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1l54taa8j | maths | vector-algebra | algebra-and-modulus-of-vectors | Let A, B, C be three points whose position vectors respectively are
<p>$$\overrightarrow a = \widehat i + 4\widehat j + 3\widehat k$$</p>
<p>$$\overrightarrow b = 2\widehat i + \alpha \widehat j + 4\widehat k,\,\alpha \in R$$</p>
<p>$$\overrightarrow c = 3\widehat i - 2\widehat j + 5\widehat k$$</p>
<p>If $$\alpha$... | [{"identifier": "A", "content": "$${{\\sqrt {82} } \\over 2}$$"}, {"identifier": "B", "content": "$${{\\sqrt {62} } \\over 2}$$"}, {"identifier": "C", "content": "$${{\\sqrt {69} } \\over 2}$$"}, {"identifier": "D", "content": "$${{\\sqrt {66} } \\over 2}$$"}] | ["A"] | null | $\overrightarrow{A B} \| \overrightarrow{A C}$ if
<br/><br/>
$\frac{1}{2}=\frac{\alpha-4}{-6}=\frac{1}{2}$
<br/><br/>
$\Rightarrow \alpha=1$
<br/><br/>
$\vec{a}, \vec{b}, \vec{c}$ are non-collinear for $\alpha=2$ (smallest positive integer)
<br/><br/>
Mid point of $B C=M\left(\frac{5}{2}, 0, \frac{9}{2}\right)$
<br/><b... | mcq | jee-main-2022-online-29th-june-evening-shift | 4,301 |
1l6m59jzo | maths | vector-algebra | algebra-and-modulus-of-vectors | <p>Let the vectors $$\vec{a}=(1+t) \hat{i}+(1-t) \hat{j}+\hat{k}, \vec{b}=(1-t) \hat{i}+(1+t) \hat{j}+2 \hat{k}$$ and $$\vec{c}=t \hat{i}-t \hat{j}+\hat{k}, t \in \mathbf{R}$$ be such that for $$\alpha, \beta, \gamma \in \mathbf{R}, \alpha \vec{a}+\beta \vec{b}+\gamma \vec{c}=\overrightarrow{0} \Rightarrow \alpha=\beta... | [{"identifier": "A", "content": "a non-empty finite set"}, {"identifier": "B", "content": "equal to $$\\mathbf{N}$$"}, {"identifier": "C", "content": "equal to $$\\mathbf{R}-\\{0\\}$$"}, {"identifier": "D", "content": "equal to $$\\mathbf{R}$$"}] | ["C"] | null | <p>Clearly $$\overrightarrow a $$, $$\overrightarrow b $$, $$\overrightarrow c $$ are non-coplanar</p>
<p>$$\left| {\matrix{
{1 + t} & {1 - t} & 1 \cr
{1 - t} & {1 + t} & 2 \cr
t & { - t} & 1 \cr
} } \right| \ne 0$$</p>
<p>$$ \Rightarrow (1 + t)(1 + t + 2t) - (1 - t)(1 - t - 2t) + 1({t^2} - t - t - {t^... | mcq | jee-main-2022-online-28th-july-morning-shift | 4,302 |
1ldybks1h | maths | vector-algebra | algebra-and-modulus-of-vectors | <p>Let PQR be a triangle. The points A, B and C are on the sides QR, RP and PQ respectively such that <br/><br/>$${{QA} \over {AR}} = {{RB} \over {BP}} = {{PC} \over {CQ}} = {1 \over 2}$$. Then $${{Area(\Delta PQR)} \over {Area(\Delta ABC)}}$$ is equal to :</p> | [{"identifier": "A", "content": "$$\\frac{5}{2}$$"}, {"identifier": "B", "content": "4"}, {"identifier": "C", "content": "2"}, {"identifier": "D", "content": "3"}] | ["D"] | null | Let the position vector of $P, Q, R$ be $\overrightarrow{0}, \overrightarrow{a}, \overrightarrow{b}$
<br><br>$\Rightarrow$ Position vector of $A=\frac{2 \overrightarrow{a}+\overrightarrow{b}}{3}, $
<br><br>Position vector of $B=\frac{2 \overrightarrow{b}}{3}$ and
<br><br>Position vector of $C=\frac{\overrightarrow{a}... | mcq | jee-main-2023-online-24th-january-morning-shift | 4,303 |
lgnwmpwc | maths | vector-algebra | algebra-and-modulus-of-vectors | Let $\mathrm{ABCD}$ be a quadrilateral. If $\mathrm{E}$ and $\mathrm{F}$ are the mid points of the diagonals $\mathrm{AC}$ and $\mathrm{BD}$ respectively and $(\overrightarrow{A B}-\overrightarrow{B C})+(\overrightarrow{A D}-\overrightarrow{D C})=k \overrightarrow{F E}$, then $k$ is equal to : | [{"identifier": "A", "content": "-2"}, {"identifier": "B", "content": "4"}, {"identifier": "C", "content": "-4"}, {"identifier": "D", "content": "2"}] | ["C"] | null | <p>Let the position vectors of $A, B, C,$ and $D$ be $\vec{a}, \vec{b}, \vec{c},$ and $\vec{d}$, respectively.</p>
<p>Then the position vector of $E$ is:</p>
<p>$$
\vec{E} = \frac{\vec{a} + \vec{c}}{2}
$$</p>
<p>And the position vector of $F$ is:</p>
<p>$$
\vec{F} = \frac{\vec{b} + \vec{d}}{2}
$$</p>
<p>Now, we are giv... | mcq | jee-main-2023-online-15th-april-morning-shift | 4,304 |
1lguu8s7l | maths | vector-algebra | algebra-and-modulus-of-vectors | <p>For any vector $$\vec{a}=a_{1} \hat{i}+a_{2} \hat{j}+a_{3} \hat{k}$$, with $$10\left|a_{i}\right|<1, i=1,2,3$$, consider the following statements :</p>
<p>(A): $$\max \left\{\left|a_{1}\right|,\left|a_{2}\right|,\left|a_{3}\right|\right\} \leq|\vec{a}|$$</p>
<p>(B) : $$|\vec{a}| \leq 3 \max \left\{\left|a_{1}\rig... | [{"identifier": "A", "content": "Only (B) is true"}, {"identifier": "B", "content": "Only (A) is true"}, {"identifier": "C", "content": "Neither (A) nor (B) is true"}, {"identifier": "D", "content": "Both (A) and (B) are true"}] | ["D"] | null | We have,
<br/><br/>$$
\begin{aligned}
& 10\left|a_i\right|<1, i=1,2,3 \\\\
& \text { Let } \left|a_1\right| \geq\left|a_2\right| \geq\left|a_3\right| \\\\
& |\vec{a}|=\sqrt{a_1^2+a_2^2+a_3^2} \geq \sqrt{a_1^2} \\\\
& \therefore|\vec{a}| \geq\left|a_1\right| \text { or } \max \left\{\left|a_1\right|,\left|a_2\right|,\le... | mcq | jee-main-2023-online-11th-april-morning-shift | 4,305 |
1lgvq2r23 | maths | vector-algebra | algebra-and-modulus-of-vectors | <p>If the points $$\mathrm{P}$$ and $$\mathrm{Q}$$ are respectively the circumcenter and the orthocentre of a $$\triangle \mathrm{ABC}$$, then $$\overrightarrow{\mathrm{PA}}+\overrightarrow{\mathrm{PB}}+\overrightarrow{\mathrm{PC}}$$ is
equal to :</p> | [{"identifier": "A", "content": "$$\\overrightarrow {QP} $$"}, {"identifier": "B", "content": "$$\\overrightarrow {PQ} $$"}, {"identifier": "C", "content": "$$2\\overrightarrow {PQ} $$"}, {"identifier": "D", "content": "$$2\\overrightarrow {QP} $$"}] | ["B"] | null | 1. **Circumcenter $ P $**:
<br/><br/>The circumcenter of a triangle is equidistant from the vertices of the triangle. It is the center of the circumcircle, the circle that passes through all three vertices of the triangle.
<br/><br/>2. **Orthocenter $ Q $**:
<br/><br/>The orthocenter of a triangle is the point of in... | mcq | jee-main-2023-online-10th-april-evening-shift | 4,306 |
1lgxt48t5 | maths | vector-algebra | algebra-and-modulus-of-vectors | <p>An arc PQ of a circle subtends a right angle at its centre O. The mid point of the arc PQ is R. If $$\overrightarrow {OP} = \overrightarrow u ,\overrightarrow {OR} = \overrightarrow v $$, and $$\overrightarrow {OQ} = \alpha \overrightarrow u + \beta \overrightarrow v $$, then $$\alpha ,{\beta ^2}$$ are the roots... | [{"identifier": "A", "content": "$${x^2} + x - 2 = 0$$"}, {"identifier": "B", "content": "$$3{x^2} + 2x - 1 = 0$$"}, {"identifier": "C", "content": "$$3{x^2} - 2x - 1 = 0$$"}, {"identifier": "D", "content": "$${x^2} - x - 2 = 0$$"}] | ["D"] | null | An arc $P Q$ of a circle subtends a right angle at its centre $O$. The mid-point of an $\operatorname{arc} P Q$ is $R$. So, $P R=R Q$
<br><br>Also given that, $$\overrightarrow {OP} = \overrightarrow u ,\overrightarrow {OR} = \overrightarrow v $$, and $$\overrightarrow {OQ} = \alpha \overrightarrow u + \beta \over... | mcq | jee-main-2023-online-10th-april-morning-shift | 4,307 |
1lgzzzo1k | maths | vector-algebra | algebra-and-modulus-of-vectors | <p>If the points with position vectors $$\alpha \hat{i}+10 \hat{j}+13 \hat{k}, 6 \hat{i}+11 \hat{j}+11 \hat{k}, \frac{9}{2} \hat{i}+\beta \hat{j}-8 \hat{k}$$ are collinear, then $$(19 \alpha-6 \beta)^{2}$$ is equal to :</p> | [{"identifier": "A", "content": "16"}, {"identifier": "B", "content": "49"}, {"identifier": "C", "content": "36"}, {"identifier": "D", "content": "25"}] | ["C"] | null | Given : Points with position vectors
<br/><br/>$$
\alpha \hat{i}+10 \hat{j}+13 \hat{k}, 6 \hat{i}+11 \hat{j}+11 \hat{k}
$$
<br/><br/>and $\frac{9}{2} \hat{i}+\beta \hat{j}-8 \hat{k}$ are collinear.
<br/><br/>So, $\frac{\alpha-6}{6-\frac{9}{2}}=\frac{10-11}{11-\beta}=\frac{13-11}{11+8}$
<br/><br/>$$
\begin{aligned}
& \R... | mcq | jee-main-2023-online-8th-april-morning-shift | 4,308 |
jaoe38c1lsco116r | maths | vector-algebra | algebra-and-modulus-of-vectors | <p>The position vectors of the vertices $$\mathrm{A}, \mathrm{B}$$ and $$\mathrm{C}$$ of a triangle are $$2 \hat{i}-3 \hat{j}+3 \hat{k}, 2 \hat{i}+2 \hat{j}+3 \hat{k}$$ and $$-\hat{i}+\hat{j}+3 \hat{k}$$ respectively. Let $$l$$ denotes the length of the angle bisector $$\mathrm{AD}$$ of $$\angle \mathrm{BAC}$$ where $$... | [{"identifier": "A", "content": "45"}, {"identifier": "B", "content": "50"}, {"identifier": "C", "content": "42"}, {"identifier": "D", "content": "49"}] | ["A"] | null | <p>$$\begin{aligned}
& \mathrm{AB}=5 \\
& \mathrm{AC}=5
\end{aligned}$$</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lt1v5c9e/0c2c210a-ab17-46c0-813d-6d826d2adf6f/7c1dc320-d40e-11ee-b9d5-0585032231f0/file-1lt1v5c9f.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image... | mcq | jee-main-2024-online-27th-january-evening-shift | 4,309 |
jaoe38c1lseyalpj | maths | vector-algebra | algebra-and-modulus-of-vectors | <p>Let $$\vec{a}, \vec{b}$$ and $$\vec{c}$$ be three non-zero vectors such that $$\vec{b}$$ and $$\vec{c}$$ are non-collinear. If $$\vec{a}+5 \vec{b}$$ is collinear with $$\vec{c}, \vec{b}+6 \vec{c}$$ is collinear with $$\vec{a}$$ and $$\vec{a}+\alpha \vec{b}+\beta \vec{c}=\overrightarrow{0}$$, then $$\alpha+\beta$$ is... | [{"identifier": "A", "content": "30"}, {"identifier": "B", "content": "$$-$$30"}, {"identifier": "C", "content": "$$-$$25"}, {"identifier": "D", "content": "35"}] | ["D"] | null | <p>$$\begin{aligned}
& \vec{a}+5 \vec{b}=\lambda \vec{c} \\
& \vec{b}+6 \vec{c}=\mu \vec{a}
\end{aligned}$$</p>
<p>Eliminating $$\vec{a}$$</p>
<p>$$\begin{aligned}
& \lambda \overrightarrow{\mathrm{c}}-5 \overrightarrow{\mathrm{b}}=\frac{6}{\mu} \overrightarrow{\mathrm{c}}+\frac{1}{\mu} \overrightarrow{\mathrm{b}} \\
&... | mcq | jee-main-2024-online-29th-january-morning-shift | 4,310 |
sQ5ibg4HiGWlSCmT | maths | vector-algebra | scalar-and-vector-triple-product | If $$\overrightarrow a \,\,,\,\,\overrightarrow b \,\,,\,\,\overrightarrow c $$ are vectors such that $$\left[ {\overrightarrow a \,\overrightarrow b \,\overrightarrow c } \right] = 4$$ then $$\left[ {\overrightarrow a \, \times \overrightarrow b \,\,\overrightarrow b \times \,\overrightarrow c \,\,\overrightarrow c \... | [{"identifier": "A", "content": "$$16$$ "}, {"identifier": "B", "content": "$$64$$ "}, {"identifier": "C", "content": "$$4$$ "}, {"identifier": "D", "content": "$$8$$ "}] | ["A"] | null | We have, $$\left[ {\overrightarrow a \times \overrightarrow b \,\,\overrightarrow b \times \overrightarrow c \,\,\overrightarrow c \times \overrightarrow a } \right]$$
<br><br>$$ = \left( {\overrightarrow a \times \overrightarrow b } \right).\,\,\left\{ {\left( {\overrightarrow b \times \overrightarrow c } \right)... | mcq | aieee-2002 | 4,311 |
D4Bcevo00Vrh7Gww | maths | vector-algebra | scalar-and-vector-triple-product | If $$\overrightarrow u \,,\overrightarrow v $$ and $$\overrightarrow w $$ are three non-coplanar vectors, then $$\,\left( {\overrightarrow u + \overrightarrow v - \overrightarrow w } \right).\left( {\overrightarrow u - \overrightarrow v } \right) \times \left( {\overrightarrow v - \overrightarrow w} \right)$$ equal... | [{"identifier": "A", "content": "$$3\\overrightarrow u .\\overrightarrow v \\times \\overrightarrow w $$ "}, {"identifier": "B", "content": "$$0$$"}, {"identifier": "C", "content": "$$\\overrightarrow u .\\overrightarrow v \\times \\overrightarrow w $$ "}, {"identifier": "D", "content": "$$\\overrightarrow u .\\overr... | ["C"] | null | $$\left( {\overrightarrow u + \overrightarrow v - \overrightarrow w } \right).\left( {\overrightarrow u \times \overrightarrow v - \overrightarrow u \times \overrightarrow w - \overrightarrow v \times \overrightarrow v + \overrightarrow v \times \overrightarrow w } \right)$$
<br><br>$$ = \left( {\overrightarro... | mcq | aieee-2003 | 4,312 |
jZuo6aOIGIMwAj1z | maths | vector-algebra | scalar-and-vector-triple-product | If $${\overrightarrow a ,\overrightarrow b ,\overrightarrow c }$$ are non-coplanar vectors and $$\lambda $$ is a real number, then the vectors $${\overrightarrow a + 2\overrightarrow b + 3\overrightarrow c ,\,\,\lambda \overrightarrow b + 4\overrightarrow c }$$ and $$\left( {2\lambda - 1} \right)\overrightarrow c $... | [{"identifier": "A", "content": "no value of $$\\lambda $$ "}, {"identifier": "B", "content": "all except one value of $$\\lambda $$ "}, {"identifier": "C", "content": "all except two values of $$\\lambda $$ "}, {"identifier": "D", "content": "all values of $$\\lambda $$ "}] | ["C"] | null | Vectors $$\overrightarrow a + 2\overrightarrow b + 3\overrightarrow c ,\lambda \overrightarrow b + 4\overrightarrow c ,\,\,\,$$ <br><br>and $$\left( {2\lambda - 1} \right)\overrightarrow c $$ are
<br><br>coplanar if $$\left| {\matrix{
1 & 2 & 3 \cr
0 & \lambda & 4 \cr
0 & 0 & ... | mcq | aieee-2004 | 4,313 |
BRq9yBuutoyT3OZ3 | maths | vector-algebra | scalar-and-vector-triple-product | Let $$\overrightarrow a \,\, = \,\,\widehat i - \widehat k,\,\,\,\,\,\overrightarrow b \,\,\, = \,\,\,x\widehat i + \widehat j\,\,\, + \,\,\,\left( {1 - x} \right)\widehat k$$ and $$\overrightarrow c \,\, = \,\,y\widehat i + x\widehat j + \left( {1 + x - y} \right)\widehat k.$$ Then $$\left[ {\overrightarrow a ,\overri... | [{"identifier": "A", "content": "only $$y$$ "}, {"identifier": "B", "content": "only $$x$$ "}, {"identifier": "C", "content": "both $$x$$ and $$y$$ "}, {"identifier": "D", "content": "neither $$x$$ nor $$y$$ "}] | ["D"] | null | $$\overrightarrow a = \widehat j - \widehat k,\overrightarrow b = x\widehat i + \overrightarrow j + \left( {1 - x} \right)\widehat k$$
<br><br>and $$\overrightarrow c = y\widehat i + x\widehat j + \left( {1 + x - y} \right)\widehat k$$
<br><br>$$\left[ {\overrightarrow a \,\overrightarrow b \,\overrightarrow c } \... | mcq | aieee-2005 | 4,314 |
uM4lvmwUCURnQ18y | maths | vector-algebra | scalar-and-vector-triple-product | If $$\overrightarrow a ,\overrightarrow b ,\overrightarrow c $$ are non coplanar vectors and $$\lambda $$ is a real number then <br/><br/>$$\left[ {\lambda \left( {\overrightarrow a + \overrightarrow b } \right)\,\,\,\,\,\,\,\,{\lambda ^2}\overrightarrow b \,\,\,\,\,\,\,\,\lambda \overrightarrow c } \right] = \left[ {... | [{"identifier": "A", "content": "exactly one value of $$\\lambda $$ "}, {"identifier": "B", "content": "no value of $$\\lambda $$"}, {"identifier": "C", "content": "exactly three values of $$\\lambda $$"}, {"identifier": "D", "content": "exactly two values of $$\\lambda $$"}] | ["B"] | null | $$\left[ {\lambda \left( {\overrightarrow a + \overrightarrow b } \right){\lambda ^2}\overrightarrow b \,\,\,\lambda \overrightarrow c } \right] = \left[ {\overrightarrow a \,\,\overrightarrow b + \overrightarrow c \,\,\overrightarrow b } \right]$$
<br><br>$$ \Rightarrow {\lambda ^4}\left[ {\overrightarrow a + \over... | mcq | aieee-2005 | 4,315 |
zsaWIZzGBykVkw8A | maths | vector-algebra | scalar-and-vector-triple-product | If $$\left( {\overrightarrow a \times \overrightarrow b } \right) \times \overrightarrow c = \overrightarrow a \times \left( {\overrightarrow b \times \overrightarrow c } \right)$$ where $${\overrightarrow a ,\overrightarrow b }$$ and $${\overrightarrow c }$$ are any three vectors such that $$\overrightarrow a .\ov... | [{"identifier": "A", "content": "inclined at an angle of $${\\pi \\over 3}$$ between them "}, {"identifier": "B", "content": "inclined at an angle of $${\\pi \\over 6}$$ between them "}, {"identifier": "C", "content": "perpendicular "}, {"identifier": "D", "content": "parallel "}] | ["D"] | null | $$\left( {\overrightarrow a \times \overrightarrow b } \right) \times \overrightarrow c $$
<br><br>$$\,\,\,\,\,\,\,\,\,\,\,\,\, = \overrightarrow a \times \left( {\overrightarrow b \times \overrightarrow c } \right),\overrightarrow a .\overrightarrow b \ne 0,\,\,\overrightarrow b .\overrightarrow c \ne 0$$
<br><br... | mcq | aieee-2006 | 4,316 |
lWNMlgYnWH2QteNt | maths | vector-algebra | scalar-and-vector-triple-product | Let $$\overrightarrow a = \widehat i + \widehat j + \widehat k,\overrightarrow b = \widehat i - \widehat j + 2\widehat k$$ and $$\overrightarrow c = x\widehat i + \left( {x - 2} \right)\widehat j - \widehat k\,\,.$$ If the vectors $$\overrightarrow c $$ lies in the plane of $$\overrightarrow a $$ and $$\overrightar... | [{"identifier": "A", "content": "$$-4$$ "}, {"identifier": "B", "content": "$$-2$$"}, {"identifier": "C", "content": "$$0$$ "}, {"identifier": "D", "content": "$$1.$$"}] | ["B"] | null | Given $$\overrightarrow a = \widehat i + \widehat j + \widehat k,\overrightarrow b = \widehat i - \widehat j + 2\widehat k$$
<br><br>and $$\overrightarrow c = x\widehat i + \left( {x - 2} \right)\widehat j - \widehat k$$
<br><br>If $$\overrightarrow c $$ lies in the plane of $$\overrightarrow a $$ and $$\overrighta... | mcq | aieee-2007 | 4,317 |
QgpDQ7Vm5ZN1ORQI | maths | vector-algebra | scalar-and-vector-triple-product | The vector $$\overrightarrow a = \alpha \widehat i + 2\widehat j + \beta \widehat k$$ lies in the plane of the vectors
<br/>$$\overrightarrow b = \widehat i + \widehat j$$ and $$\overrightarrow c = \widehat j + \widehat k$$ and bisects the angle between $$\overrightarrow b $$ and $$\overrightarrow c $$.Then which ... | [{"identifier": "A", "content": "$$\\alpha = 2,\\,\\,\\beta = 2$$ "}, {"identifier": "B", "content": "$$\\alpha = 1,\\,\\,\\beta = 2$$"}, {"identifier": "C", "content": "$$\\alpha = 2,\\,\\,\\beta = 1$$"}, {"identifier": "D", "content": "$$\\alpha = 1,\\,\\,\\beta = 1$$"}] | ["D"] | null | As $$\overrightarrow a $$ lies in the plane of $$\overrightarrow b $$ and $$\overrightarrow c $$
<br><br>$$\therefore$$ $$\overrightarrow a = \overrightarrow b + \lambda \overrightarrow c $$
<br><br>$$ \Rightarrow \alpha \widehat i + 2\widehat j + \beta \widehat k = \widehat i + \widehat j + \lambda \left( {\widehat ... | mcq | aieee-2008 | 4,318 |
hZS5SQ5907JAkQoO | maths | vector-algebra | scalar-and-vector-triple-product | If $$\overrightarrow u ,\overrightarrow v ,\overrightarrow w $$ are non-coplanar vectors and $$p,q$$ are real numbers, then the equality $$\left[ {3\overrightarrow u \,\,p\overrightarrow v \,\,p\overrightarrow w } \right] - \left[ {p\overrightarrow v \,\,\overrightarrow w \,\,q\overrightarrow u } \right] - \left[ {2\ov... | [{"identifier": "A", "content": "exactly two values of $$(p,q)$$"}, {"identifier": "B", "content": "more than two but not all values of $$(p,q)$$ "}, {"identifier": "C", "content": "all values of $$(p,q)$$ "}, {"identifier": "D", "content": "exactly one value of $$(p,q)$$"}] | ["D"] | null | $$\left[ {3\overrightarrow u \,\,p\overrightarrow v \,\,p\overrightarrow \omega } \right] - \left[ {p\overrightarrow v \,\,\overrightarrow \omega \,\,q\overrightarrow u } \right] - \left[ {2\overrightarrow \omega \,\,q\overrightarrow v \,\,q\overrightarrow u } \right] = 0$$
<br><br>$$ \Rightarrow \left( {3{p^2} - pq... | mcq | aieee-2009 | 4,319 |
FVPtJqT2k8NUVozx | maths | vector-algebra | scalar-and-vector-triple-product | Let $$\overrightarrow a = \widehat j - \widehat k$$ and $$\overrightarrow c = \widehat i - \widehat j - \widehat k.$$ Then the vector $$\overrightarrow b $$ satisfying $$\overrightarrow a \times \overrightarrow b + \overrightarrow c = \overrightarrow 0 $$ and $$\overrightarrow a .\overrightarrow b = 3$$ : | [{"identifier": "A", "content": "$$2\\widehat i - \\widehat j + 2\\widehat k$$ "}, {"identifier": "B", "content": "$$\\widehat i - \\widehat j - 2\\widehat k$$"}, {"identifier": "C", "content": "$$\\widehat i + \\widehat j - 2\\widehat k$$"}, {"identifier": "D", "content": "$$-\\widehat i +\\widehat j - 2\\widehat k$$"... | ["D"] | null | $$\overrightarrow c = \overrightarrow b \times \overrightarrow a $$
<br><br>$$ \Rightarrow \overrightarrow b .\overrightarrow c = \overrightarrow b .\left( {\overrightarrow b \times \overrightarrow a } \right) \Rightarrow \overrightarrow b .\overrightarrow c = 0$$
<br><br>$$ \Rightarrow \left( {{b_1}\widehat i + {... | mcq | aieee-2010 | 4,320 |
tPY4NnwJZCoCQIE5 | maths | vector-algebra | scalar-and-vector-triple-product | If $$\overrightarrow a = {1 \over {\sqrt {10} }}\left( {3\widehat i + \widehat k} \right)$$ and $$\overrightarrow b = {1 \over 7}\left( {2\widehat i + 3\widehat j - 6\widehat k} \right),$$ then the value
<br/><br>of $$\left( {2\overrightarrow a - \overrightarrow b } \right)\left[ {\left( {\overrightarrow a \times ... | [{"identifier": "A", "content": "$$-3$$ "}, {"identifier": "B", "content": "$$5$$ "}, {"identifier": "C", "content": "$$3$$ "}, {"identifier": "D", "content": "$$-5$$ "}] | ["D"] | null | We have $$\overrightarrow a .\overrightarrow b = 0,\,\,\overrightarrow a .\overrightarrow a = 1,\,\,\overrightarrow b .\overrightarrow b = 1$$
<br><br>$$\left( {2\overrightarrow a - \overrightarrow b } \right).\left[ {\left( {\overrightarrow a \times \overrightarrow b } \right) \times \left( {\overrightarrow a + ... | mcq | aieee-2011 | 4,321 |
fnnbNmLCR5jL4JWe | maths | vector-algebra | scalar-and-vector-triple-product | If $$\left[ {\overrightarrow a \times \overrightarrow b \,\,\,\,\overrightarrow b \times \overrightarrow c \,\,\,\,\overrightarrow c \times \overrightarrow a } \right] = \lambda {\left[ {\overrightarrow a\,\,\,\,\,\,\,\, \overrightarrow b \,\,\,\,\,\,\,\,\overrightarrow c } \right]^2}$$ then $$\lambda $$ is equal to... | [{"identifier": "A", "content": "$$0$$ "}, {"identifier": "B", "content": "$$1$$"}, {"identifier": "C", "content": "$$2$$"}, {"identifier": "D", "content": "$$3$$"}] | ["B"] | null | $$L.H.S$$ $$ = \left( {\overrightarrow a \times \overrightarrow b } \right).\left[ {\left( {\overrightarrow b \times \overrightarrow c } \right) \times \left( {\overrightarrow c \times \overrightarrow a } \right)} \right]$$
<br><br>$$ = \left( {\overrightarrow a \times \overrightarrow b } \right).\left[ {\left( {\o... | mcq | jee-main-2014-offline | 4,322 |
GHozmbvUXpiFybyD | maths | vector-algebra | scalar-and-vector-triple-product | Let $$\overrightarrow a ,\overrightarrow b $$ and $$\overrightarrow c $$ be three non-zero vectors such that no two of them are collinear and <br/><br/>$$\left( {\overrightarrow a \times \overrightarrow b } \right) \times \overrightarrow c = {1 \over 3}\left| {\overrightarrow b } \right|\left| {\overrightarrow c } \r... | [{"identifier": "A", "content": "$${2 \\over 3}$$"}, {"identifier": "B", "content": "$${{ - 2\\sqrt 3 } \\over 3}$$ "}, {"identifier": "C", "content": "$${{ 2\\sqrt 2 } \\over 3}$$"}, {"identifier": "D", "content": "$${{ - \\sqrt 2 } \\over 3}$$ "}] | ["C"] | null | $$\left( {\overrightarrow a \times \overrightarrow b } \right) \times \overrightarrow c = {1 \over 3}\left| {\overrightarrow b } \right|\left| {\overrightarrow c } \right|\overrightarrow a $$
<br><br>$$ \Rightarrow - \overrightarrow c \times \left( {\overrightarrow a \times \overrightarrow b } \right) = {1 \over 3... | mcq | jee-main-2015-offline | 4,323 |
6uIl4ZYYXZz8Y3pd | maths | vector-algebra | scalar-and-vector-triple-product | Let $$\overrightarrow a ,\overrightarrow b $$ and $$\overrightarrow c $$ be three unit vectors such that $$\overrightarrow a \times \left( {\overrightarrow b \times \overrightarrow c } \right) = {{\sqrt 3 } \over 2}\left( {\overrightarrow b + \overrightarrow c } \right).$$ If $${\overrightarrow b }$$ is not parallel... | [{"identifier": "A", "content": "$${{2\\pi } \\over 3}$$ "}, {"identifier": "B", "content": "$${{5\\pi } \\over 6}$$"}, {"identifier": "C", "content": "$${{3\\pi } \\over 4}$$"}, {"identifier": "D", "content": "$${{\\pi } \\over 2}$$"}] | ["B"] | null | $$\overrightarrow a \times \left( {\overrightarrow b \times \overrightarrow c } \right) = {{\sqrt 3 } \over 2}\left( {\overrightarrow b + \overrightarrow c } \right)$$
<br><br>$$ \Rightarrow \left( {\overrightarrow a .\overrightarrow c } \right)\overrightarrow b - \left( {\overrightarrow a .\overrightarrow b } \rig... | mcq | jee-main-2016-offline | 4,324 |
EtDY30nGqeWkJ7RGPLaJz | maths | vector-algebra | scalar-and-vector-triple-product | Let $$\overrightarrow a $$, $$\overrightarrow b $$ and $$\overrightarrow c $$ be three unit vectors, out of which vectors $$\overrightarrow b $$ and $$\overrightarrow c $$ are non-parallel. If $$\alpha $$ and $$\beta $$ are the angles which vector $$\overrightarrow a $$ makes with vectors $$\overrightarrow b $$ and $$... | [{"identifier": "A", "content": "90<sup>o</sup>"}, {"identifier": "B", "content": "30<sup>o</sup>"}, {"identifier": "C", "content": "45<sup>o</sup>"}, {"identifier": "D", "content": "60<sup>o</sup>"}] | ["B"] | null | $$\left( {\overrightarrow a .\overrightarrow c } \right)\overrightarrow b - \left( {\overrightarrow a .\overrightarrow b } \right).\overrightarrow c = {1 \over 2}\overrightarrow b $$
<br><br>$$ \because $$ $$\overrightarrow b \,\,$$ & $$\overrightarrow c \,\,$$ are linearly independent
<br><br>$$ \th... | mcq | jee-main-2019-online-12th-january-evening-slot | 4,325 |
K0mbLDiKg5frblM18B3rsa0w2w9jxaz0egd | maths | vector-algebra | scalar-and-vector-triple-product | Let $$\alpha $$ $$ \in $$ R and the three vectors <br/><br/>$$\overrightarrow a = \alpha \widehat i + \widehat j + 3\widehat k$$, $$\overrightarrow b = 2\widehat i + \widehat j - \alpha \widehat k$$ <br/><br/>and $$\overrightarrow c = \alpha \widehat i - 2\widehat j + 3\widehat k$$. <br/><br/>Then the set
S = {$$\al... | [{"identifier": "A", "content": "contains exactly two numbers only one of which is positive"}, {"identifier": "B", "content": "is singleton"}, {"identifier": "C", "content": "contains exactly two positive numbers"}, {"identifier": "D", "content": "is empty"}] | ["D"] | null | Since these vectors are coplanar then,<br><br>
$$\left| {\matrix{
\alpha & 1 & 3 \cr
2 & 1 & { - \alpha } \cr
\alpha & { - 2} & 3 \cr
} } \right| = 0$$<br><br>
Now, $$\alpha (3 - 2\alpha ) - 1\left( {6 + {\alpha ^2}} \right) + 3\left( { - 4 - \alpha } \right) = 0$$<br><br>
$$ ... | mcq | jee-main-2019-online-12th-april-evening-slot | 4,326 |
0v0x2Efs8pBEclG5EP3rsa0w2w9jx6glnbj | maths | vector-algebra | scalar-and-vector-triple-product | If the volume of parallelopiped formed by the vectors $$\widehat i + \lambda \widehat j + \widehat k$$, $$\widehat j + \lambda \widehat k$$ and $$\lambda \widehat i + \widehat k$$ is minimum, then $$\lambda $$ is
equal to : | [{"identifier": "A", "content": "$$ - {1 \\over {\\sqrt 3 }}$$"}, {"identifier": "B", "content": "$${\\sqrt 3 }$$"}, {"identifier": "C", "content": "$$-{\\sqrt 3 }$$"}, {"identifier": "D", "content": "$$ {1 \\over {\\sqrt 3 }}$$"}] | ["D"] | null | $$V = \left[ {\overrightarrow a \overrightarrow b \overrightarrow c } \right] = \left| {\matrix{
1 & \lambda & 1 \cr
0 & 1 & \lambda \cr
\lambda & 0 & 1 \cr
} } \right|$$<br><br>
$$ \Rightarrow 1 - \lambda \left( { - {\lambda ^2}} \right) + 1.\left( {0 - \lambda } \right) = {... | mcq | jee-main-2019-online-12th-april-morning-slot | 4,327 |
XR9wDoMoZhk3UqVQLWL6Q | maths | vector-algebra | scalar-and-vector-triple-product | The sum of the distinct real values of $$\mu $$, for which the vectors, $$\mu \widehat i + \widehat j + \widehat k,$$ $$\widehat i + \mu \widehat j + \widehat k,$$ $$\widehat i + \widehat j + \mu \widehat k$$ are co-planar, is : | [{"identifier": "A", "content": "2"}, {"identifier": "B", "content": "$$-$$1"}, {"identifier": "C", "content": "0"}, {"identifier": "D", "content": "1"}] | ["B"] | null | $$\left| {\matrix{
\mu & 1 & 1 \cr
1 & \mu & 1 \cr
1 & 1 & \mu \cr
} } \right| = 0$$
<br><br>$$\mu \left( {{\mu ^2} - 1} \right) - 1\left( {\mu - 1} \right) + 1\left( {1 - \mu } \right) = 0$$
<br><br>$${\mu ^3} - \mu - \mu + 1 + 1\mu = 0$$
<br><br>$${\mu ^3} - 3\mu + 2 =... | mcq | jee-main-2019-online-12th-january-morning-slot | 4,328 |
NaM2UpfKHhYSEwnbzsWn0 | maths | vector-algebra | scalar-and-vector-triple-product | Let $$\overrightarrow a $$ = $$\widehat i - \widehat j$$, $$\overrightarrow b $$ = $$\widehat i + \widehat j + \widehat k$$ and $$\overrightarrow c $$
<br/><br/>be a vector such that $$\overrightarrow a $$ × $$\overrightarrow c $$ + $$\overrightarrow b $$ = $$\overrightarrow 0 $$ <br/><br/>and $$\overrightarrow a $$ .... | [{"identifier": "A", "content": "8"}, {"identifier": "B", "content": "$$19 \\over 2$$"}, {"identifier": "C", "content": "9"}, {"identifier": "D", "content": "$$17 \\over 2$$"}] | ["B"] | null | Given that,
<br><br>$$\overrightarrow a \times \overrightarrow c + \overrightarrow b = \overrightarrow 0 $$
<br><br>$$ \Rightarrow $$ $$\overrightarrow a \times \left( {\overrightarrow a \times \overrightarrow c } \right) + \overrightarrow a \times \overrightarrow b = \overrightarrow 0 $$
<br><br>$$ ... | mcq | jee-main-2019-online-9th-january-morning-slot | 4,329 |
l8CGSsy1iXuyJWw16U7k9k2k5gpqnac | maths | vector-algebra | scalar-and-vector-triple-product | Let the volume of a parallelopiped whose
coterminous edges are given by
<br/><br>$$\overrightarrow u = \widehat i + \widehat j + \lambda \widehat k$$, $$\overrightarrow v = \widehat i + \widehat j + 3\widehat k$$ and
<br/><br>$$\overrightarrow w = 2\widehat i + \widehat j + \widehat k$$ be 1 cu. unit. If $$\theta $$... | [{"identifier": "A", "content": "$${7 \\over {6\\sqrt 3 }}$$"}, {"identifier": "B", "content": "$${7 \\over {6\\sqrt 6 }}$$"}, {"identifier": "C", "content": "$${5 \\over 7}$$"}, {"identifier": "D", "content": "$${5 \\over {3\\sqrt 3 }}$$"}] | ["A"] | null | Volume of parallelopiped = 1
<br><br>$$\left| {\left[ {\matrix{
{\overrightarrow u } & {\overrightarrow v } & {\overrightarrow w } \cr
} } \right]} \right|$$ = 1
<br><br>$$ \Rightarrow $$ $$\left| {\matrix{
1 & 1 & \lambda \cr
1 & 1 & 3 \cr
2 & 1 & 1 \cr
} } \rig... | mcq | jee-main-2020-online-8th-january-morning-slot | 4,330 |
uH7b90Y6mDcfmjaFSn7k9k2k5iu6y6w | maths | vector-algebra | scalar-and-vector-triple-product | If the vectors, $$\overrightarrow p = \left( {a + 1} \right)\widehat i + a\widehat j + a\widehat k$$,
<br/><br>
$$\overrightarrow q = a\widehat i + \left( {a + 1} \right)\widehat j + a\widehat k$$ and
<br/><br>
$$\overrightarrow r = a\widehat i + a\widehat j + \left( {a + 1} \right)\widehat k\left( {a \in R} \right)... | [] | null | 1 | $$ \because $$ $$\overrightarrow p$$, $$\overrightarrow q$$, $$\overrightarrow r$$ are coplanar
<br><br>$$ \therefore $$ $$\left[ {\matrix{
{\overrightarrow p } & {\overrightarrow q } & {\overrightarrow r } \cr
} } \right]$$ = 0
<br><br>$$ \Rightarrow $$ $$\left| {\matrix{
{a + 1} & a & a \cr... | integer | jee-main-2020-online-9th-january-morning-slot | 4,331 |
9QNY5X8lrChVbAhY95jgy2xukf7gthwr | maths | vector-algebra | scalar-and-vector-triple-product | Let x<sub>0</sub> be the point of Local maxima of $$f(x) = \overrightarrow a .\left( {\overrightarrow b \times \overrightarrow c } \right)$$, where <br/>$$\overrightarrow a = x\widehat i - 2\widehat j + 3\widehat k$$, $$\overrightarrow b = - 2\widehat i + x\widehat j - \widehat k$$, $$\overrightarrow c = 7\widehat... | [{"identifier": "A", "content": "14"}, {"identifier": "B", "content": "-30"}, {"identifier": "C", "content": "-4"}, {"identifier": "D", "content": "-22"}] | ["D"] | null | $$f(x) = \overrightarrow a \,.\,(\overrightarrow b \times \overrightarrow c )$$<br><br>$$ = \left[ {\overrightarrow a \,\overrightarrow b \,\overrightarrow c } \right]$$<br><br>$$ = \left| {\matrix{
x & { - 2} & 3 \cr
{ - 2} & x & { - 1} \cr
7 & { - 2} & x \cr
} } \right|$$<br>... | mcq | jee-main-2020-online-4th-september-morning-slot | 4,332 |
WGSrrJfn4qBtQkI2kKjgy2xukfg6dkb0 | maths | vector-algebra | scalar-and-vector-triple-product | If the volume of a parallelopiped, whose<br/> coterminus edges are given by the <br/>vectors $$\overrightarrow a = \widehat i + \widehat j + n\widehat k$$, <br/>$$\overrightarrow b = 2\widehat i + 4\widehat j - n\widehat k$$ and <br/>$$\overrightarrow c = \widehat i + n\widehat j + 3\widehat k$$ ($$n \ge 0$$), is 1... | [{"identifier": "A", "content": "n = 7"}, {"identifier": "B", "content": "$$\\overrightarrow b .\\overrightarrow c = 10$$"}, {"identifier": "C", "content": "$$\\overrightarrow a .\\overrightarrow c = 17$$"}, {"identifier": "D", "content": "n = 9"}] | ["B"] | null | We know, Volume(V) = $$\left[ {\overrightarrow a \overrightarrow b \overrightarrow c } \right]$$
<br><br>$$ \Rightarrow $$ 158 = $$\left| {\matrix{
1 & 1 & n \cr
2 & 4 & { - n} \cr
1 & n & 3 \cr
} } \right|$$
<br><br>$$ \Rightarrow $$ (12 + n<sup>2</sup>) – (6 + n) + n(2n–4)=158... | mcq | jee-main-2020-online-5th-september-morning-slot | 4,333 |
MgYtxj9nMPDMzcRPl01klrie7jw | maths | vector-algebra | scalar-and-vector-triple-product | Let three vectors $$\overrightarrow a ,\overrightarrow b $$ and $$\overrightarrow c $$ be such that $$\overrightarrow c $$ is coplanar <br/>with $$\overrightarrow a $$ and $$\overrightarrow b $$,
$$\overrightarrow a .\overrightarrow c $$ = 7 and
$$\overrightarrow b $$ is perpendicular to $$\overrightarrow c $$, ... | [] | null | 75 | $$\overrightarrow c = \lambda (\overrightarrow b \times (\overrightarrow a \times \overrightarrow b ))$$<br><br>$$ = \lambda ((\overrightarrow b \,.\,\overrightarrow b )\overrightarrow a - (\overrightarrow b \,.\,\overrightarrow a )\overrightarrow b )$$<br><br>$$ = \lambda (5( - \widehat i + \widehat j + \widehat k... | integer | jee-main-2021-online-24th-february-morning-slot | 4,334 |
obXnX2oiZZRcp8NgLj1klugp3ss | maths | vector-algebra | scalar-and-vector-triple-product | If $$\overrightarrow a $$ and $$\overrightarrow b $$ are perpendicular, then <br/>$$\overrightarrow a \times \left( {\overrightarrow a \times \left( {\overrightarrow a \times \left( {\overrightarrow a \times \overrightarrow b } \right)} \right)} \right)$$ is equal to : | [{"identifier": "A", "content": "$${1 \\over 2}|\\overrightarrow a {|^4}\\overrightarrow b $$"}, {"identifier": "B", "content": "$$\\overrightarrow 0 $$"}, {"identifier": "C", "content": "$$\\overrightarrow a \\times \\overrightarrow b $$"}, {"identifier": "D", "content": "$$|\\overrightarrow a {|^4}\\overrightarrow b... | ["D"] | null | $$\overrightarrow a \,.\,\overrightarrow b = 0$$<br><br>$$\overrightarrow a \times (\overrightarrow a \times \overrightarrow b ) = (\overrightarrow a \,.\,\overrightarrow b )\overrightarrow a - (\overrightarrow a \,.\,\overrightarrow a )\overrightarrow b = - |\overrightarrow a {|^2}\overrightarrow b $$<br><br>Now... | mcq | jee-main-2021-online-26th-february-morning-slot | 4,335 |
XOTDvrbroQnWGWAp5G1kmizgckj | maths | vector-algebra | scalar-and-vector-triple-product | Let $$\overrightarrow c $$ be a vector perpendicular to the vectors, $$\overrightarrow a $$ = $$\widehat i$$ + $$\widehat j$$ $$-$$ $$\widehat k$$ and <br/>$$\overrightarrow b $$ = $$\widehat i$$ + 2$$\widehat j$$ + $$\widehat k$$. If $$\overrightarrow c \,.\,\left( {\widehat i + \widehat j + 3\widehat k} \right)$$ = 8... | [] | null | 28 | $$\overrightarrow a \times \overrightarrow b = \left| {\matrix{
{\widehat i} & {\widehat j} & {\widehat k} \cr
1 & 1 & { - 1} \cr
1 & 2 & 1 \cr
} } \right| = (3, - 2,1)$$<br><br>$$\overrightarrow c \bot \overrightarrow a ,\overrightarrow c \bot \overrightarrow b \Rightarrow... | integer | jee-main-2021-online-16th-march-evening-shift | 4,336 |
Z7BG0LplvDpHW9S7EN1kmjafjsy | maths | vector-algebra | scalar-and-vector-triple-product | Let $$\overrightarrow a $$ = 2$$\widehat i$$ $$-$$ 3$$\widehat j$$ + 4$$\widehat k$$ and $$\overrightarrow b $$ = 7$$\widehat i$$ + $$\widehat j$$ $$-$$ 6$$\widehat k$$.<br/><br/>If $$\overrightarrow r $$ $$\times$$ $$\overrightarrow a $$ = $$\overrightarrow r $$ $$\times$$ $$\overrightarrow b $$, $$\overrightarrow r $... | [{"identifier": "A", "content": "10"}, {"identifier": "B", "content": "8"}, {"identifier": "C", "content": "13"}, {"identifier": "D", "content": "12"}] | ["D"] | null | $$\overrightarrow a = (2, - 3,4)$$, $$\overrightarrow b = (7,1, - 6)$$<br><br>$$\overrightarrow r \times \overrightarrow a - \overrightarrow r \times \overrightarrow b = 0$$<br><br>$$\overrightarrow r \times (\overrightarrow a - \overrightarrow b ) = 0$$<br><br>$$\overrightarrow r = \lambda (\overrightarrow a ... | mcq | jee-main-2021-online-17th-march-morning-shift | 4,337 |
ALfCwHSHBoNiDqWmYY1kmjckog0 | maths | vector-algebra | scalar-and-vector-triple-product | If $$\overrightarrow a = \alpha \widehat i + \beta \widehat j + 3\widehat k$$,<br/><br/>$$\overrightarrow b = - \beta \widehat i - \alpha \widehat j - \widehat k$$ and <br/><br/>$$\overrightarrow c = \widehat i - 2\widehat j - \widehat k$$<br/><br/>such that $$\overrightarrow a \,.\,\overrightarrow b = 1$$ and $$\... | [] | null | 2 | $$\overrightarrow a .\overrightarrow b = 1 \Rightarrow - \alpha \beta - \alpha \beta - 3 = 1$$<br><br>$$ \Rightarrow \alpha \beta = - 2$$ .... (i)<br><br>$$\overrightarrow b .\overrightarrow c = - 3 \Rightarrow - \beta + 2\alpha + 1 = - 3$$<br><br>$$2\alpha - \beta = - 4$$ ..... (ii)<br><br>Solving (i) &... | integer | jee-main-2021-online-17th-march-morning-shift | 4,338 |
7hj16RzMi655FEX1h41kmkmjktz | maths | vector-algebra | scalar-and-vector-triple-product | Let O be the origin. Let $$\overrightarrow {OP} = x\widehat i + y\widehat j - \widehat k$$ and $$\overrightarrow {OQ} = - \widehat i + 2\widehat j + 3x\widehat k$$, x, y$$\in$$R, x > 0, be such that $$\left| {\overrightarrow {PQ} } \right| = \sqrt {20} $$ and the vector $$\overrightarrow {OP} $$ is perpendicular ... | [{"identifier": "A", "content": "2"}, {"identifier": "B", "content": "9"}, {"identifier": "C", "content": "7"}, {"identifier": "D", "content": "1"}] | ["B"] | null | $$\overrightarrow {OP} = x\widehat i + y\widehat j - \widehat k\,$$
<br/><br/>$$\overrightarrow {OP} \bot \overrightarrow {OQ} $$<br><br>$$\overrightarrow {OQ} = - \widehat i + 2\widehat j + 3x\widehat k$$<br><br>$$\overrightarrow {PQ} = \left( { - 1 - x} \right)\widehat i + \left( {2 - y} \right)\widehat j + \lef... | mcq | jee-main-2021-online-17th-march-evening-shift | 4,339 |
1krpwmm6x | maths | vector-algebra | scalar-and-vector-triple-product | Let $$\overrightarrow a = 2\widehat i + \widehat j - 2\widehat k$$ and $$\overrightarrow b = \widehat i + \widehat j$$. If $$\overrightarrow c $$ is a vector such that $$\overrightarrow a .\,\overrightarrow c = \left| {\overrightarrow c } \right|,\left| {\overrightarrow c - \overrightarrow a } \right| = 2\sqrt 2 $$... | [{"identifier": "A", "content": "$${2 \\over 3}$$"}, {"identifier": "B", "content": "4"}, {"identifier": "C", "content": "3"}, {"identifier": "D", "content": "$${3 \\over 2}$$"}] | ["D"] | null | $$\left| {\overrightarrow a } \right| = 3 = a;\overrightarrow a \,.\,\overrightarrow c = c$$<br><br>Now, $$\left| {\overrightarrow c - \overrightarrow a } \right| = 2\sqrt 2 $$<br><br>$$ \Rightarrow {c^2} + {a^2} - 2\overrightarrow c \,.\,\overrightarrow a = 8$$<br><br>$$ \Rightarrow {c^2} + 9 - 2(c) = 8$$<br><br>$$... | mcq | jee-main-2021-online-20th-july-morning-shift | 4,340 |
1krtcv0nx | maths | vector-algebra | scalar-and-vector-triple-product | Let a vector $${\overrightarrow a }$$ be coplanar with vectors $$\overrightarrow b = 2\widehat i + \widehat j + \widehat k$$ and $$\overrightarrow c = \widehat i - \widehat j + \widehat k$$. If $${\overrightarrow a}$$ is perpendicular to $$\overrightarrow d = 3\widehat i + 2\widehat j + 6\widehat k$$, and $$\left| {... | [{"identifier": "A", "content": "$$-$$42"}, {"identifier": "B", "content": "$$-$$40"}, {"identifier": "C", "content": "$$-$$29"}, {"identifier": "D", "content": "$$-$$38"}] | ["A"] | null | $$\overrightarrow a = \lambda \overrightarrow b + \mu \overrightarrow c = \widehat i(2\lambda + \mu ) + \widehat j(\lambda - \mu ) + \widehat k(\lambda + \mu )$$<br><br>$$\overrightarrow a \,.\,\overrightarrow d = 0 = 3(2\lambda + \mu ) + 2(\lambda - \mu ) + 6(\lambda + \mu )$$<br><br>$$ \Rightarrow 14\lambda... | mcq | jee-main-2021-online-22th-july-evening-shift | 4,341 |
1krthnmq3 | maths | vector-algebra | scalar-and-vector-triple-product | Let three vectors $$\overrightarrow a $$, $$\overrightarrow b $$ and $$\overrightarrow c $$ be such that $$\overrightarrow a \times \overrightarrow b = \overrightarrow c $$, $$\overrightarrow b \times \overrightarrow c = \overrightarrow a $$ and $$\left| {\overrightarrow a } \right| = 2$$. Then which one of the fol... | [{"identifier": "A", "content": "$$\\overrightarrow a \\times \\left( {(\\overrightarrow b + \\overrightarrow c ) \\times (\\overrightarrow b \\times \\overrightarrow c )} \\right) = \\overrightarrow 0 $$"}, {"identifier": "B", "content": "Projection of $$\\overrightarrow a $$ on $$(\\overrightarrow b \\times \\ove... | ["D"] | null | (1) $$\overrightarrow a \times \left( {(\overrightarrow b + \overrightarrow c ) \times (\overrightarrow b \times \overrightarrow c )} \right)$$<br><br>$$ = \overrightarrow a ( - \overrightarrow b \times \overrightarrow c + \overrightarrow c \times \overrightarrow b ) = - 2\left( {\overrightarrow a \times (\over... | mcq | jee-main-2021-online-22th-july-evening-shift | 4,342 |
1krvuv970 | maths | vector-algebra | scalar-and-vector-triple-product | Let the vectors<br/><br/>$$(2 + a + b)\widehat i + (a + 2b + c)\widehat j - (b + c)\widehat k,(1 + b)\widehat i + 2b\widehat j - b\widehat k$$ and $$(2 + b)\widehat i + 2b\widehat j + (1 - b)\widehat k$$, $$a,b,c, \in R$$<br><br/> be co-planar. Then which of the following is true?</br> | [{"identifier": "A", "content": "2b = a + c"}, {"identifier": "B", "content": "3c = a + b"}, {"identifier": "C", "content": "a = b + 2c"}, {"identifier": "D", "content": "2a = b + c"}] | ["A"] | null | If the vectors are co-planar,<br><br>$$\left| {\matrix{
{a + b + 2} & {a + 2b + c} & { - b - c} \cr
{b + 1} & {2b} & { - b} \cr
{b + 2} & {2b} & {1 - b} \cr
} } \right| = 0$$<br><br>Now, $${R_3} \to {R_3} - {R_2},{R_1} \to {R_1} - {R_2}$$<br><br>So, $$\left| {\matrix{
{a + 1}... | mcq | jee-main-2021-online-25th-july-morning-shift | 4,343 |
1kryf0e1y | maths | vector-algebra | scalar-and-vector-triple-product | Let $$\overrightarrow a $$, $$\overrightarrow b $$ and $$\overrightarrow c $$ be three vectors such that $$\overrightarrow a $$ = $$\overrightarrow b $$ $$\times$$ ($$\overrightarrow b $$ $$\times$$ $$\overrightarrow c $$). If magnitudes of the vectors $$\overrightarrow a $$, $$\overrightarrow b $$ and $$\overrightarro... | [{"identifier": "A", "content": "$$\\sqrt 3 + 1$$"}, {"identifier": "B", "content": "2"}, {"identifier": "C", "content": "1"}, {"identifier": "D", "content": "$${{\\sqrt 3 + 1} \\over {\\sqrt 3 }}$$"}] | ["B"] | null | $$\overrightarrow a = \left( {\overrightarrow b .\,\overrightarrow c } \right)\overrightarrow b - \left( {\overrightarrow b \,.\,\overrightarrow b } \right)\overrightarrow c $$<br><br>$$ = 1.2\cos \theta \overrightarrow b - \overrightarrow c $$<br><br>$$ \Rightarrow \overrightarrow a = 2\cos \theta \overrightarrow ... | mcq | jee-main-2021-online-27th-july-evening-shift | 4,344 |
1kryfi5s7 | maths | vector-algebra | scalar-and-vector-triple-product | Let $$\overrightarrow a = \widehat i - \alpha \widehat j + \beta \widehat k$$, $$\overrightarrow b = 3\widehat i + \beta \widehat j - \alpha \widehat k$$ and $$\overrightarrow c = -\alpha \widehat i - 2\widehat j + \widehat k$$, where $$\alpha$$ and $$\beta$$ are integers. If $$\overrightarrow a \,.\,\overrightarr... | [] | null | 9 | $$\overrightarrow a = (1, - \alpha ,\beta )$$<br><br>$$\overrightarrow b = (3,\beta , - \alpha )$$<br><br>$$\overrightarrow c = ( - \alpha , - 2,1);\alpha ,\beta \in I$$<br><br>$$\overrightarrow a \,.\,\overrightarrow b = - 1 \Rightarrow 3 - \alpha \beta - \alpha \beta = - 1$$<br><br>$$ \Rightarrow \alpha \bet... | integer | jee-main-2021-online-27th-july-evening-shift | 4,345 |
1ktbebrpm | maths | vector-algebra | scalar-and-vector-triple-product | Let $$\overrightarrow a = \widehat i + \widehat j + \widehat k$$ and $$\overrightarrow b = \widehat j - \widehat k$$. If $$\overrightarrow c $$ is a vector such that $$\overrightarrow a \times \overrightarrow c = \overrightarrow b $$ and $$\overrightarrow a .\overrightarrow c = 3$$, then $$\overrightarrow a .(\ove... | [{"identifier": "A", "content": "$$-$$2"}, {"identifier": "B", "content": "$$-$$6"}, {"identifier": "C", "content": "6"}, {"identifier": "D", "content": "2"}] | ["A"] | null | $$\left| {\overrightarrow a } \right| = \sqrt 3 $$; $$\overrightarrow a .\overrightarrow c = 3$$; $$\overrightarrow a \times \overrightarrow b = - 2\widehat i + \widehat j + \widehat k$$, $$\overrightarrow a \times \overrightarrow c = \overrightarrow b $$<br><br>Cross with $$\overrightarrow a $$,<br><br>$$\overri... | mcq | jee-main-2021-online-26th-august-morning-shift | 4,346 |
1ktk3sgt1 | maths | vector-algebra | scalar-and-vector-triple-product | Let $$\overrightarrow a ,\overrightarrow b ,\overrightarrow c $$ three vectors mutually perpendicular to each other and have same magnitude. If a vector $${ \overrightarrow r } $$ satisfies.
<br/><br/>$$\overrightarrow a \times \{ (\overrightarrow r - \overrightarrow b ) \times \overrightarrow a \} + \overrightar... | [{"identifier": "A", "content": "$${1 \\over 3}(\\overrightarrow a + \\overrightarrow b + \\overrightarrow c )$$"}, {"identifier": "B", "content": "$${1 \\over 3}(2\\overrightarrow a + \\overrightarrow b - \\overrightarrow c )$$"}, {"identifier": "C", "content": "$${1 \\over 2}(\\overrightarrow a + \\overrightarro... | ["C"] | null | Suppose $$\overrightarrow r = x\overrightarrow a + y\overrightarrow b + 2\overrightarrow c $$<br><br>and $$\left| {\overrightarrow a } \right| = \left| {\overrightarrow b } \right| = \left| {\overrightarrow c } \right| = k$$<br><br>$$\overrightarrow a \times \{ (\overrightarrow r - \overrightarrow b ) \times \over... | mcq | jee-main-2021-online-31st-august-evening-shift | 4,347 |
1l58a2gs8 | maths | vector-algebra | scalar-and-vector-triple-product | <p>If $$\overrightarrow a \,.\,\overrightarrow b = 1,\,\overrightarrow b \,.\,\overrightarrow c = 2$$ and $$\overrightarrow c \,.\,\overrightarrow a = 3$$, then the value of $$\left[ {\overrightarrow a \times \left( {\overrightarrow b \times \overrightarrow c } \right),\,\overrightarrow b \times \left( {\overrigh... | [{"identifier": "A", "content": "0"}, {"identifier": "B", "content": "$$ - 6\\overrightarrow a \\,.\\,\\left( {\\overrightarrow b \\times \\overrightarrow c } \\right)$$"}, {"identifier": "C", "content": "$$ - 12\\overrightarrow c \\,.\\,\\left( {\\overrightarrow a \\times \\overrightarrow b } \\right)$$"}, {"identif... | ["A"] | null | <p>$$\because$$ $$\overrightarrow a \times \left( {\overrightarrow b \times \overrightarrow c } \right) = 3\overrightarrow b - \overrightarrow c = \overrightarrow u $$</p>
<p>$$\overrightarrow b \times \left( {\overrightarrow c \times \overrightarrow a } \right) = \overrightarrow c - 2\overrightarrow a = \overr... | mcq | jee-main-2022-online-26th-june-morning-shift | 4,348 |
1l5w0bd5u | maths | vector-algebra | scalar-and-vector-triple-product | <p>Let a vector $$\overrightarrow c $$ be coplanar with the vectors $$\overrightarrow a = - \widehat i + \widehat j + \widehat k$$ and $$\overrightarrow b = 2\widehat i + \widehat j - \widehat k$$. If the vector $$\overrightarrow c $$ also satisfies the conditions $$\overrightarrow c \,.\,\left[ {\left( {\overrighta... | [{"identifier": "A", "content": "24"}, {"identifier": "B", "content": "29"}, {"identifier": "C", "content": "35"}, {"identifier": "D", "content": "42"}] | ["C"] | null | <p>Given,</p>
<p>$$\overrightarrow a = - \widehat i + \widehat j + \widehat k$$</p>
<p>$$\overrightarrow b = 2\widehat i + \widehat j - \widehat k$$</p>
<p>and let $$\overrightarrow c = x\widehat i + y\widehat j + z\widehat k$$</p>
<p>Now, $$\overrightarrow a + \overrightarrow b = \widehat i + 2\widehat j$$</p>
<... | mcq | jee-main-2022-online-30th-june-morning-shift | 4,349 |
1l6jc594i | maths | vector-algebra | scalar-and-vector-triple-product | <p>$$
\text { Let } \vec{a}=2 \hat{i}-\hat{j}+5 \hat{k} \text { and } \vec{b}=\alpha \hat{i}+\beta \hat{j}+2 \hat{k} \text {. If }((\vec{a} \times \vec{b}) \times \hat{i}) \cdot \hat{k}=\frac{23}{2} \text {, then }|\vec{b} \times 2 \hat{j}|
$$ is equal to :</p> | [{"identifier": "A", "content": "4"}, {"identifier": "B", "content": "5"}, {"identifier": "C", "content": "$$\\sqrt{21}$$"}, {"identifier": "D", "content": "$$\\sqrt{17}$$"}] | ["B"] | null | <p>Given, $$\overrightarrow a = 2\widehat i - \widehat j + 5\widehat k$$ and $$\overrightarrow b = \alpha \widehat i + \beta \widehat j + 2\widehat k$$</p>
<p>Also, $$\left( {\left( {\overrightarrow a \times \overrightarrow b } \right) \times i} \right)\,.\,\widehat k = {{23} \over 2}$$</p>
<p>$$ \Rightarrow \left( ... | mcq | jee-main-2022-online-27th-july-morning-shift | 4,350 |
1l6p23cfm | maths | vector-algebra | scalar-and-vector-triple-product | <p>Let $$\overrightarrow{\mathrm{a}}=3 \hat{i}+\hat{j}$$ and $$\overrightarrow{\mathrm{b}}=\hat{i}+2 \hat{j}+\hat{k}$$. Let $$\overrightarrow{\mathrm{c}}$$ be a vector satisfying $$\overrightarrow{\mathrm{a}} \times(\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}})=\overrightarrow{\mathrm{b}}+\lambda \ove... | [{"identifier": "A", "content": "$$-$$5"}, {"identifier": "B", "content": "5"}, {"identifier": "C", "content": "1"}, {"identifier": "D", "content": "$$-$$1"}] | ["A"] | null | <p>$$\overrightarrow a = 3\widehat i + \widehat j$$ & $$\overrightarrow b = \widehat i + 2\widehat j + \widehat k$$</p>
<p>$$\overrightarrow a \times (\overrightarrow b \times \overrightarrow c ) = (\overrightarrow a \,.\,\overrightarrow c )\overrightarrow b - (\overrightarrow a \,.\,\overrightarrow b )\overrighta... | mcq | jee-main-2022-online-29th-july-morning-shift | 4,351 |
1ldoob9wo | maths | vector-algebra | scalar-and-vector-triple-product | <p>Let $$\vec{v}=\alpha \hat{i}+2 \hat{j}-3 \hat{k}, \vec{w}=2 \alpha \hat{i}+\hat{j}-\hat{k}$$ and $$\vec{u}$$ be a vector such that $$|\vec{u}|=\alpha>0$$. If the minimum value of the scalar triple product $$\left[ {\matrix{
{\overrightarrow u } & {\overrightarrow v } & {\overrightarrow w } \cr
} } ... | [] | null | 3501 | $\vec{v} \times \vec{w}=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ \alpha & 2 & -3 \\ 2 \alpha & 1 & -1\end{array}\right|=\hat{i}-5 \alpha \hat{j}-3 \alpha \hat{k}$
<br/><br/>$$\left[ {\matrix{
{\overrightarrow u } & {\overrightarrow v } & {\overrightarrow w } \cr
} } \right] = \overrightarrow u .\l... | integer | jee-main-2023-online-1st-february-morning-shift | 4,352 |
ldqwqsa2 | maths | vector-algebra | scalar-and-vector-triple-product | Let $\lambda \in \mathbb{R}, \vec{a}=\lambda \hat{i}+2 \hat{j}-3 \hat{k}, \vec{b}=\hat{i}-\lambda \hat{j}+2 \hat{k}$.
<br/><br/>If $((\vec{a}+\vec{b}) \times(\vec{a} \times \vec{b})) \times(\vec{a}-\vec{b})=8 \hat{i}-40 \hat{j}-24 \hat{k}$,<br/><br/> then $|\lambda(\vec{a}+\vec{b}) \times(\vec{a}-\vec{b})|^2$ is equal ... | [{"identifier": "A", "content": "136"}, {"identifier": "B", "content": "140"}, {"identifier": "C", "content": "144"}, {"identifier": "D", "content": "132"}] | ["B"] | null | <p>$$\left( {\overrightarrow a \times \left( {\overrightarrow a \times \overrightarrow b } \right) + \overrightarrow b \times \left( {\overrightarrow a \times \overrightarrow b } \right)} \right) \times \left( {\overrightarrow a - \overrightarrow b } \right)$$</p>
<p>$$ = \left( {\overrightarrow a \left( {\overrig... | mcq | jee-main-2023-online-30th-january-evening-shift | 4,353 |
1ldr70m58 | maths | vector-algebra | scalar-and-vector-triple-product | <p>If $$\overrightarrow a ,\overrightarrow b ,\overrightarrow c $$ are three non-zero vectors and $$\widehat n$$ is a unit vector perpendicular to $$\overrightarrow c $$ such that $$\overrightarrow a = \alpha \overrightarrow b - \widehat n,(\alpha \ne 0)$$ and $$\overrightarrow b \,.\overrightarrow c = 12$$, then $... | [{"identifier": "A", "content": "15"}, {"identifier": "B", "content": "9"}, {"identifier": "C", "content": "6"}, {"identifier": "D", "content": "12"}] | ["D"] | null | <p>$$\widehat n = \alpha \overrightarrow b - \overrightarrow a $$</p>
<p>$$\overrightarrow c \times \left( {\overrightarrow a \times \overrightarrow b } \right) = \left( {\overrightarrow c \,.\,\overrightarrow b } \right)\overrightarrow a - \left( {\overrightarrow c \,.\,\overrightarrow a } \right)\overrightarrow b... | mcq | jee-main-2023-online-30th-january-morning-shift | 4,354 |
1ldsx2tkq | maths | vector-algebra | scalar-and-vector-triple-product | <p>Let $$\overrightarrow a $$, $$\overrightarrow b $$ and $$\overrightarrow c $$ be three non-zero non-coplanar vectors. Let the position vectors of four points $$A,B,C$$ and $$D$$ be $$\overrightarrow a - \overrightarrow b + \overrightarrow c ,\lambda \overrightarrow a - 3\overrightarrow b + 4\overrightarrow c , -... | [] | null | 2 | $\overline{A B}=(\lambda-1) \bar{a}-2 \bar{b}+3 \bar{c}$
<br/><br/>
$$
\overline{A C}=2 \bar{a}+3 \bar{b}-4 \bar{c}
$$<br/><br/>$$
\overline{A D}=\bar{a}-3 \bar{b}+5 \bar{c}
$$<br/><br/>$$
\left|\begin{array}{ccc}
\lambda-1 & -2 & 3 \\
-2 & 3 & -4 \\
1 & -3 & 5
\end{array}\right|=0
$$<br/><br/>$$
\Rightarrow(\lambda-1)... | integer | jee-main-2023-online-29th-january-morning-shift | 4,355 |
1ldu5mzow | maths | vector-algebra | scalar-and-vector-triple-product | <p>Let $$\overrightarrow a = - \widehat i - \widehat j + \widehat k,\overrightarrow a \,.\,\overrightarrow b = 1$$ and $$\overrightarrow a \times \overrightarrow b = \widehat i - \widehat j$$. Then $$\overrightarrow a - 6\overrightarrow b $$ is equal to :</p> | [{"identifier": "A", "content": "$$3\\left( {\\widehat i + \\widehat j + \\widehat k} \\right)$$"}, {"identifier": "B", "content": "$$3\\left( {\\widehat i - \\widehat j - \\widehat k} \\right)$$"}, {"identifier": "C", "content": "$$3\\left( {\\widehat i + \\widehat j - \\widehat k} \\right)$$"}, {"identifier": "D", "c... | ["A"] | null | $$
\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}=(\hat{\mathrm{i}}-\hat{\mathrm{j}})
$$<br/><br/>
Taking cross product with $\vec{a}$<br/><br/>
$$
\begin{aligned}
& \Rightarrow \vec{a} \times(\vec{a} \times \vec{b})=\vec{a} \times(\hat{i}-\hat{j}) \\\\
& \Rightarrow (\vec{a} \cdot \vec{b}) \vec{a}-(\v... | mcq | jee-main-2023-online-25th-january-evening-shift | 4,356 |
1ldu5p9lo | maths | vector-algebra | scalar-and-vector-triple-product | <p>If the four points, whose position vectors are $$3\widehat i - 4\widehat j + 2\widehat k,\widehat i + 2\widehat j - \widehat k, - 2\widehat i - \widehat j + 3\widehat k$$ and $$5\widehat i - 2\alpha \widehat j + 4\widehat k$$ are coplanar, then $$\alpha$$ is equal to :</p> | [{"identifier": "A", "content": "$${{73} \\over {17}}$$"}, {"identifier": "B", "content": "$$ - {{73} \\over {17}}$$"}, {"identifier": "C", "content": "$$ - {{107} \\over {17}}$$"}, {"identifier": "D", "content": "$${{107} \\over {17}}$$"}] | ["A"] | null | Let $\mathrm{A}:(3,-4,2) \quad \mathrm{C}:(-2,-1,3)$<br/><br/>
$$
\text { B : }(1,2,-1) \quad \text { D: }(5,-2 \alpha, 4)
$$<br/><br/>
A, B, C, D are coplanar points, then<br/><br/>
$$
\begin{aligned}
& \Rightarrow\left|\begin{array}{ccc}
1-3 & 2+4 & -1-2 \\
-2-3 & -1+4 & 3-2 \\
5-3 & -2 \alpha+4 & 4-2
\end{array}\rig... | mcq | jee-main-2023-online-25th-january-evening-shift | 4,357 |
1ldv26m8w | maths | vector-algebra | scalar-and-vector-triple-product | <p>Let $$\overrightarrow a $$, $$\overrightarrow b $$ and $$\overrightarrow c $$ be three non zero vectors such that $$\overrightarrow b $$ . $$\overrightarrow c $$ = 0 and $$\overrightarrow a \times (\overrightarrow b \times \overrightarrow c ) = {{\overrightarrow b - \overrightarrow c } \over 2}$$. If $$\overright... | [{"identifier": "A", "content": "$$\\frac{1}{2}$$"}, {"identifier": "B", "content": "$$-\\frac{1}{4}$$"}, {"identifier": "C", "content": "$$\\frac{1}{4}$$"}, {"identifier": "D", "content": "$$\\frac{3}{4}$$"}] | ["C"] | null | $\vec{b}(\vec{a} \cdot \vec{c})-\vec{c}(\vec{a} \cdot \vec{b})=\frac{\vec{b}-\vec{c}}{2}$ $\vec{a} \cdot \vec{c}=\frac{1}{2}, \quad \vec{a} \cdot \vec{b}=\frac{1}{2}$
<br/><br/>
$$
\begin{aligned}
(\vec{a} \times \vec{b}) \cdot(\vec{c} \times \vec{d}) & =(\vec{b} \cdot \vec{d})(\vec{a} \cdot \vec{c})-(\vec{a} \cdot \ve... | mcq | jee-main-2023-online-25th-january-morning-shift | 4,358 |
1ldybqbds | maths | vector-algebra | scalar-and-vector-triple-product | <p>Let $$\overrightarrow u = \widehat i - \widehat j - 2\widehat k,\overrightarrow v = 2\widehat i + \widehat j - \widehat k,\overrightarrow v .\,\overrightarrow w = 2$$ and $$\overrightarrow v \times \overrightarrow w = \overrightarrow u + \lambda \overrightarrow v $$. Then $$\overrightarrow u .\,\overrightarrow... | [{"identifier": "A", "content": "$$ - {2 \\over 3}$$"}, {"identifier": "B", "content": "$${3 \\over 2}$$"}, {"identifier": "C", "content": "2"}, {"identifier": "D", "content": "1"}] | ["D"] | null | $$
\begin{aligned}
&\begin{aligned}
& \vec{v} \times \vec{w}=(\vec{u}+\lambda \vec{v})=\hat{i}-\hat{j}-2 \hat{k}+\lambda(2 \hat{i}+\hat{j}-\hat{k}) \\\\
& =(2 \lambda+1) \hat{i}+(\lambda-1) \hat{j}-(2+\lambda) \hat{k} \\
&
\end{aligned}\\
&\begin{aligned}
& \text { Now, } \vec{v} \cdot(\vec{v} \times \vec{w})=0 \\\\
& ... | mcq | jee-main-2023-online-24th-january-morning-shift | 4,359 |
lgnw7ha3 | maths | vector-algebra | scalar-and-vector-triple-product | Let $S$ be the set of all $(\lambda, \mu)$ for which the vectors $\lambda \hat{i}-\hat{j}+\hat{k}, \hat{i}+2 \hat{j}+\mu \hat{k}$ and $3 \hat{i}-4 \hat{j}+5 \hat{k}$, where $\lambda-\mu=5$, are coplanar, then $\sum\limits_{(\lambda, \mu) \in S} 80\left(\lambda^2+\mu^2\right)$ is equal to : | [{"identifier": "A", "content": "2370"}, {"identifier": "B", "content": "2130"}, {"identifier": "C", "content": "2210"}, {"identifier": "D", "content": "2290"}] | ["D"] | null | Step 1: Given condition for coplanarity
<br/><br/>For three vectors to be coplanar, their scalar triple product must be zero. We have the vectors A, B, and C, and we know the given relation between λ and μ:
<br/><br/>$$A = \lambda \hat{i} - \hat{j} + \hat{k}$$
<br/><br/>$$B = \hat{i} + 2 \hat{j} + \mu \hat{k}$$
<br/><... | mcq | jee-main-2023-online-15th-april-morning-shift | 4,360 |
1lgrellln | maths | vector-algebra | scalar-and-vector-triple-product | <p>Let $$a, b, c$$ be three distinct real numbers, none equal to one. If the vectors $$a \hat{i}+\hat{\mathrm{j}}+\hat{\mathrm{k}}, \hat{\mathrm{i}}+b \hat{j}+\hat{\mathrm{k}}$$ and $$\hat{\mathrm{i}}+\hat{\mathrm{j}}+c \hat{\mathrm{k}}$$ are coplanar, then $$\frac{1}{1-a}+\frac{1}{1-b}+\frac{1}{1-c}$$ is equal to :</p... | [{"identifier": "A", "content": "$$-$$2"}, {"identifier": "B", "content": "1"}, {"identifier": "C", "content": "$$-$$1"}, {"identifier": "D", "content": "2"}] | ["B"] | null | $$
\left|\begin{array}{lll}
a & 1 & 1 \\\\
1 & \mathrm{~b} & 1 \\\\
1 & 1 & \mathrm{c}
\end{array}\right|=0
$$
<br/><br/>$$
\mathrm{C}_2 \rightarrow \mathrm{C}_2-\mathrm{C}_1, \mathrm{C}_3 \rightarrow \mathrm{C}_3-\mathrm{C}_1
$$
<br/><br/>$$
\begin{aligned}
& \left|\begin{array}{lll}
a & 1-a & 1-a \\
1 & b-1 & 0 \\
1 ... | mcq | jee-main-2023-online-12th-april-morning-shift | 4,361 |
1lgremt4q | maths | vector-algebra | scalar-and-vector-triple-product | <p>Let $$\lambda \in \mathbb{Z}, \vec{a}=\lambda \hat{i}+\hat{j}-\hat{k}$$ and $$\vec{b}=3 \hat{i}-\hat{j}+2 \hat{k}$$. Let $$\vec{c}$$ be a vector such that $$(\vec{a}+\vec{b}+\vec{c}) \times \vec{c}=\overrightarrow{0}, \vec{a} \cdot \vec{c}=-17$$ and $$\vec{b} \cdot \vec{c}=-20$$. Then $$|\vec{c} \times(\lambda \hat{... | [{"identifier": "A", "content": "53"}, {"identifier": "B", "content": "62"}, {"identifier": "C", "content": "49"}, {"identifier": "D", "content": "46"}] | ["D"] | null | The given vectors are :
<br/><br/>$$\vec{a} = \lambda \hat{i} + \hat{j} - \hat{k}$$
<br/><br/>$$\vec{b} = 3\hat{i} - \hat{j} + 2\hat{k}$$
<br/><br/>We are given that $(\vec{a} + \vec{b} + \vec{c}) \times \vec{c} = 0$ which implies $(\vec{a} + \vec{b}) \times \vec{c} = 0$. So, $\vec{c}$ is in the direction of $\vec{a... | mcq | jee-main-2023-online-12th-april-morning-shift | 4,362 |
1lgsujylg | maths | vector-algebra | scalar-and-vector-triple-product | <p>If four distinct points with position vectors $$\vec{a}, \vec{b}, \vec{c}$$ and $$\vec{d}$$ are coplanar, then $$[\vec{a} \,\,\vec{b} \,\,\vec{c}]$$ is equal to :</p> | [{"identifier": "A", "content": "$$[\\vec{d} \\,\\,\\,\\,\\,\\vec{b} \\,\\,\\,\\,\\,\\vec{a}]+[\\vec{a} \\,\\,\\,\\,\\,\\vec{c} \\,\\,\\,\\,\\,\\vec{d}]+[\\vec{d} \\,\\,\\,\\,\\,\\vec{b} \\,\\,\\,\\,\\,\\vec{c}]$$"}, {"identifier": "B", "content": "$$[\\vec{b} \\,\\,\\,\\,\\,\\vec{c} \\,\\,\\,\\,\\,\\vec{d}]+[\\vec{d} ... | ["D"] | null | $$
\begin{aligned}
& {[\vec{b}-\vec{a} \,\,\,\,\,\vec{c}-\vec{a} \,\,\,\,\,\vec{d}-\vec{a}]=0} \\\\
& (\vec{b}-\vec{a}) \cdot[(\vec{c}-\vec{a}) \times(\vec{d}-\vec{a})]=0 \\\\
& (\vec{b}-\vec{a}) \cdot(\vec{c} \times \vec{d}-\vec{c} \times \vec{a}-\vec{a} \times \vec{d})=0 \\\\
& {[\vec{b}\,\,\,\,\, \vec{c} \,\,\,\,\,\... | mcq | jee-main-2023-online-11th-april-evening-shift | 4,363 |
1lgyl6v9x | maths | vector-algebra | scalar-and-vector-triple-product | <p>Let the vectors $$\vec{u}_{1}=\hat{i}+\hat{j}+a \hat{k}, \vec{u}_{2}=\hat{i}+b \hat{j}+\hat{k}$$ and $$\vec{u}_{3}=c \hat{i}+\hat{j}+\hat{k}$$ be coplanar. If the vectors $$\vec{v}_{1}=(a+b) \hat{i}+c \hat{j}+c \hat{k}, \vec{v}_{2}=a \hat{i}+(b+c) \hat{j}+a \hat{k}$$ and $$\vec{v}_{3}=b \hat{i}+b \hat{j}+(c+a) \hat{... | [{"identifier": "A", "content": "12"}, {"identifier": "B", "content": "6"}, {"identifier": "C", "content": "0"}, {"identifier": "D", "content": "4"}] | ["A"] | null | Since, $\vec{u}_1, \vec{u}_2, \vec{u}_3$ are coplanar.
<br/><br/>So, $\left[\begin{array}{lll}\vec{u}_1 & \vec{u}_2 & \vec{u}_3\end{array}\right]=0$
<br/><br/>$$
\begin{aligned}
& \Rightarrow\left|\begin{array}{lll}
1 & 1 & a \\
1 & b & 1 \\
c & 1 & 1
\end{array}\right|=0 \\\\
& \Rightarrow 1(b-1)-1(1-c)+a(1-b c)=0 \\\... | mcq | jee-main-2023-online-8th-april-evening-shift | 4,364 |
1lh21ilhv | maths | vector-algebra | scalar-and-vector-triple-product | <p>Let the position vectors of the points A, B, C and D be
$$5 \hat{i}+5 \hat{j}+2 \lambda \hat{k}, \hat{i}+2 \hat{j}+3 \hat{k},-2 \hat{i}+\lambda \hat{j}+4 \hat{k}$$ and $$-\hat{i}+5 \hat{j}+6 \hat{k}$$. Let the set $$S=\{\lambda \in \mathbb{R}$$ :
the points A, B, C and D are coplanar $$\}$$.
<br/><br/>Then $$\sum_\... | [{"identifier": "A", "content": "$$\\frac{37}{2}$$"}, {"identifier": "B", "content": "25"}, {"identifier": "C", "content": "13"}, {"identifier": "D", "content": "41"}] | ["D"] | null | Given, position vectors of the points $A, B, C$ and $D$ be
<br/><br/>$5 \hat{\mathbf{i}}+5 \hat{\mathbf{j}}+2 \lambda \hat{\mathbf{k}}, \hat{\mathbf{i}}+2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}},-2 \hat{\mathbf{i}}+\lambda \hat{\mathbf{j}}+4 \hat{\mathbf{k}}$ and $-\hat{\mathbf{i}}+5 \hat{\mathbf{j}}+6 \hat{\mathbf{k}}$
<... | mcq | jee-main-2023-online-6th-april-morning-shift | 4,365 |
1lh2y5tet | maths | vector-algebra | scalar-and-vector-triple-product | <p>Let the vectors $$\vec{a}, \vec{b}, \vec{c}$$ represent three coterminous edges of a parallelopiped of volume V. Then the volume of the parallelopiped, whose coterminous edges are represented by $$\vec{a}, \vec{b}+\vec{c}$$ and $$\vec{a}+2 \vec{b}+3 \vec{c}$$ is equal to :</p> | [{"identifier": "A", "content": "3 V"}, {"identifier": "B", "content": "2 V"}, {"identifier": "C", "content": "6 V"}, {"identifier": "D", "content": "V"}] | ["D"] | null | Given that the volume $V$ of the parallelepiped formed by the vectors $\vec{a}$, $\vec{b}$, and $\vec{c}$ is represented by the scalar triple product $[\vec{a},\vec{b},\vec{c}]$, which is the determinant of the 3 x 3 matrix with vectors $\vec{a}$, $\vec{b}$, and $\vec{c}$ as its rows (or columns).
<br/><br/>When the v... | mcq | jee-main-2023-online-6th-april-evening-shift | 4,366 |
1lh2yfbhy | maths | vector-algebra | scalar-and-vector-triple-product | <p>The sum of all values of $$\alpha$$, for which the points whose position vectors are $$\hat{i}-2 \hat{j}+3 \hat{k}, 2 \hat{i}-3 \hat{j}+4 \hat{k},(\alpha+1) \hat{i}+2 \hat{k}$$ and $$9 \hat{i}+(\alpha-8) \hat{j}+6 \hat{k}$$ are coplanar, is equal to :</p> | [{"identifier": "A", "content": "6"}, {"identifier": "B", "content": "4"}, {"identifier": "C", "content": "$$-$$2"}, {"identifier": "D", "content": "2"}] | ["D"] | null | Let $\overrightarrow{O A}=\hat{\mathbf{i}}-2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}$
<br/><br/>$$
\begin{aligned}
& \overrightarrow{O B}=2 \hat{\mathbf{i}}-3 \hat{\mathbf{j}}+4 \hat{\mathbf{k}} \\\\
& \overrightarrow{O C}=(a+1) \hat{\mathbf{i}}+2 \hat{\mathbf{k}}
\end{aligned}
$$
<br/><br/>and $ \overrightarrow{O D}=9 \ha... | mcq | jee-main-2023-online-6th-april-evening-shift | 4,367 |
lsbl7k96 | maths | vector-algebra | scalar-and-vector-triple-product | Let $\overrightarrow{\mathrm{a}}=\hat{i}+2 \hat{j}+\hat{k}, $
<br/>$\overrightarrow{\mathrm{b}}=3(\hat{i}-\hat{j}+\hat{k})$.
<br/>Let $\overrightarrow{\mathrm{c}}$ be the vector such that $\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{c}}=\overrightarrow{\mathrm{b}}$ and $\vec{a} \cdot \vec{c}=3$.
<br/> T... | [{"identifier": "A", "content": "32"}, {"identifier": "B", "content": "36"}, {"identifier": "C", "content": "24"}, {"identifier": "D", "content": "20"}] | ["C"] | null | <p>$$\begin{aligned}
& \vec{a} \cdot[(\vec{c} \times \vec{b})-\vec{b}-\vec{c}] \\
& \vec{a} \cdot(\vec{c} \times \vec{b})-\vec{a} \cdot \vec{b}-\vec{a} \cdot \vec{c} \quad \text{..... (i)}
\end{aligned}$$</p>
<p>$$\begin{aligned}
& \text { given } \vec{a} \times \vec{c}=\vec{b} \\
& \Rightarrow(\vec{a} \times \vec{c}) ... | mcq | jee-main-2024-online-27th-january-morning-shift | 4,368 |
lv2er3tg | maths | vector-algebra | scalar-and-vector-triple-product | <p>Let $$\vec{a}=\hat{i}+\hat{j}+\hat{k}, \vec{b}=2 \hat{i}+4 \hat{j}-5 \hat{k}$$ and $$\vec{c}=x \hat{i}+2 \hat{j}+3 \hat{k}, x \in \mathbb{R}$$.
If $$\vec{d}$$ is the unit vector in the direction of $$\vec{b}+\vec{c}$$ such that $$\vec{a} \cdot \vec{d}=1$$, then $$(\vec{a} \times \vec{b}) \cdot \vec{c}$$ is equal to<... | [{"identifier": "A", "content": "3"}, {"identifier": "B", "content": "9"}, {"identifier": "C", "content": "11"}, {"identifier": "D", "content": "6"}] | ["C"] | null | <p>$$\vec{a}=\hat{i}+\hat{j}+\hat{k}, \vec{b}=2 \hat{i}+4 \hat{j}-5 \hat{k}, \vec{c}=x \hat{i}+2 \hat{j}+3 \hat{k}, x \in R$$<?p>
<p>$$\text { also, } \vec{b}+\vec{c}=(x+2) \hat{i}+6 \hat{j}-2 \hat{k}$$</p>
<p>$$\vec{d} \text { is the unit vector in the direction of } \vec{b}+\vec{c}$$</p>
<p>$$\begin{aligned}
& |\vec{... | mcq | jee-main-2024-online-4th-april-evening-shift | 4,369 |
ntSo8ltNI3kPao4H | maths | vector-algebra | scalar-or-dot-product-of-two-vectors-and-its-applications | If $$\left| {\overrightarrow a } \right| = 5,\left| {\overrightarrow b } \right| = 4,\left| {\overrightarrow c } \right| = 3$$ thus what will be the value of $$\left| {\overrightarrow a .\overrightarrow b + \overrightarrow b .\overrightarrow c + \overrightarrow c .\overrightarrow a } \right|,$$ given that $$\overrigh... | [{"identifier": "A", "content": "$$25$$"}, {"identifier": "B", "content": "$$50$$ "}, {"identifier": "C", "content": "$$-25$$"}, {"identifier": "D", "content": "$$-50$$"}] | ["A"] | null | We have, $$\overrightarrow a + \overrightarrow 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 } \right|^2} + {\left|... | mcq | aieee-2002 | 4,370 |
LrCwfitL3Reaie1v | maths | vector-algebra | scalar-or-dot-product-of-two-vectors-and-its-applications | $$\overrightarrow a \,,\overrightarrow b \,,\overrightarrow c $$ are $$3$$ vectors, such that <br/><br/>$$\overrightarrow a + \overrightarrow b + \overrightarrow c = 0$$ , $$\left| {\overrightarrow a } \right| = 1\,\,\,\left| {\overrightarrow b } \right| = 2,\,\,\,\left| {\overrightarrow c } \right| = 3,$$,
<br/><b... | [{"identifier": "A", "content": "$$1$$"}, {"identifier": "B", "content": "$$0$$"}, {"identifier": "C", "content": "$$-7$$ "}, {"identifier": "D", "content": "$$7$$"}] | ["C"] | null | $$\overrightarrow a + \overrightarrow b + \overrightarrow c = 0$$
<br><br>$$ \Rightarrow \left( {\overrightarrow a + \overrightarrow b + \overrightarrow c } \right).\left( {\overrightarrow a + \overrightarrow b + \overrightarrow c } \right) = 0$$
<br><br>$${\left| {\overrightarrow a } \right|^2} + {\left| {\ove... | mcq | aieee-2003 | 4,371 |
WBKKBa1dT8vFmppP | maths | vector-algebra | scalar-or-dot-product-of-two-vectors-and-its-applications | A particle acted on by constant forces $$4\widehat i + \widehat j - 3\widehat k$$ and $$3\widehat i + \widehat j - \widehat k$$ is displaced from the point $$\widehat i + 2\widehat j + 3\widehat k$$ to the point $$\,5\widehat i + 4\widehat j + \widehat k.$$ The total work done by the forces is : | [{"identifier": "A", "content": "$$50$$ units "}, {"identifier": "B", "content": "$$20$$ units "}, {"identifier": "C", "content": "$$30$$ units "}, {"identifier": "D", "content": "$$40$$ units "}] | ["D"] | null | The work done by a force on a particle is given by the dot product of the force and the displacement vector of the particle. The displacement vector can be found by subtracting the initial position from the final position:
<br/><br/>
$$\mathbf{displacement} = \mathbf{final\ position} - \mathbf{initial\ position} = (5\w... | mcq | aieee-2004 | 4,372 |
3XEW0uiZJ9xlVtmR | maths | vector-algebra | scalar-or-dot-product-of-two-vectors-and-its-applications | Let $$\overrightarrow u ,\overrightarrow v ,\overrightarrow w $$ be such that $$\left| {\overrightarrow u } \right| = 1,\,\,\,\left| {\overrightarrow v } \right|2,\,\,\,\left| {\overrightarrow w } \right|3.$$ If the projection $${\overrightarrow v }$$ along $${\overrightarrow u }$$ is equal to that of $${\overrightarro... | [{"identifier": "A", "content": "$$14$$ "}, {"identifier": "B", "content": "$${\\sqrt {7} }$$"}, {"identifier": "C", "content": "$${\\sqrt {14} }$$ "}, {"identifier": "D", "content": "$$2$$"}] | ["C"] | null | Projection of $$\overrightarrow v $$ along $$\overrightarrow u = {{\overrightarrow v .\overrightarrow u } \over {\left| {\overrightarrow u } \right|}} = {{\overrightarrow v .\overrightarrow u } \over 2}$$
<br><br>projection of $$\overrightarrow w $$ along $$\overrightarrow u = {{\overrightarrow w .\overrightarrow u ... | mcq | aieee-2004 | 4,373 |
R1ZCf7lzVetrtUe4 | maths | vector-algebra | scalar-or-dot-product-of-two-vectors-and-its-applications | The values of a, for which the points $$A, B, C$$ with position vectors $$2\widehat i - \widehat j + \widehat k,\,\,\widehat i - 3\widehat j - 5\widehat k$$ and $$a\widehat i - 3\widehat j + \widehat k$$ respectively are the vertices of a right angled triangle with $$C = {\pi \over 2}$$ are : | [{"identifier": "A", "content": "$$2$$ and $$1$$ "}, {"identifier": "B", "content": "$$-2$$ and $$-1$$ "}, {"identifier": "C", "content": "$$-2$$ and $$1$$ "}, {"identifier": "D", "content": "$$2$$ and $$-1$$ "}] | ["A"] | null | $$\overrightarrow {CA} = \left( {2 - a} \right)\widehat i + 2\widehat j;$$
<br><br>$$\overrightarrow {CB} = \left( {1 - a} \right)\widehat i - 6\widehat k$$
<br><br>$$\overrightarrow {CA} .\overrightarrow {CB} = 0$$
<br><br>$$\,\,\,\,\,\,\,\, \Rightarrow \left( {2 - a} \right)\left( {1 - a} \right) = 0$$
<br><br>$$ ... | mcq | aieee-2006 | 4,374 |
DRDbNWOhlsTgTTBc | maths | vector-algebra | scalar-or-dot-product-of-two-vectors-and-its-applications | If the vectors $$\overrightarrow a = \widehat i - \widehat j + 2\widehat k,\,\,\,\,\,\overrightarrow b = 2\widehat i + 4\widehat j + \widehat k\,\,\,$$ and $$\,\overrightarrow c = \lambda \widehat i + \widehat j + \mu \widehat k$$ are mutually orthogonal, then $$\,\left( {\lambda ,\mu } \right)$$ is equal to : | [{"identifier": "A", "content": "$$(2, -3)$$"}, {"identifier": "B", "content": "$$(-2, 3)$$"}, {"identifier": "C", "content": "$$(3, -2)$$"}, {"identifier": "D", "content": "$$(-3, 2)$$"}] | ["D"] | null | Since, $$\overrightarrow a ,\overrightarrow b $$ and $$\overrightarrow c $$ are mutually orthogonal
<br><br> $$\overrightarrow a .\overrightarrow b = 0,\,\,\overrightarrow b .\overrightarrow c = 0,\,\,\overrightarrow c .\overrightarrow a = 0$$
<br><br>$$ \Rightarrow 2\lambda + 4 + \mu = 0\,\,\,\,\,\,\,\,\,\,\,...... | mcq | aieee-2010 | 4,375 |
LYE5VbPrgMavFhGX | maths | vector-algebra | scalar-or-dot-product-of-two-vectors-and-its-applications | Let $$\overrightarrow a $$ and $$\overrightarrow b $$ be two unit vectors. If the vectors $$\,\overrightarrow c = \widehat a + 2\widehat b$$ and $$\overrightarrow d = 5\widehat a - 4\widehat b$$ are perpendicular to each other, then the angle between $$\overrightarrow a $$ and $$\overrightarrow b $$ is : | [{"identifier": "A", "content": "$${\\pi \\over 6}$$ "}, {"identifier": "B", "content": "$${\\pi \\over 2}$$"}, {"identifier": "C", "content": "$${\\pi \\over 3}$$"}, {"identifier": "D", "content": "$${\\pi \\over 4}$$"}] | ["C"] | null | Let $$\overrightarrow c = \widehat a + 2\widehat b$$ and $$\overrightarrow d = 5\widehat a - 4\widehat b$$
<br><br>Since $$\overrightarrow c $$ and $$\overrightarrow d $$ are perpendicular to each other
<br><br>$$\therefore$$ $$\overrightarrow c .\overrightarrow d = 0 \Rightarrow \left( {\widehat a + 2\widehat b} \... | mcq | aieee-2012 | 4,376 |
J6fh5bJMpbtUOm1J | maths | vector-algebra | scalar-or-dot-product-of-two-vectors-and-its-applications | Let $$ABCD$$ be a parallelogram such that $$\overrightarrow {AB} = \overrightarrow q ,\overrightarrow {AD} = \overrightarrow p $$ and $$\angle BAD$$ be an acute angle. If $$\overrightarrow r $$ is the vector that coincide with the altitude directed from the vertex $$B$$ to the side $$AD,$$ then $$\overrightarrow r $... | [{"identifier": "A", "content": "$$\\overrightarrow r = 3\\overrightarrow q - {{3\\left( {\\overrightarrow p .\\overrightarrow q } \\right)} \\over {\\left( {\\overrightarrow p .\\overrightarrow p } \\right)}}\\overrightarrow p $$ "}, {"identifier": "B", "content": "$$\\overrightarrow r = - \\overrightarrow q + {{... | ["B"] | null | Let $$ABCD$$ be a parallelogram such that
<br><br>$$\overrightarrow {AB} = \overrightarrow q ,\overrightarrow {AD} = \overrightarrow p $$ and $$\angle BAD$$ be an acute angle.
<br><br>We have
<br><br><img class="question-image" src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734263647/exam_images/vxui8byefbr... | mcq | aieee-2012 | 4,377 |
MyLIse0cRI3zW4o3T2Zji | maths | vector-algebra | scalar-or-dot-product-of-two-vectors-and-its-applications | In a triangle ABC, right angled at the vertex A, if the position vectors of A, B and C are respectively 3$$\widehat i$$ + $$\widehat j$$ $$-$$ $$\widehat k$$, $$-$$$$\widehat i$$ + 3$$\widehat j$$ + p$$\widehat k$$ and 5$$\widehat i$$ + q$$\widehat j$$ $$-$$ 4$$\widehat k$$, then the point (p, q) lies
on a line : | [{"identifier": "A", "content": "parallel to x-axis. "}, {"identifier": "B", "content": "parallel to y-axis."}, {"identifier": "C", "content": "making an acute angle with the positive direction of x-axis."}, {"identifier": "D", "content": "making an obtuse angle with the positive direction of x-axis. "}] | ["C"] | null | Given,
<br><br>$$\overrightarrow A = 3\widehat i + \widehat j - \widehat k$$
<br><br>$$\overrightarrow B = - \widehat i + 3\widehat j - p\widehat k$$
<br><br>$$\overrightarrow C = 5\widehat i + 9\widehat j - 4\widehat k$$
<br><br>$$ \therefore $$ $$\overrightarrow {AB} = - 4\widehat i + 2\wideha... | mcq | jee-main-2016-online-9th-april-morning-slot | 4,378 |
CknOj9CXeWNGkcOX | maths | vector-algebra | scalar-or-dot-product-of-two-vectors-and-its-applications | Let $$\overrightarrow u $$ be a vector coplanar with the vectors $$\overrightarrow a = 2\widehat i + 3\widehat j - \widehat k$$ and $$\overrightarrow b = \widehat j + \widehat k$$. If $$\overrightarrow u $$ is perpendicular to $$\overrightarrow a $$ and $$\overrightarrow u .\overrightarrow b = 24$$, then $${\left| {... | [{"identifier": "A", "content": "336"}, {"identifier": "B", "content": "315"}, {"identifier": "C", "content": "256"}, {"identifier": "D", "content": "84"}] | ["A"] | null | You should know that, when $$\overrightarrow u $$ is coplanar with $$\overrightarrow a $$ and $$\overrightarrow b $$ then we can write $$\overrightarrow u = x\overrightarrow a + y\overrightarrow b $$
<br><br>Here, $$\overrightarrow u $$ is perpendicular with $$\overrightarrow a $$ then,
<br><br>$$\overrightarrow u... | mcq | jee-main-2018-offline | 4,379 |
U1UOPS1llBEOw9OxMNW7E | maths | vector-algebra | scalar-or-dot-product-of-two-vectors-and-its-applications | Let $$\sqrt 3 \widehat i + \widehat j,$$ $$\widehat i + \sqrt 3 \widehat j$$ and $$\beta \widehat i + \left( {1 - \beta } \right)\widehat j$$ respectively be the position vectors of the points A, B and C with respect to the origin O. If the distance of C from the bisector of the acute angle between OA and OB is ... | [{"identifier": "A", "content": "4"}, {"identifier": "B", "content": "1"}, {"identifier": "C", "content": "2"}, {"identifier": "D", "content": "3"}] | ["B"] | null | Angle bisector is x $$-$$ y = 0
<br><br>$$ \Rightarrow $$ $${{\left| {\beta - \left( {1 - \beta } \right)} \right|} \over {\sqrt 2 }} = {3 \over {\sqrt 2 }}$$
<br><br>$$ \Rightarrow $$ $$\left| {2\beta - 1} \right| = 3$$
<br><br>$$ \Rightarrow $$ $$\beta $$ = 2 or $$-$$ 1 | mcq | jee-main-2019-online-11th-january-evening-slot | 4,380 |
wXePTqL6zttwwsSMnl3rsa0w2w9jwy0f6ft | maths | vector-algebra | scalar-or-dot-product-of-two-vectors-and-its-applications | Let A (3, 0, –1), B(2, 10, 6) and C(1, 2, 1) be the vertices of a triangle and M be the midpoint of AC. If G
divides BM in the ratio, 2 : 1, then cos ($$\angle $$GOA) (O being the origin) is equal to : | [{"identifier": "A", "content": "$${1 \\over {\\sqrt {15} }}$$"}, {"identifier": "B", "content": "$${1 \\over {6\\sqrt {10} }}$$"}, {"identifier": "C", "content": "$${1 \\over {\\sqrt {30} }}$$"}, {"identifier": "D", "content": "$${1 \\over {2\\sqrt {15} }}$$"}] | ["A"] | null | G is the centroid of $$\Delta $$ABC<br><br>
<picture><source media="(max-width: 320px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266720/exam_images/loddnxhki4jx9azlbmpr.webp"><source media="(max-width: 500px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265343/exam_images/jmidgr... | mcq | jee-main-2019-online-10th-april-morning-slot | 4,381 |
5UABRSZyRW07gN53tq18hoxe66ijvww95a3 | maths | vector-algebra | scalar-or-dot-product-of-two-vectors-and-its-applications | If a unit vector $$\overrightarrow a $$ makes angles $$\pi $$/3 with $$\widehat i$$ , $$\pi $$/ 4
with $$\widehat j$$ and $$\theta $$$$ \in $$(0, $$\pi $$) with $$\widehat k$$, then a value of $$\theta $$
is :- | [{"identifier": "A", "content": "$${{5\\pi } \\over {6}}$$"}, {"identifier": "B", "content": "$${{5\\pi } \\over {12}}$$"}, {"identifier": "C", "content": "$${{2\\pi } \\over {3}}$$"}, {"identifier": "D", "content": "$${{\\pi } \\over {4}}$$"}] | ["C"] | null | A unit vector $$\overrightarrow a $$ makes angles $$\pi $$/3 with $$\widehat i$$
<br><br>$$ \therefore $$ $$\alpha $$ = $$\pi $$/3
<br><br> and $$\pi $$/ 4
with $$\widehat j$$
<br><br>$$ \therefore $$ $$\beta $$ = $$\pi $$/ 4
<br><br>and $$\theta $$$$ \in $$(0, $$\pi $$) with $$\widehat k$$
<br><br>$$ \therefore $$ $$... | mcq | jee-main-2019-online-9th-april-evening-slot | 4,382 |
6C4xzfmrghEfZNGqxuFTE | maths | vector-algebra | scalar-or-dot-product-of-two-vectors-and-its-applications | Let $$\overrightarrow a = 2\widehat i + {\lambda _1}\widehat j + 3\widehat k,\,\,$$ $$\overrightarrow b = 4\widehat i + \left( {3 - {\lambda _2}} \right)\widehat j + 6\widehat k,$$ and $$\overrightarrow c = 3\widehat i + 6\widehat j + \left( {{\lambda _3} - 1} \right)\widehat k$$ be three vectors such that $$\o... | [{"identifier": "A", "content": "(1, 5, 1)"}, {"identifier": "B", "content": "(1, 3, 1)"}, {"identifier": "C", "content": "$$\\left( { - {1 \\over 2},4,0} \\right)$$"}, {"identifier": "D", "content": "$$\\left( {{1 \\over 2},4, - 2} \\right)$$"}] | ["C"] | null | Given $$\overrightarrow b = 2\overrightarrow a $$
<br><br>$$ \therefore $$ $$4\widehat i + \left( {3 - {\lambda _2}} \right)\widehat j + 6\widehat k = 4\widehat i + 2{\lambda _1}\widehat j + 6\widehat k$$
<br><br>$$ \Rightarrow 3 - {\lambda _2} = 2{\lambda _1} \Rightarrow 2{\lambda _1} + {\lambda _2} = 3\,\,...(1)$$
<... | mcq | jee-main-2019-online-10th-january-morning-slot | 4,383 |
fzUXOCvAXq7qZfRqzTtn7 | maths | vector-algebra | scalar-or-dot-product-of-two-vectors-and-its-applications | Let $$\overrightarrow a = \widehat i + \widehat j + \sqrt 2 \widehat k,$$ $$\overrightarrow b = {b_1}\widehat i + {b_2}\widehat j + \sqrt 2 \widehat k$$, $$\overrightarrow c = 5\widehat i + \widehat j + \sqrt 2 \widehat k$$ be three vectors such that the projection vector of $$\overrightarrow b $$ on $$\o... | [{"identifier": "A", "content": "$$\\sqrt {32} $$"}, {"identifier": "B", "content": "6"}, {"identifier": "C", "content": "$$\\sqrt {22} $$"}, {"identifier": "D", "content": "4"}] | ["B"] | null | Projection of $$\overrightarrow b $$ on $$\overrightarrow a $$ is $$\overrightarrow a $$
<br><br>$$ \therefore $$ $${{\overrightarrow b \cdot \overrightarrow a } \over {\left| {\overrightarrow a } \right|}} = \left| {\overrightarrow a } \right|$$
<br><br>$$ \Rightarrow $$ $${{{b_1} + {b_2} ... | mcq | jee-main-2019-online-9th-january-evening-slot | 4,384 |
gBxihkaJRK9XmHPQAQ7k9k2k5e2n780 | maths | vector-algebra | scalar-or-dot-product-of-two-vectors-and-its-applications | A vector $$\overrightarrow a = \alpha \widehat i + 2\widehat j + \beta \widehat k\left( {\alpha ,\beta \in R} \right)$$ lies in the plane of the vectors, $$\overrightarrow b = \widehat i + \widehat j$$ and $$\overrightarrow c = \widehat i - \widehat j + 4\widehat k$$. If $$\overrightarrow a $$ bisects the angle be... | [{"identifier": "A", "content": "$$\\overrightarrow a .\\widehat i + 3 = 0$$"}, {"identifier": "B", "content": "$$\\overrightarrow a .\\widehat k - 4 = 0$$"}, {"identifier": "C", "content": "$$\\overrightarrow a .\\widehat i + 1 = 0$$"}, {"identifier": "D", "content": "$$\\overrightarrow a .\\widehat k + 2 = 0$$"}] | ["B"] | null | Angle bisector $$\overrightarrow a = \lambda \left( {\widehat b + \widehat c} \right)$$
<br><br>= $$\lambda \left( {{{\widehat i + \widehat j} \over {\sqrt 2 }} + {{\widehat i - \widehat j + 4\widehat k} \over {3\sqrt 2 }}} \right)$$
<br><br>$$ \Rightarrow $$ $$\overrightarrow a = {\lambda \over {3\sqrt 2 }}\left( {... | mcq | jee-main-2020-online-7th-january-morning-slot | 4,385 |
wxnynWl5nmUW9g7btjjgy2xukewt1roa | maths | vector-algebra | scalar-or-dot-product-of-two-vectors-and-its-applications | Let $$\overrightarrow a $$, $$\overrightarrow b $$ and $$\overrightarrow c $$ be three unit vectors such that
<br/>$${\left| {\overrightarrow a - \overrightarrow b } \right|^2}$$ + $${\left| {\overrightarrow a - \overrightarrow c } \right|^2}$$ = 8.
<br/><br/>Then $${\left| {\overrightarrow a + 2\overrightarrow b } ... | [] | null | 2 | Given, $$\left| {\overrightarrow a } \right| = \left| {\overrightarrow b } \right| = \left| {\overrightarrow c } \right| = 1$$
<br><br>$${\left| {\overrightarrow a - \overrightarrow b } \right|^2}$$ + $${\left| {\overrightarrow a - \overrightarrow c } \right|^2}$$ = 8
<br><br>$$ \Rightarrow $$ $${\left| {\overrightar... | integer | jee-main-2020-online-2nd-september-morning-slot | 4,386 |
qU27zRsydVCHN6lGyOjgy2xukf3zyzqz | maths | vector-algebra | scalar-or-dot-product-of-two-vectors-and-its-applications | Let a, b c $$ \in $$ R be such that a<sup>2</sup>
+ b<sup>2</sup>
+ c<sup>2</sup>
= 1. If <br/>$$a\cos \theta = b\cos \left( {\theta + {{2\pi } \over 3}} \right) = c\cos \left( {\theta + {{4\pi } \over 3}} \right)$$,
<br/>where
$${\theta = {\pi \over 9}}$$, then the angle between the vectors
$$a\widehat i + b\w... | [{"identifier": "A", "content": "0"}, {"identifier": "B", "content": "$${{\\pi \\over 9}}$$"}, {"identifier": "C", "content": "$${{{2\\pi } \\over 3}}$$"}, {"identifier": "D", "content": "$${{\\pi \\over 2}}$$"}] | ["D"] | null | Let, $$\overrightarrow {{a_1}} = a\widehat i + b\widehat j + c\widehat k$$<br><br>and $$\overrightarrow {{a_2}} = b\widehat i + c\widehat j + a\widehat k$$<br><br>We know, Angle between two vectors<br><br>$$\cos \alpha = {{\overrightarrow {{a_1}} \,.\,\overrightarrow {{a_2}} } \over {|\overrightarrow {{a_1}} \,|.|\,... | mcq | jee-main-2020-online-3rd-september-evening-slot | 4,387 |
IjCSOdBC2AtfLTHw6Cjgy2xukfqgbdp5 | maths | vector-algebra | scalar-or-dot-product-of-two-vectors-and-its-applications | Let the vectors $$\overrightarrow a $$, $$\overrightarrow b $$, $$\overrightarrow c $$
be such that
<br/>$$\left| {\overrightarrow a } \right| = 2$$, $$\left| {\overrightarrow b } \right| = 4$$
and $$\left| {\overrightarrow c } \right| = 4$$. If the projection of
<br/>$$\overrightarrow b $$
on $$\overrightarrow a $$... | [] | null | 6 | Projection of $$\overrightarrow b $$
on $$\overrightarrow a $$
= Projection of $$\overrightarrow c $$
on $$\overrightarrow a $$
<br><br>$$ \Rightarrow $$ $${{\overrightarrow b .\overrightarrow a } \over {\left| {\overrightarrow a } \right|}} = {{\overrightarrow c .\overrightarrow a } \over {\left| {\overrightarrow a ... | integer | jee-main-2020-online-5th-september-evening-slot | 4,388 |
oBUoyZ4YjQ6r5Z8sqVjgy2xukg4n5m60 | maths | vector-algebra | scalar-or-dot-product-of-two-vectors-and-its-applications | If $$\overrightarrow x $$ and $$\overrightarrow y $$ be two non-zero vectors such that
$$\left| {\overrightarrow x + \overrightarrow y } \right| = \left| {\overrightarrow x } \right|$$ and $${2\overrightarrow x + \lambda \overrightarrow y }$$ is perpendicular to $${\overrightarrow y }$$,
then the value of $$\lambda $... | [] | null | 1 | $$\left| {\overrightarrow x + \overrightarrow y } \right| = \left| {\overrightarrow x } \right|$$
<br>Squaring both sides we get
<br><br>$${\left| {\overrightarrow x } \right|^2} + 2\overrightarrow x .\overrightarrow y + {\left| {\overrightarrow y } \right|^2} = {\left| {\overrightarrow x } \right|^2}$$
<br><br>$$ \R... | integer | jee-main-2020-online-6th-september-evening-slot | 4,389 |
hnJwMJTUDbFeYVoTKR1kmknwag0 | maths | vector-algebra | scalar-or-dot-product-of-two-vectors-and-its-applications | Let $$\overrightarrow x $$ be a vector in the plane containing vectors $$\overrightarrow a = 2\widehat i - \widehat j + \widehat k$$ and $$\overrightarrow b = \widehat i + 2\widehat j - \widehat k$$. If the vector $$\overrightarrow x $$ is perpendicular to $$\left( {3\widehat i + 2\widehat j - \widehat k} \right)$$ a... | [] | null | 486 | Let, $$\overrightarrow x = k(\overrightarrow a + \lambda \overrightarrow b )$$<br><br>$$\overrightarrow x$$ is perpendicular to $$3\widehat i + 2\widehat j - \widehat k$$<br><br><b>I.</b> k{(2 + $$\lambda$$)3 + (2$$\lambda$$ $$-$$ 1)2 + (1 $$-$$ $$\lambda$$)($$-$$1) = 0<br><br>$$ \Rightarrow $$ 8$$\lambda$$ + 3 = 0<... | integer | jee-main-2021-online-17th-march-evening-shift | 4,390 |
uSK38CUUVnLZXt4bdl1kmm3d8s3 | maths | vector-algebra | scalar-or-dot-product-of-two-vectors-and-its-applications | In a triangle ABC, if $$|\overrightarrow {BC} | = 8,|\overrightarrow {CA} | = 7,|\overrightarrow {AB} | = 10$$, then the projection of the vector $$\overrightarrow {AB} $$ on $$\overrightarrow {AC} $$ is equal to : | [{"identifier": "A", "content": "$${{25} \\over 4}$$"}, {"identifier": "B", "content": "$${{127} \\over 20}$$"}, {"identifier": "C", "content": "$${{85} \\over 14}$$"}, {"identifier": "D", "content": "$${{115} \\over 16}$$"}] | ["C"] | null | <picture><source media="(max-width: 1728px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266979/exam_images/tvnba426ygdio3l2ckwo.webp"><source media="(max-width: 320px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264573/exam_images/vlfqrjsuvhpnf9m6eq5m.webp"><source media="(max-wi... | mcq | jee-main-2021-online-18th-march-evening-shift | 4,391 |
1krq02yea | maths | vector-algebra | scalar-or-dot-product-of-two-vectors-and-its-applications | Let $$\overrightarrow a $$, $$\overrightarrow b $$, $$\overrightarrow c $$ be three mutually perpendicular vectors of the same magnitude and equally inclined at an angle $$\theta$$, with the vector $$\overrightarrow a $$ + $$\overrightarrow b $$ + $$\overrightarrow c $$. Then 36cos<sup>2</sup>2$$\theta$$ is equal to __... | [] | null | 4 | $${\left| {\overrightarrow a + \overrightarrow b + \overrightarrow c } \right|^2} = {\left| {\overrightarrow a } \right|^2} + {\left| {\overrightarrow b } \right|^2} + {\left| {\overrightarrow c } \right|^2} + 2(\overrightarrow a \,.\,\overrightarrow b + \overrightarrow a \,.\,\overrightarrow c + \overrightarrow b ... | integer | jee-main-2021-online-20th-july-morning-shift | 4,392 |
1krrv1lee | maths | vector-algebra | scalar-or-dot-product-of-two-vectors-and-its-applications | In a triangle ABC, if $$\left| {\overrightarrow {BC} } \right| = 3$$, $$\left| {\overrightarrow {CA} } \right| = 5$$ and $$\left| {\overrightarrow {BA} } \right| = 7$$, then the projection of the vector $$\overrightarrow {BA} $$ on $$\overrightarrow {BC} $$ is equal to : | [{"identifier": "A", "content": "$${{19} \\over 2}$$"}, {"identifier": "B", "content": "$${{13} \\over 2}$$"}, {"identifier": "C", "content": "$${{11} \\over 2}$$"}, {"identifier": "D", "content": "$${{15} \\over 2}$$"}] | ["C"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264070/exam_images/ep6a24u2iigywhjyduzo.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 20th July Evening Shift Mathematics - Vector Algebra Question 131 English Explanation"> <br><br>Pr... | mcq | jee-main-2021-online-20th-july-evening-shift | 4,393 |
1krrw91t0 | maths | vector-algebra | scalar-or-dot-product-of-two-vectors-and-its-applications | For p > 0, a vector $${\overrightarrow v _2} = 2\widehat i + (p + 1)\widehat j$$ is obtained by rotating the vector $${\overrightarrow v _1} = \sqrt 3 p\widehat i + \widehat j$$ by an angle $$\theta$$ about origin in counter clockwise direction. If $$\tan \theta = {{\left( {\alpha \sqrt 3 - 2} \right)} \over {\lef... | [] | null | 6 | <picture><source media="(max-width: 320px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265929/exam_images/pfsd2oapdy3p6fqplgh1.webp"><source media="(max-width: 500px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264150/exam_images/yosxjcnh7ins9tydvyvk.webp"><img src="https://res.c... | integer | jee-main-2021-online-20th-july-evening-shift | 4,394 |
1krzrafms | maths | vector-algebra | scalar-or-dot-product-of-two-vectors-and-its-applications | If $$\left( {\overrightarrow a + 3\overrightarrow b } \right)$$ is perpendicular to $$\left( {7\overrightarrow a - 5\overrightarrow b } \right)$$ and $$\left( {\overrightarrow a - 4\overrightarrow b } \right)$$ is perpendicular to $$\left( {7\overrightarrow a - 2\overrightarrow b } \right)$$, then the angle between... | [] | null | 60 | $$\left( {\overrightarrow a + 3\overrightarrow b } \right) \bot \left( {7\overrightarrow a - 5\overrightarrow b } \right)$$<br><br>$$ \therefore $$ $$\left( {\overrightarrow a + 3\overrightarrow b } \right)\,.\,\left( {7\overrightarrow a - 5\overrightarrow b } \right) = 0$$<br><br>$$ \Rightarrow $$ $$7{\left| {\ove... | integer | jee-main-2021-online-25th-july-evening-shift | 4,395 |
1ktd1se61 | maths | vector-algebra | scalar-or-dot-product-of-two-vectors-and-its-applications | A hall has a square floor of dimension 10 m $$\times$$ 10 m (see the figure) and vertical walls. If the angle GPH between the diagonals AG and BH is $${\cos ^{ - 1}}{1 \over 5}$$, then the height of the hall (in meters) is :<br/><br/><img src="data:image/png;base64,UklGRkQOAABXRUJQVlA4IDgOAADwVwCdASocARIBPm00l0ekIyKhJX... | [{"identifier": "A", "content": "5"}, {"identifier": "B", "content": "2$$\\sqrt {10} $$"}, {"identifier": "C", "content": "5$$\\sqrt {3} $$"}, {"identifier": "D", "content": "5$$\\sqrt {2} $$"}] | ["D"] | null | $$A(\widehat j)\,.\,B(10\widehat i)$$<br><br>$$H(h\widehat j + 10\widehat k)$$<br><br>$$G(10\widehat i + h\widehat j + 10\widehat k)$$<br><br>$$\overrightarrow {AG} = 10\widehat i + h\widehat j + 10\widehat k$$<br><br>$$\overrightarrow {BH} = - 10\widehat i + h\widehat j + 10\widehat k$$<br><br>$$\cos \theta = {{\o... | mcq | jee-main-2021-online-26th-august-evening-shift | 4,396 |
1ktd3i9x4 | maths | vector-algebra | scalar-or-dot-product-of-two-vectors-and-its-applications | If the projection of the vector $$\widehat i + 2\widehat j + \widehat k$$ on the sum of the two vectors $$2\widehat i + 4\widehat j - 5\widehat k$$ and $$ - \lambda \widehat i + 2\widehat j + 3\widehat k$$ is 1, then $$\lambda$$ is equal to __________. | [] | null | 5 | $$\overrightarrow a = \widehat i + 2\widehat j + \widehat k$$<br><br>$$\overrightarrow b = (2 - \lambda )\widehat i + 6\widehat j - 2\widehat k$$<br><br>$${{\overrightarrow a \,.\,\overrightarrow b } \over {|\overrightarrow b |}} = 1,\overrightarrow a \,.\,\overrightarrow b = 12 - \lambda $$<br><br>$$\left( {\overri... | integer | jee-main-2021-online-26th-august-evening-shift | 4,397 |
1ktip5iva | maths | vector-algebra | scalar-or-dot-product-of-two-vectors-and-its-applications | Let $$\overrightarrow a $$ and $$\overrightarrow b $$ be two vectors <br/>such that $$\left| {2\overrightarrow a + 3\overrightarrow b } \right| = \left| {3\overrightarrow a + \overrightarrow b } \right|$$ and the angle between $$\overrightarrow a $$ and $$\overrightarrow b $$ is 60$$^\circ$$. If $${1 \over 8}\overrig... | [{"identifier": "A", "content": "4"}, {"identifier": "B", "content": "6"}, {"identifier": "C", "content": "5"}, {"identifier": "D", "content": "8"}] | ["C"] | null | $${\left| {3\overrightarrow a + \overrightarrow b } \right|^2} = {\left| {2\overrightarrow a + 3\overrightarrow b } \right|^2}$$<br><br>$$\left( {3\overrightarrow a + \overrightarrow b } \right).\left( {3\overrightarrow a + \overrightarrow b } \right) = \left( {2\overrightarrow a + 3\overrightarrow b } \right).\le... | mcq | jee-main-2021-online-31st-august-morning-shift | 4,398 |
1l5ainfwa | maths | vector-algebra | scalar-or-dot-product-of-two-vectors-and-its-applications | <p>Let $$\overrightarrow a = {a_1}\widehat i + {a_2}\widehat j + {a_3}\widehat k$$ $${a_i} > 0$$, $$i = 1,2,3$$ be a vector which makes equal angles with the coordinate axes OX, OY and OZ. Also, let the projection of $$\overrightarrow a $$ on the vector $$3\widehat i + 4\widehat j$$ be 7. Let $$\overrightarrow b $$... | [{"identifier": "A", "content": "$$\\sqrt 7 $$"}, {"identifier": "B", "content": "$$\\sqrt 2 $$"}, {"identifier": "C", "content": "2"}, {"identifier": "D", "content": "7"}] | ["B"] | null | <p>$${\cos ^2}\alpha + {\cos ^2}\beta + {\cos ^2}\gamma = 1 \Rightarrow {\cos ^2}\alpha = {1 \over 3} \Rightarrow \cos \alpha = {1 \over {\sqrt 3 }}$$</p>
<p>$$\overrightarrow a = {\lambda \over 3}(\widehat i + \widehat j + \widehat k),\,\lambda > 0$$</p>
<p>$${\lambda \over {\sqrt 3 }}{{(\widehat i + \widehat... | mcq | jee-main-2022-online-25th-june-morning-shift | 4,399 |
1l6nnts5x | maths | vector-algebra | scalar-or-dot-product-of-two-vectors-and-its-applications | <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> | [{"identifier": "A", "content": "$$\\left(-\\infty,-\\frac{4}{3}\\right)$$"}, {"identifier": "B", "content": "$$\\Phi $$"}, {"identifier": "C", "content": "$$\\left(-\\frac{4}{3}, 0\\right)$$"}, {"identifier": "D", "content": "$$\\left(\\frac{12}{7}, \\infty\\right)$$"}] | ["B"] | null | <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>$$ \Rightarrow a{({\log _e}b)^2} - 12 + 6a({\log _e}b) > 0$$</p>... | mcq | jee-main-2022-online-28th-july-evening-shift | 4,400 |
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