<p>$$\begin{aligned}
& \overrightarrow{\mathrm{p}} \times \overrightarrow{\mathrm{b}}-\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{b}}=\overrightarrow{0} \\
& (\overrightarrow{\mathrm{p}}-\overrightarrow{\mathrm{c}}) \times \overrightarrow{\mathrm{b}}=\overrightarrow{0} \\
& \overright... |
<p>The distance of the point $$Q(0,2,-2)$$ form the line passing through the point $$P(5,-4, 3)$$ and perpendicular to the lines $$\vec{r}=(-3 \hat{i}+2 \hat{k})+\lambda(2 \hat{i}+3 \hat{j}+5 \hat{k}), \lambda \in \mathbb{R}$$ and $$\vec{r}=(\hat{i}-2 \hat{j}+\hat{k})+\mu(-\hat{i}+3 \hat{j}+2 \hat{k}), \mu \in \mathbb{... | ["A"]
Explanation:
<p>A vector in the direction of the required line can be obtained by cross product of</p>
<p>$$\begin{aligned}
& \left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
2 & 3 & 5 \\
-1 & 3 & 2
\end{array}\right| \\\\
& =-9 \hat{i}-9 \hat{j}+9 \hat{k}
\end{aligned}$$</... |
<p>Let $$\vec{a}$$ and $$\vec{b}$$ be two vectors such that $$|\vec{a}|=1,|\vec{b}|=4$$, and $$\vec{a} \cdot \vec{b}=2$$. If $$\vec{c}=(2 \vec{a} \times \vec{b})-3 \vec{b}$$ and the angle between $$\vec{b}$$ and $$\vec{c}$$ is $$\alpha$$, then $$192 \sin ^2 \alpha$$ is equal to ________.</p>
Options:
[] | 48
Explanation:
<p>$$\begin{aligned}
& \overrightarrow{\mathrm{b}} \cdot \overrightarrow{\mathrm{c}}=(2 \overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}) \cdot \overrightarrow{\mathrm{b}}-3|\mathrm{b}|^2 \\
& |\mathrm{~b}||c| \cos \alpha=-3|\mathrm{~b}|^2 \\
& |\mathrm{c}| \cos \alpha=-12 \text {, as }|\... |
<p>Let $$\overrightarrow{O A}=\vec{a}, \overrightarrow{O B}=12 \vec{a}+4 \vec{b} \text { and } \overrightarrow{O C}=\vec{b}$$, where O is the origin. If S is the parallelogram with adjacent sides OA and OC, then $$\mathrm{{{area\,of\,the\,quadrilateral\,OA\,BC} \over {area\,of\,S}}}$$ is equal to _________.</p>
Option... | ["C"]
Explanation:
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lsr8xnpy/836457b4-b43d-438c-b5ce-783fa67cdc3d/c728ff60-ce37-11ee-9412-cd4f9c6f2c40/file-6y3zli1lsr8xnpz.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/6y3zli1lsr8xnpy/836457b4-b43d-438c-b5ce-783fa67cdc3... |
<p>Let $$\vec{a}=\hat{i}+\alpha \hat{j}+\beta \hat{k}, \alpha, \beta \in \mathbb{R}$$. Let a vector $$\vec{b}$$ be such that the angle between $$\vec{a}$$ and $$\vec{b}$$ is $$\frac{\pi}{4}$$ and $$|\vec{b}|^2=6$$. If $$\vec{a} \cdot \vec{b}=3 \sqrt{2}$$, then the value of $$\left(\alpha^2+\beta^2\right)|\vec{a} \times... | ["B"]
Explanation:
<p>$$\begin{aligned}
& |\overrightarrow{\mathrm{b}}|^2=6 ;|\overrightarrow{\mathrm{a}}||\overrightarrow{\mathrm{b}}| \cos \theta=3 \sqrt{2} \\
& |\overrightarrow{\mathrm{a}}|^2|\overrightarrow{\mathrm{b}}|^2 \cos ^2 \theta=18 \\
& |\overrightarrow{\mathrm{a}}|^2=6
\end{aligned}$$</p>
<p>Also $$1+\al... |
<p>Let $$\vec{a}$$ and $$\vec{b}$$ be two vectors such that $$|\vec{b}|=1$$ and $$|\vec{b} \times \vec{a}|=2$$. Then $$|(\vec{b} \times \vec{a})-\vec{b}|^2$$ is equal to</p>
Options:
[{"identifier": "A", "content": "1"}, {"identifier": "B", "content": "3"}, {"identifier": "C", "content": "5"}, {"identifier": "D", "con... | ["C"]
Explanation:
<p>To find the value of $$|(\vec{b} \times \vec{a})-\vec{b}|^2$$, we can use properties of vector operations and magnitudes. Given $$|\vec{b}| = 1$$ and $$|\vec{b} \times \vec{a}| = 2$$, let's break down the calculation step by step:</p>
<p>Firstly, we observe that the cross product of two vectors $... |
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