Vector – JEE Advanced Previous Year Questions with Solutions

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Q. (A) If $\vec{a}, \vec{b}, \vec{c}$ and $\vec{d}$ are unit vectors such that $(\vec{a} \times \vec{b}) \cdot(\vec{c} \times \vec{d})=1$ and $\vec{a} \cdot \vec{c}=\frac{1}{2},$ then $-$

(A) $\overrightarrow{\mathrm{a}}, \overrightarrow{\mathrm{b}}, \overrightarrow{\mathrm{c}}$ are non-coplanar

(B) $\overrightarrow{\mathrm{b}}, \overrightarrow{\mathrm{c}}, \overrightarrow{\mathrm{d}}$ are non-coplanar

(C) $\overrightarrow{\mathrm{b}}, \overrightarrow{\mathrm{d}}$ arenon-parallel

(D) $\overrightarrow{\mathrm{a}}, \overrightarrow{\mathrm{d}}$ are parallel and $\overrightarrow{\mathrm{b}}, \overrightarrow{\mathrm{c}}$ are parallel

(B) Match the statements / expressions given in Column I with the values given in Column II.

[JEE 2009, 3+8]

Sol. ((A) C (B) (A) Q,S              (B) P,R,S,T             (C) T             (D) R )

(A) Let angle between $\overrightarrow{\mathrm{a}}$ and $\overrightarrow{\mathrm{b}}$ be $\theta_{1}, \overrightarrow{\mathrm{c}}$ and $\overrightarrow{\mathrm{d}}$ be $\theta_{2}$ and $\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}$ and $\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{d}}$ be $\theta$

since $(\vec{a} \times \vec{b}) .(\vec{c} \times \vec{d})=1$

$\sin \theta_{1} \sin \theta_{2} \cos \theta=1$

$\theta_{1}=90, \theta_{2}=90, \theta=0$

$\overrightarrow{\mathrm{a}} \perp \overrightarrow{\mathrm{b}}, \overrightarrow{\mathrm{c}} \perp \overrightarrow{\mathrm{d}},(\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}) \|(\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{d}})$

so $\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}=\mathrm{k}(\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{d}})$

and $\quad(\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}) \cdot \overrightarrow{\mathrm{c}}=\mathrm{k}(\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{d}}) \cdot \overrightarrow{\mathrm{c}}$

Case I $(\vec{a} \times \vec{b}) \cdot \vec{d}=k(\vec{c} \times \vec{d}) . \vec{d} \quad \because[\vec{a} \vec{b} \vec{c}]=0$

Case II $\overrightarrow{\mathrm{b}} \cdot(\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}})=\mathrm{k} \overrightarrow{\mathrm{b}} \cdot(\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{d}}) \therefore[\overrightarrow{\mathrm{b}} \overrightarrow{\mathrm{c}} \cdot \overrightarrow{\mathrm{d}}]=0$

(A) and (B) are in correct

let $\overrightarrow{\mathrm{b}} \| \overrightarrow{\mathrm{d}} \Rightarrow \overrightarrow{\mathrm{b}}=\pm \overrightarrow{\mathrm{d}}$

As $(\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}) \cdot(\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{d}})=1$

$\Rightarrow(\vec{a} \times \vec{b}) \cdot(\vec{c} \times \vec{b})=\pm 1$

$[\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}} \overrightarrow{\mathrm{c}} \overrightarrow{\mathrm{b}}]=\pm 1$

$[\overrightarrow{\mathrm{c}} \quad \overrightarrow{\mathrm{b}} \quad \overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}]=\pm 1$

$\overrightarrow{\mathrm{c}} \cdot[\overrightarrow{\mathrm{b}} \times(\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}})]=\pm 1$

$\overrightarrow{\mathrm{c}} \cdot[\overrightarrow{\mathrm{a}}-(\overrightarrow{\mathrm{b}} \cdot \overrightarrow{\mathrm{a}}) \overrightarrow{\mathrm{b}}]=\pm 1$

$\overrightarrow{\mathrm{c} .} \overrightarrow{\mathrm{a}}=\pm 1 \quad \because \quad \overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}=0$

Which is contradiction so option (c) is correct let option (d) is correct

$\overrightarrow{\mathrm{d}}=\pm \overrightarrow{\mathrm{a}}$ and $\overrightarrow{\mathrm{c}}=\pm \overrightarrow{\mathrm{b}}$

$(\vec{a} \times \vec{b}) \cdot(\vec{c} \times \vec{d})=1$

$(\vec{a} \times \vec{b}) \cdot(\vec{b} \times \vec{a})=\pm 1$

so (d) incorrect

(B) $(\mathrm{A}) 2 \sin ^{2} \theta+\sin ^{2} 2 \theta=2$

$\sin ^{2} 2 \theta=2 \cos ^{2} \theta$

$4 \sin ^{2} \theta \cos ^{2} \theta=2 \cos ^{2} \theta$

$\cos ^{2} \theta=0$ or $\sin ^{2} \theta=\frac{1}{2}$

$\theta=\frac{\pi}{2}$ or $\sin 0=\pm \frac{1}{\sqrt{2}}$

$\theta=\frac{\pi}{2}, \pm \frac{\pi}{4}$

(B) $\mathrm{f}(\mathrm{x})=\left[\frac{6 \mathrm{x}}{\pi}\right] \cos \left[\frac{3 \mathrm{x}}{\pi}\right]$

possible point of discontinuty of $\left[\frac{6 \mathrm{x}}{\pi}\right]$

$\frac{6 \mathrm{x}}{\pi}=\mathrm{n}, \mathrm{n} \in \mathrm{I}$

$x=\frac{n \pi}{6}$

$\Rightarrow \mathrm{x}=\frac{\pi}{6}, \frac{\pi}{3}, \frac{\pi}{2}, \pi$

Lt $f(x)=1 \cos 0=1$

$\mathrm{x} \rightarrow \frac{\pi^{+}}{6}$

$\mathrm{Lt}-\mathrm{f}(\mathrm{x})=0 \cos 0=0$

$\mathrm{X} \rightarrow \frac{\pi^{-}}{6}$

disconti at $x=\frac{\pi}{6}$

similarly $x=\frac{\pi}{3}, \frac{\pi}{2}, \pi$

(C) Here $\mathrm{v}=\left|\begin{array}{ccc}{1} & {1} & {0} \\ {1} & {2} & {0} \\ {1} & {1} & {\pi}\end{array}\right|=\pi \mathrm{cm}^{3}$

(d) Given $\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}+\sqrt{3} \overrightarrow{\mathrm{c}}=0$

$\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}=-\sqrt{3} \overrightarrow{\mathrm{c}}$

$|\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}|^{2}=|\sqrt{3} \overrightarrow{\mathrm{c}}|^{2}$

$a^{2}+b^{2}+2 \vec{a} \cdot \vec{b}=3 c^{2}$

$2+\cos \theta=3$

$\cos \theta=\frac{1}{2}$

$\theta=\frac{\pi}{3}$

Q. (A) Two adjacent sides of a parallelogram ABCD are given by $\overrightarrow{\mathrm{AB}}=2 \hat{\mathrm{i}}+10 \hat{\mathrm{j}}+11 \hat{\mathrm{k}}$ and $\overrightarrow{\mathrm{AD}}=-\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+2 \hat{\mathrm{k}} .$ The side $\mathrm{AD}$ is rotated by an acute angle $\alpha$ in the plane of the parallelogram so that AD becomes AD’. If AD’ makes a right angle with the side AB, then the cosine of the angle $\alpha$ is given by –

(A) $\frac{8}{9}$

(B) $\frac{\sqrt{17}}{9}$

(C) $\frac{1}{9}$

(D) $\frac{4 \sqrt{5}}{9}$

(B) If $\vec{a}$ and $\vec{b}$ are vectors in space given by $\vec{a}=\frac{\hat{i}-2 \hat{j}}{\sqrt{5}}$ and $\vec{b}=\frac{2 \hat{i}+\hat{j}+3 \hat{k}}{\sqrt{14}},$ then the value of $(2 \vec{a}+\vec{b}) \cdot[(\vec{a} \times \vec{b}) \times(\vec{a}-2 \vec{b})]$ is

[JEE 2010, 5+3]

Sol. ( (A) B (B) 5 )

Q. (A) Let $\vec{a}=\hat{i}+\hat{j}+\hat{k}, \vec{b}=\hat{i}-\hat{j}+\hat{k}$ and $\vec{c}=\hat{i}-\hat{j}-\hat{k}$ be three vectors. A vector $\vec{v}$ in the plane of $\overrightarrow{\mathrm{a}}$ and $\overrightarrow{\mathrm{b}},$ whose projection on $\overrightarrow{\mathrm{c}}$ is $\frac{1}{\sqrt{3}},$ is given by

(A) $\hat{\mathrm{i}}-3 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}$

(B) $-3 \hat{\mathrm{i}}-3 \hat{\mathrm{j}}-\hat{\mathrm{k}}$

(C) $3 \hat{\mathrm{i}}-\hat{\mathrm{j}}+3 \hat{\mathrm{k}}$

(D) $\hat{\mathrm{i}}+3 \hat{\mathrm{j}}-3 \hat{\mathrm{k}}$

(B) The vector(s) which is/are coplanar with vectors $\hat{\mathrm{i}}+\hat{\mathrm{j}}+2 \hat{\mathrm{k}}$ and $\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\hat{\mathrm{k}},$ and perpendicular to the vector $\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}$ is/are

(A) $\hat{\mathrm{j}}-\hat{\mathrm{k}}$

(B) $-\hat{\hat{\mathrm{i}}}+\hat{\hat{\mathrm{j}}}$

(C) $\hat{\mathrm{i}}-\hat{\mathrm{j}}$

$(D)-\hat{j}+\hat{k}$

(C) Let $\vec{a}=-\hat{i}-\hat{k}, \vec{b}=-\hat{i}+\hat{j}$ and $\vec{c}=\hat{i}+2 \hat{j}+3 \hat{k}$ be three given vectors. If $\vec{r}$ is a vector such that $\overrightarrow{\mathrm{r}} \times \overrightarrow{\mathrm{b}}=\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{b}}$ and $\overrightarrow{\mathrm{r} .} \overrightarrow{\mathrm{a}}=0,$ then the value of $\overrightarrow{\mathrm{r}} . \overrightarrow{\mathrm{b}}$ is

[JEE 2011, 3+4+4]

Sol. ( (A) C (B) A,D (C) 9 )

$\overrightarrow{\mathrm{v}}=\mathrm{x} \overrightarrow{\mathrm{a}}+\mathrm{y} \overrightarrow{\mathrm{b}}$

$=\hat{\mathbf{i}}(\mathrm{x}+\mathrm{y})+\hat{\mathrm{j}}(\mathrm{x}-\mathrm{y})+\hat{\mathrm{k}}(\mathrm{x}+\mathrm{y}) \ldots .(\mathrm{i})$

Given $, \overrightarrow{\mathrm{v}} . \hat{\mathrm{c}}=\frac{1}{\sqrt{3}}$

$\Rightarrow \frac{x+y-x+y-x-y}{\sqrt{3}}=\frac{1}{\sqrt{3}}$

$\mathrm{y}-\mathrm{x}=1$

$\Rightarrow \mathrm{x}-\mathrm{y}=-1$

using (ii) in (i) we get $\overrightarrow{\mathrm{v}}=(\mathrm{x}+\mathrm{y}) \hat{\mathrm{i}}-\hat{\mathrm{j}}+(\mathrm{x}+\mathrm{y}) \hat{\mathrm{k}}$

(b) $\vec{a}=\hat{i}+\hat{j}+2 \hat{k}$

$\overrightarrow{\mathrm{b}}=\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\hat{\mathrm{k}}$

$\overrightarrow{\mathrm{c}}=\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}$

$\overrightarrow{\mathrm{v}}=\lambda((\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}) \times \overrightarrow{\mathrm{c}})=\lambda((\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{c}}) \overrightarrow{\mathrm{b}}-(\overrightarrow{\mathrm{b}} \cdot \overrightarrow{\mathrm{c}}) \overrightarrow{\mathrm{a}}$

$\overrightarrow{\mathrm{v}}=\lambda[4(\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\hat{\mathrm{k}})-4(\hat{\mathrm{i}}+\hat{\mathrm{j}}+2 \hat{\mathrm{k}})]$

$\overrightarrow{\mathrm{v}}=4 \lambda(\hat{j}-\hat{\mathrm{k}})$

(c) $\overrightarrow{\mathrm{a}}=-\hat{\mathrm{i}}-\hat{\mathrm{k}}$

$\overrightarrow{\mathrm{b}}=-\hat{\mathrm{i}}+\hat{\mathrm{j}}$

$\overrightarrow{\mathrm{c}}=\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}$

$\overrightarrow{\mathrm{r}} \times \overrightarrow{\mathrm{b}}=\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{b}}$

Taking cross product by $\overrightarrow{\mathrm{a}}$

$(\overrightarrow{\mathrm{r}} \times \overrightarrow{\mathrm{b}}) \times \overrightarrow{\mathrm{a}}=(\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{b}}) \times \overrightarrow{\mathrm{a}}$

$\Rightarrow(\overrightarrow{\mathrm{r}} \cdot \overrightarrow{\mathrm{a}}) \overrightarrow{\mathrm{b}}-(\overrightarrow{\mathrm{b}} \cdot \overrightarrow{\mathrm{a}}) \overrightarrow{\mathrm{r}}=(\overrightarrow{\mathrm{c}} \cdot \overrightarrow{\mathrm{a}}) \overrightarrow{\mathrm{b}})-(\overrightarrow{\mathrm{b}} \cdot \overrightarrow{\mathrm{a}}) \overrightarrow{\mathrm{c}}$

$\Rightarrow 0-\overrightarrow{\mathrm{r}}=(-1-3)(-\hat{\mathrm{i}}+\hat{\mathrm{j}})-(1)(\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}})$

$\overrightarrow{\mathrm{r}}=-3 \hat{\mathrm{i}}+6 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}$

$\overrightarrow{\mathrm{r}} \cdot \overrightarrow{\mathrm{b}}=3+6=9$

Q. (A) If $\vec{a}, \vec{b}$ and $\vec{c}$ are unit vectors satisfying $|\vec{a}-\vec{b}|^{2}+|\vec{b}-\vec{c}|^{2}+|\vec{c}-\vec{a}|^{2}=9,$ then

$|2 \overrightarrow{\mathrm{a}}+5 \overrightarrow{\mathrm{b}}+5 \overrightarrow{\mathrm{c}}| \text { is }$

(B) If $\vec{a}$ and $\vec{b}$ are vectors such that $|\vec{a}+\vec{b}|=\sqrt{29}$ and $\overrightarrow{\mathrm{a}} \times(2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+4 \hat{\mathrm{j}}+4 \hat{\mathrm{k}})=(2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+4 \hat{\mathrm{k}}) \times \overrightarrow{\mathrm{b}},$ then a possible value of $(\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}) \cdot(-7 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}})$ is

(A) 0 (B) 3 (C) 4 (D) 8

[JEE 2012, 4+3]

Sol. ( (A) 3 (B) C)

$|\vec{a}-\vec{b}|^{2}+|\vec{b}-\vec{c}|^{2}+|\vec{c}-\vec{a}|^{2}=9$

$\Rightarrow 6-2 \Sigma \overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}=9$

$\Rightarrow \Sigma \overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}=-\frac{3}{2}$

$|\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}}|^{2} \geq 0$

$\Sigma \overrightarrow{\mathrm{a}}^{2}+2 \Sigma \overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}} \geq 0$

$\sum \overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}} \geq-\frac{3}{2}$

for equality $|\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}}|=0$

$\Rightarrow \overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}}=0$

$5 \overrightarrow{\mathrm{b}}+5 \overrightarrow{\mathrm{c}}=-5 \overrightarrow{\mathrm{a}}$

$2 \vec{a}+5 \vec{b}+5 \vec{c}=-3 \vec{a}$

$|2 \vec{a}+5 \vec{b}+5 \vec{c}|=3|\vec{a}|=3$

(b) $(\vec{a}+\vec{b}) \times(2 \hat{i}+3 \hat{j}+4 \hat{k})=0$

$\Rightarrow \overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}=\lambda(2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+4 \hat{\mathrm{k}})$

$|\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}|=\sqrt{29} \Rightarrow|\lambda|=1$

$\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}=(2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+4 \hat{\mathrm{k}})$

$(\vec{a}+\vec{b}) \cdot(-7 \hat{i}+2 \hat{j}+3 \hat{k})=-14+6+12=4$

Q. Let $\overrightarrow{\mathrm{PR}}=3 \hat{\mathrm{i}}+\hat{\mathrm{j}}-2 \hat{\mathrm{k}}$ and $\overline{\mathrm{SQ}}=\hat{\mathrm{i}}-3 \hat{\mathrm{j}}-4 \hat{\mathrm{k}}$ determine diagonals of a parallelogram PQRS and $\overrightarrow{\mathrm{PT}}=\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}$ be another vector. Then the volume of the parallelepiped determined by the vectors $\overrightarrow{\mathrm{PT}}, \overrightarrow{\mathrm{PQ}}$ and $\overrightarrow{\mathrm{PS}}$ is

(A) 5          (B) 20          (C) 10               (D) 30

Sol. (C)

$\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}=\overrightarrow{\mathrm{PR}} \& \overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{b}}=\overrightarrow{\mathrm{QS}}$

$\overrightarrow{\mathrm{a}}=\frac{\overrightarrow{\mathrm{PR}}+\overrightarrow{\mathrm{QS}}}{2} \& \overrightarrow{\mathrm{b}}=\frac{\overrightarrow{\mathrm{PR}}-\overrightarrow{\mathrm{QS}}}{2}$

$\overrightarrow{\mathrm{a}}=2 \hat{\mathrm{i}}-\hat{\mathrm{j}}-3 \hat{\mathrm{k}} \& \overrightarrow{\mathrm{b}}=\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\hat{\mathrm{k}}$

Volume $=\left|\begin{array}{ccc}{2} & {-1} & {-3} \\ {1} & {2} & {1} \\ {1} & {2} & {3}\end{array}\right|$

$2(4)+(3-1)-3(2-2)$

$8+2=10$

Q. Consider the set of eight vectors $\mathrm{V}=\{\mathrm{a} \hat{\mathrm{i}}+\mathrm{b} \hat{\mathrm{j}}+\mathrm{c} \hat{\mathrm{k}}: \mathrm{a}, \mathrm{b}, \mathrm{c} \in\{-1,1\}\} .$ Three non- coplanar vectors can be chosen from $\mathrm{V}$ in $2^{\mathrm{p}}$ ways. Then $\mathrm{p}$ is

Sol. 5

The 8 vectors will represent $\mathrm{O}$ is at the centre of cube $\overrightarrow{\mathrm{OA}}, \overrightarrow{\mathrm{OB}} \ldots \ldots \overrightarrow{\mathrm{OD}}, \overrightarrow{\mathrm{OP}}, \ldots \ldots . \overrightarrow{\mathrm{OS}}$

ABCDPQRS any three out of these 8 will be coplanar when two of them are

There are 4 pairs of collinear vectors $\overrightarrow{\mathrm{OA}} \& \overrightarrow{\mathrm{OR}}, \overrightarrow{\mathrm{OB}} \& \overrightarrow{\mathrm{OS}}, \overrightarrow{\mathrm{OC}} \& \overrightarrow{\mathrm{OP}}, \overrightarrow{\mathrm{OD}} \& \overrightarrow{\mathrm{OQ}}$

(it will generate $4 \times 6=24$ set of coplanar vectors) rest of the combination of 3 vectors will form three edges of a tetrahedron so they will be not coplanar. So number of non-coplanar vectors $^{8} \mathrm{C}_{3}-4.6=32$

Q. Match List-I with List-II and select the correct answer using the code given

below the lists.

Sol. (C)

(P) Given $[\overrightarrow{\mathrm{a}} \overrightarrow{\mathrm{b}} \overrightarrow{\mathrm{c}}]=2$

$[2(\vec{a} \times \vec{b}) 3(\vec{b} \times \vec{c})(\vec{c} \times \vec{a})]=6[\vec{a} \times \vec{b} \quad \vec{b} \times \vec{c} \vec{c} \times \vec{a}]$

$=6[\vec{a} \vec{b} \vec{c}]^{2}=24$

(Q) Given $[\overrightarrow{\mathrm{a}} \overrightarrow{\mathrm{b}} \overrightarrow{\mathrm{c}}]=5$

$[3(\vec{a}+\vec{b})(\vec{b}+\vec{c}) 2(\vec{c}+\vec{a})]=12[\vec{a} \vec{b} \vec{c}]=60$

(R) Given $\frac{1}{2}|\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}|=20$

$\Rightarrow|\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}|=40$

$\left|\frac{1}{2}(2 \vec{a}+3 \vec{b}) \times(\vec{a}-\vec{b})\right|=\frac{1}{2}|0+3 \vec{b} \times \vec{a}-2 \vec{a} \times \vec{b}|$

$=\frac{1}{2}|-5 \overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}|=\frac{5}{2}|\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}|=\frac{5}{2} \cdot 40=100$

(S) Given $|\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}|=30$

$|(\vec{a}+\vec{b}) \times \vec{a}|=|0+\vec{b} \times \vec{a}|=30$

Q. Let $\overrightarrow{\mathrm{x}}, \overrightarrow{\mathrm{y}}$ and $\overrightarrow{\mathrm{z}}$ be three vectors each of magnitude $\sqrt{2}$ and the angle between each pair of them is $\frac{\pi}{3} .$ If $\overrightarrow{\mathrm{a}}$ is a nonzero vector perpendicular to $\overrightarrow{\mathrm{x}}$ and $\overrightarrow{\mathrm{y}} \times \overrightarrow{\mathrm{z}}$ and $\overrightarrow{\mathrm{b}}$ is nonzero vector perpendicular to $\overrightarrow{\mathrm{y}}$ and $\overrightarrow{\mathrm{z}} \times \overrightarrow{\mathrm{x}},$ then

(A) $\overrightarrow{\mathrm{b}}=(\overrightarrow{\mathrm{b}} \cdot \overrightarrow{\mathrm{z}})(\overrightarrow{\mathrm{z}}-\overrightarrow{\mathrm{x}})$

(B) $\overrightarrow{\mathrm{a}}=(\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{y}})(\overrightarrow{\mathrm{y}}-\overrightarrow{\mathrm{z}})$

(C) $\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}=-(\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{y}})(\overrightarrow{\mathrm{b}} \cdot \overrightarrow{\mathrm{z}})$

$(D) \quad \vec{a}=(\vec{a} \cdot \vec{y})(\vec{z}-\vec{y})$

Sol. (A,B,C)

Given that $|\overrightarrow{\mathrm{x}}|=|\overrightarrow{\mathrm{y}}|=|\overrightarrow{\mathrm{z}}|=\sqrt{2}$

and angle between each pair is $\frac{\pi}{3}$

$\therefore \overrightarrow{\mathrm{x}} \cdot \overrightarrow{\mathrm{y}}=\overrightarrow{\mathrm{y} .} \cdot \overrightarrow{\mathrm{z}}=\overrightarrow{\mathrm{z}} \cdot \overrightarrow{\mathrm{x}}=\sqrt{2} \cdot \sqrt{2} \cdot \frac{1}{2}=1$

Now $\vec{a}$ is $\perp$ to $\vec{x} \&(\vec{y} \times \vec{z})$

Let $\overrightarrow{\mathrm{a}}=\lambda(\overrightarrow{\mathrm{x}} \times(\overrightarrow{\mathrm{y}} \times \overrightarrow{\mathrm{z}}))$

$=\lambda((\overrightarrow{\mathrm{x}} \cdot \overrightarrow{\mathrm{z}}) \overrightarrow{\mathrm{y}}-(\overrightarrow{\mathrm{x}} \cdot \overrightarrow{\mathrm{y}}) \overrightarrow{\mathrm{z}})=\lambda(\overrightarrow{\mathrm{y}}-\overrightarrow{\mathrm{z}})$

$\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{y}}=\lambda(\overrightarrow{\mathrm{y}} \cdot \overrightarrow{\mathrm{y}}-\overrightarrow{\mathrm{y}} \cdot \overrightarrow{\mathrm{z}})=\lambda(2-1)=\lambda$

$\Rightarrow \overrightarrow{\mathrm{a}}=(\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{y}})(\overrightarrow{\mathrm{y}}-\overrightarrow{\mathrm{z}})$

Now let $\overrightarrow{\mathrm{b}}=\mu(\overrightarrow{\mathrm{y}} \times(\overrightarrow{\mathrm{z}} \times \overrightarrow{\mathrm{x}}))=\mu(\overrightarrow{\mathrm{z}}-\overrightarrow{\mathrm{x}})$

$\overrightarrow{\mathrm{b}} \cdot \overrightarrow{\mathrm{z}}=\mu(2-1)=\mu$

$\Rightarrow \overrightarrow{\mathrm{b}}=(\overrightarrow{\mathrm{b}} \cdot \overrightarrow{\mathrm{z}})(\overrightarrow{\mathrm{z}}-\overrightarrow{\mathrm{x}})$

Now $\vec{a} \cdot \vec{b}=(\vec{a} \cdot \vec{y})(\vec{y}-\vec{z}) \cdot(\vec{b} \cdot \vec{z})(\vec{z}-\vec{x})$

$=(\vec{a} \cdot \vec{y})(\vec{b} \cdot \vec{z})(\vec{y} \cdot \vec{z}-\vec{y} \cdot \vec{x}-\vec{z} \cdot \vec{z}+\vec{z} \cdot \vec{x})$

$=(\vec{a} \cdot \vec{y})(\vec{b} \cdot \vec{z})(1-1-2+1)$

$=-(\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{y}})(\overrightarrow{\mathrm{b}} \cdot \overrightarrow{\mathrm{z}})$

Q. Let $\overrightarrow{\mathrm{a}}, \overrightarrow{\mathrm{b}},$ and $\overrightarrow{\mathrm{c}}$ be three non-coplanar unit vectors such the angle between every pair of them is $\frac{\pi}{3} \cdot$ If $\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}}=\mathrm{p} \overrightarrow{\mathrm{a}}+\mathrm{qb}+\mathrm{r} \overrightarrow{\mathrm{c}},$ where $\mathrm{p}, \mathrm{q}$ and $\mathrm{r}$ are scalars, then the value of

$\frac{\mathrm{p}^{2}+2 \mathrm{q}^{2}+\mathrm{r}^{2}}{\mathrm{q}^{2}}$ is

Sol. 4

Q. Let $\Delta \mathrm{PQR}$ be a triangle. Let $\overrightarrow{\mathrm{a}}=\overrightarrow{\mathrm{QR}}, \overrightarrow{\mathrm{b}}=\overrightarrow{\mathrm{RP}}$ and $\overrightarrow{\mathrm{c}}=\overrightarrow{\mathrm{PQ}} .$ If $|\overrightarrow{\mathrm{a}}|=4 \sqrt{3}$ and $\overrightarrow{\mathrm{b} . \overrightarrow{\mathrm{c}}}=24$ then which of the following is (are) true?

(A) $\frac{|\overrightarrow{\mathrm{c}}|^{2}}{2}-|\overrightarrow{\mathrm{a}}|=12$

(B) $\frac{|\overrightarrow{\mathrm{c}}|^{2}}{2}+|\overrightarrow{\mathrm{a}}|=30$

(C) $|\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{a}}|=48 \sqrt{3}$

(D) $\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}=-72$

[JEE 2015, 4M, –2M]

Sol. (A,C,D)

Q. Suppose that $\overrightarrow{\mathrm{p}}, \overrightarrow{\mathrm{q}}$ and $\overrightarrow{\mathrm{r}}$ are three non-coplanar vectors in $\mathrm{U}^{3} .$ Let the components of a vector $\overrightarrow{\mathrm{s}}$ along $\overrightarrow{\mathrm{p}}, \overrightarrow{\mathrm{q}}$ and $\overrightarrow{\mathrm{r}}$ be $4,3$ and $5,$ respectively. If the components of this vector $\overrightarrow{\mathrm{s}}$ along $(-\overrightarrow{\mathrm{p}}+\overrightarrow{\mathrm{q}}+\overrightarrow{\mathrm{r}}),(\overrightarrow{\mathrm{p}}-\overrightarrow{\mathrm{q}}+\overrightarrow{\mathrm{r}})$ and $(-\overrightarrow{\mathrm{p}}-\overrightarrow{\mathrm{q}}+\overrightarrow{\mathrm{r}})$ are $\mathrm{x}, \mathrm{y}$ and $\mathrm{z},$ respectively, then the value of $2 \mathrm{x}+\mathrm{y}+\mathrm{z}$ is

[JEE 2015, 4M, –0M]

Sol. (Bonus)

Although the language of the question is not appropriate (in complete information) and it must be declare as bonus but as per the theme of problem it must be as follows

$\overrightarrow{\mathrm{s}}=4 \overrightarrow{\mathrm{p}}+3 \overrightarrow{\mathrm{q}}+5 \overrightarrow{\mathrm{r}}$

$\overrightarrow{\mathrm{s}}=\mathrm{x}(-\overrightarrow{\mathrm{p}}+\overrightarrow{\mathrm{q}}+\overrightarrow{\mathrm{r}})+\mathrm{y}(\overrightarrow{\mathrm{p}}-\overrightarrow{\mathrm{q}}+\overrightarrow{\mathrm{r}})+\mathrm{z}(-\overrightarrow{\mathrm{p}}-\overrightarrow{\mathrm{q}}+\overrightarrow{\mathrm{r}})$

$\overrightarrow{\mathrm{s}}=\overrightarrow{\mathrm{p}}(-\mathrm{x}+\mathrm{y}-\mathrm{z})+\overrightarrow{\mathrm{q}}(\mathrm{x}-\mathrm{y}-\mathrm{z})+\overrightarrow{\mathrm{r}}(\mathrm{x}+\mathrm{y}+\mathrm{z})$

$-x+y-z=4$

$x-y-z=3$

$x+y+z=5$

$\Rightarrow \mathrm{x}=4, \mathrm{y}=\frac{9}{2}, \mathrm{z}=-\frac{7}{2}$

$\Rightarrow 2 \mathrm{x}+\mathrm{y}+\mathrm{z}=8-\frac{7}{2}+\frac{9}{2}=9$

Q. Let $\hat{\mathbf{u}}=\mathbf{u}_{1} \hat{\mathbf{i}}+\mathbf{u}_{2} \hat{\mathbf{j}}+\mathbf{u}_{3} \hat{\mathbf{k}}$ be a unit vector in $\mathbb{U}^{2}$ and $\hat{\mathbf{w}}=\frac{1}{\sqrt{6}}(\hat{\mathbf{i}}+\hat{\mathbf{j}}+2 \hat{\mathbf{k}}) .$ Given that there exists a vector $\overrightarrow{\mathrm{v}}$ in $\square^{3}$ such that $|\hat{\mathrm{u}} \times \overrightarrow{\mathrm{v}}|=1$ and $\hat{\mathrm{w}} .(\hat{\mathrm{u}} \times \overrightarrow{\mathrm{v}})=1 .$ Which of the following statement(s) is(are) correct?

(A) There is exactly one choice for such $\overrightarrow{\mathrm{v}}$

(B) There are infinitely many choice for such $\overrightarrow{\mathrm{v}}$

(C) If \hat{u lies in the xy-plane then } $\left|\mathrm{u}_{1}\right|=\left|\mathrm{u}_{2}\right|$

(D) If ú lies in the xz-plane then $2\left|u_{1}\right|=\left|u_{3}\right|$

Sol. (B,C)

Q. Let O be the origin and let PQR be an arbitrary triangle. The point $\mathrm{S}$ is such that $\overrightarrow{\mathrm{OP}} \cdot \overrightarrow{\mathrm{OP}}+\overrightarrow{\mathrm{OR}} \cdot \overrightarrow{\mathrm{OP}}=\overrightarrow{\mathrm{OR}} \cdot \overrightarrow{\mathrm{OP}}+\overrightarrow{\mathrm{OQ}} \cdot \overrightarrow{\mathrm{OS}}=\overrightarrow{\mathrm{OQ}} \cdot \overrightarrow{\mathrm{OR}}+\overrightarrow{\mathrm{OP}} \cdot \overrightarrow{\mathrm{OS}}$ Then the triangle PQR has S as its

(A) incentre

(B) orthocenter

(C) circumcentre

(D) centroid

Sol. (B)

Let position vector of $\mathrm{P}(\overrightarrow{\mathrm{p}}), \mathrm{Q}(\overrightarrow{\mathrm{q}}), \mathrm{R}(\overrightarrow{\mathrm{r}}) \& \mathrm{S}(\overrightarrow{\mathrm{r}})$ with respect to $\mathrm{O}(\overrightarrow{\mathrm{o}})$

Now, $\overrightarrow{\mathrm{OP}} \cdot \overrightarrow{\mathrm{OQ}}+\overrightarrow{\mathrm{OR}} \cdot \overrightarrow{\mathrm{OS}}=\overrightarrow{\mathrm{OR}} \cdot \overrightarrow{\mathrm{OP}}+\overrightarrow{\mathrm{OQ}} \cdot \overrightarrow{\mathrm{OS}}$

$\Rightarrow \overrightarrow{\mathrm{p}} \cdot \overrightarrow{\mathrm{q}}+\overrightarrow{\mathrm{r}} \cdot \overrightarrow{\mathrm{s}}=\overrightarrow{\mathrm{r}} \cdot \overrightarrow{\mathrm{p}}+\overrightarrow{\mathrm{q}} \cdot \overrightarrow{\mathrm{s}}$

$\Rightarrow(\overrightarrow{\mathrm{p}}-\overrightarrow{\mathrm{s}}) \cdot(\overrightarrow{\mathrm{q}}-\overrightarrow{\mathrm{r}})=0$ …..(1)

Also, $\overrightarrow{\mathrm{OR}} \cdot \overrightarrow{\mathrm{OP}}+\overrightarrow{\mathrm{OQ}} \cdot \overrightarrow{\mathrm{OS}}=\overrightarrow{\mathrm{OQ}} \cdot \overrightarrow{\mathrm{OR}}+\overrightarrow{\mathrm{OP}} \cdot \overrightarrow{\mathrm{OS}}$

$\Rightarrow \overrightarrow{\mathrm{r}} \cdot \overrightarrow{\mathrm{p}}+\overrightarrow{\mathrm{q}} \cdot \overrightarrow{\mathrm{s}}=\overrightarrow{\mathrm{q}} \cdot \overrightarrow{\mathrm{r}}+\overrightarrow{\mathrm{p}} \cdot \overrightarrow{\mathrm{s}}$

$\Rightarrow(\overrightarrow{\mathrm{r}}-\overrightarrow{\mathrm{s}}) \cdot(\overrightarrow{\mathrm{p}}-\overrightarrow{\mathrm{q}})=0$ ……(2)

Let O be the origin, and $\overrightarrow{\mathrm{OX}}, \overrightarrow{\mathrm{OY}}, \overrightarrow{\mathrm{OZ}}$ be three unit vectors in the directions of the sides $\overrightarrow{\mathrm{QR}}, \overrightarrow{\mathrm{RP}}, \overrightarrow{\mathrm{PQ}},$ respectively, of a triangle $\mathrm{PQR}$

Q. $|\overrightarrow{\mathrm{OX}} \times \overrightarrow{\mathrm{OY}}|=$

(A) sin(Q + R)

(B) sin(P + R)

(C) sin 2R

(D) sin(P + Q)

Sol. (D)

$\overrightarrow{\mathrm{OX}}=\frac{\overrightarrow{\mathrm{QR}}}{\mathrm{QR}}$

$\overrightarrow{\mathrm{OY}}=\frac{\overrightarrow{\mathrm{RP}}}{\mathrm{RP}}$

$|\overrightarrow{\mathrm{OX}} \times \overrightarrow{\mathrm{OY}}|=\sin \mathrm{R}=\sin (\mathrm{P}+\mathrm{Q})$

Q. If the triangle PQR varies, then the minimum value of cos(P + Q) + cos(Q + R) + cos(R + P) is

Sol. (B)

$-(\cos P+\cos Q+\cos R) \geq-\frac{3}{2}$ as we know $\cos P+\cos Q+$ cosR will take

its maximum value when $\mathrm{P}=\mathrm{Q}=\mathrm{R}=\frac{\pi}{3}$

Q. Let $\vec{a}$ and $\vec{b}$ be two unit vectors such that $\vec{a} . \vec{b}=0 .$ For some $\mathbf{x}, y \in \square,$ let $\vec{c}=x \vec{a}+y \vec{b}+(\vec{a} \times \vec{b})$ If $|\overrightarrow{\mathrm{c}}|=2$ and the vector $\overrightarrow{\mathrm{c}}$ is inclined at the same angle $\alpha$ to both $\overrightarrow{\mathrm{a}}$ and $\overrightarrow{\mathrm{b}},$ then the value of $8 \cos ^{2} \alpha$ is

Sol. 3

$\overrightarrow{\mathrm{c}}=\mathrm{x} \overrightarrow{\mathrm{a}}+\mathrm{y} \overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}$

$\overrightarrow{\mathrm{c}} \cdot \overrightarrow{\mathrm{a}}=\mathrm{x}$ and $\mathrm{x}=2 \cos \alpha$

$\overrightarrow{\mathrm{c}} \cdot \overrightarrow{\mathrm{b}}=\mathrm{y}$ and $\mathrm{y}=2 \cos \alpha$

Also, $|\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}|=1$

\begin{aligned} \therefore \quad & \overrightarrow{\mathrm{c}}=2 \cos (\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}})+\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}} \\ \overrightarrow{\mathrm{c}}^{2} &=4 \cos ^{2} \alpha(\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}})^{2}+(\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}})^{2}+2 \cos \alpha(\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}) \cdot(\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}) \\ & 4=8 \cos ^{2} \alpha+1 \\ & 8 \cos ^{2} \alpha=3 \end{aligned}