Vector Algebra – JEE Main Previous Year Question with Solutions

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Q. If $\overrightarrow{\mathrm{u}}, \overrightarrow{\mathrm{v}}, \overrightarrow{\mathrm{w}}$ are non-coplanar vectors and $\mathrm{p}, \mathrm{q}$ are real numbers, then the equality $[3 \vec{u} p \vec{v} p \vec{w}]-[p \vec{v} \vec{w} q \vec{u}]-[2 \vec{w} q \vec{v} q \vec{u}]=0$ holds for $:-$

(1) More than two but not all values of (p,q)

(2) All values of (p, q)

(3) Exactly one value of (p, q)

(4) Exactly two values of (p, q)

[AIEEE-2009]

Sol. (3)


Q. Let $\vec{a}=\hat{j}-\hat{k}$ and $\vec{c}=\hat{i}-\hat{j}-\hat{k} .$ Then the vector $\vec{b}$ satisfying $\vec{a} \times \vec{b}+\vec{c}=\overrightarrow{0}$ and $\vec{a} . \vec{b}=3$ is :

(1) $-\hat{\mathrm{i}}+\hat{\mathrm{j}}-2 \hat{\mathrm{k}}$

(2) $2 \hat{\mathrm{i}}-\hat{\mathrm{j}}+2 \hat{\mathrm{k}}$

(3) $\hat{\mathrm{i}}-\hat{\mathrm{j}}-2 \hat{\mathrm{k}}$

(4) $\hat{\mathrm{i}}+\hat{\mathrm{j}}-2 \hat{\mathrm{k}}$

[AIEEE-2010]

Sol. (1)

$(\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}})+\overrightarrow{\mathrm{c}}=0$

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

$\Rightarrow \overrightarrow{\mathrm{a}} \times(\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}})=-\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{c}} \Rightarrow(\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}) \overrightarrow{\mathrm{a}}-|\overrightarrow{\mathrm{a}}|^{2} \overrightarrow{\mathrm{b}}=-\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{c}}$

$\Rightarrow 3(\overrightarrow{\mathrm{j}}-\overrightarrow{\mathrm{k}})-2 \overrightarrow{\mathrm{b}}=-(-2 \mathrm{i}-\mathrm{j}-\mathrm{k})$

\[(\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{c}}=-2 \mathrm{i}-\mathrm{j}-\mathrm{k})\]

$\Rightarrow 2 \overrightarrow{\mathrm{b}}=(-2 \mathrm{i}+2 \mathrm{j}-4 \mathrm{k}) \quad \Rightarrow \overrightarrow{\mathrm{b}}=-\mathrm{i}+\mathrm{j}-2 \mathrm{k}$


Q. The vectors $\overrightarrow{\mathrm{a}}$ and $\overrightarrow{\mathrm{b}}$ are not perpendicular and $\overrightarrow{\mathrm{c}}$ and $\overrightarrow{\mathrm{d}}$ are two vectors satisfying: $\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}}=\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{d}}$ and $\overrightarrow{\mathrm{a} .} \overrightarrow{\mathrm{d}}=0 .$ Then the vector $\overrightarrow{\mathrm{d}}$ is equal to :-

$(1) \overrightarrow{\mathrm{b}}+\left(\frac{\overrightarrow{\mathrm{b}} \cdot \overrightarrow{\mathrm{c}}}{\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}}\right) \overrightarrow{\mathrm{c}}$

(2) $\overrightarrow{\mathrm{c}}-\left(\frac{\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{c}}}{\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}}\right) \overrightarrow{\mathrm{b}}$

(3) $\overrightarrow{\mathrm{b}}-\left(\frac{\overrightarrow{\mathrm{b}} \cdot \overrightarrow{\mathrm{c}}}{\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}}\right) \overrightarrow{\mathrm{c}}$

(4) $\overrightarrow{\mathrm{c}}+\left(\frac{\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{c}}}{\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}}\right) \overrightarrow{\mathrm{b}}$

[AIEEE-2011]

Sol. (2)


Q. If $\overrightarrow{\mathrm{a}}=\frac{1}{\sqrt{10}}(3 \hat{\mathrm{i}}+\hat{\mathrm{k}})$ and $\overrightarrow{\mathrm{b}}=\frac{1}{7}(2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}-6 \hat{\mathrm{k}}),$ then the value of $(2 \overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{b}}) \cdot[(\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}) \times(\overrightarrow{\mathrm{a}}+2 \overrightarrow{\mathrm{b}})]$ is 1:

(1) 5                    (2) 3                   (3) – 5                   (4) – 3

[AIEEE-2011]

Sol. (3)

$=-4 a^{2}-b^{2}+4 \vec{a} \cdot \vec{b}=-5$


Q. Let $\vec{a}, \vec{b}, \vec{c}$ be three non-zero vectors which are pairwise non-collinear. If $\vec{a}+3 \vec{b}$ is collinear with $\overrightarrow{\mathrm{c}}$ and $\overrightarrow{\mathrm{b}}+2 \overrightarrow{\mathrm{c}}$ is colliner with $\overrightarrow{\mathrm{a}},$ then $\overrightarrow{\mathrm{a}}+3 \overrightarrow{\mathrm{b}}+6 \overrightarrow{\mathrm{c}}$ is:

(1) $\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{c}}$ ( 2)$\quad \vec{a}$ ( 3)$\quad \overrightarrow{\mathrm{c}}$ ( 4) $\overrightarrow{0}$

[AIEEE-2011]

Sol. (4)


Q. Let $\hat{\mathbf{a}}$ and $\hat{\mathbf{b}}$ be two unit vectors. If the vectors $\overrightarrow{\mathbf{c}}=\hat{\mathbf{a}}+2 \hat{\mathbf{b}}$ and $\overrightarrow{\mathbf{d}}=5 \hat{\mathbf{a}}-4 \hat{\mathbf{b}}$ are perpendicular to each other, then the angle between $\hat{\mathbf{a}}$ and $\hat{\mathbf{b}}$ is :

(1) $\frac{\pi}{4}$

( 2)$\frac{\pi}{6}$

(3) $\frac{\pi}{2}$

( 4)$\frac{\pi}{3}$

[AIEEE-2012]

Sol. (4)

$\overline{\mathrm{c}} \cdot \overline{\mathrm{d}}=0 \quad \Rightarrow(\hat{\mathrm{a}}+2 \hat{\mathrm{b}}) \cdot(5 \hat{\mathrm{a}}-4 \hat{\mathrm{b}})=0$

$\Rightarrow 5-8+6 \hat{a} \cdot \hat{b}=0$

$\Rightarrow \hat{\mathrm{a}} \cdot \hat{\mathrm{b}}=1 / 2 \Rightarrow \cos \theta=1 / 2$

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


Q. Let ABCD be a parallelogram such that $\overrightarrow{\mathrm{AB}}=\overrightarrow{\mathrm{q}}, \overrightarrow{\mathrm{AD}}=\overrightarrow{\mathrm{p}}$ and $\angle \mathrm{BAD}$ be an acute angle. If $\overrightarrow{\mathrm{r}}$ is the vector that coincides with the altitude directed from the vertex $B$ to the side $A D,$ then $\overrightarrow{\mathrm{r}}$ is given by :

[AIEEE-2012]

Sol. (3)


Q. If the vectors $\overrightarrow{\mathrm{AB}}=3 \hat{\mathrm{i}}+4 \hat{\mathrm{k}}$ and $\overrightarrow{\mathrm{AC}}=5 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}+4 \hat{\mathrm{k}}$ are the sides of a triangle ABC, then the length of the median through A is :

(1) $\sqrt{18}$

(2) $\sqrt{72}$

(3) $\sqrt{33}$

(4) $\sqrt{45}$

[JEE-MAINS 2013]

Sol. (3)

$\overrightarrow{\mathrm{AD}}=\frac{\overrightarrow{\mathrm{AB}}+\overrightarrow{\mathrm{AC}}}{2}=4 \hat{\mathrm{j}}-\hat{\mathrm{j}}+4 \hat{\mathrm{k}}$

$|\overrightarrow{\mathrm{AD}}|=\sqrt{33}$


Q. Let $\overrightarrow{\mathrm{a}}=2 \hat{\mathrm{i}}-\hat{\mathrm{j}}+\hat{\mathrm{k}}, \overrightarrow{\mathrm{b}}=\hat{\mathrm{i}}+2 \hat{\mathrm{j}}-\hat{\mathrm{k}}$ and $\overrightarrow{\mathrm{c}}=\hat{\mathrm{i}}+\hat{\mathrm{j}}-2 \hat{\mathrm{k}}$ be three vectors. A vectors. A vectors of the type $\overrightarrow{\mathrm{b}}+\lambda \overrightarrow{\mathrm{c}}$ for some scalar $\lambda,$ whose projection on $\overrightarrow{\mathrm{a}}$ is of magnitude $\sqrt{\frac{2}{3}},$ is :

(1) $2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}-3 \hat{\mathrm{k}}$

(2) $2 \hat{\mathrm{i}}+\hat{\mathrm{j}}+5 \hat{\mathrm{k}}$

(3) $2 \hat{\mathrm{i}}-\hat{\mathrm{j}}+5 \hat{\mathrm{k}}$

(4) $2 \hat{i}+3 \hat{j}+3 \hat{k}$

[JEE-MAINS Online 2013]

Sol. (1)

$\frac{|(\overline{\mathrm{b}}+\lambda \overline{\mathrm{c}}) \cdot \overline{\mathrm{a}}|}{|\overline{\mathrm{a}}|}=\sqrt{\frac{2}{3}}$

$\left|[(1+\lambda) \hat{\mathrm{i}}+(2+\lambda) \hat{\mathrm{j}}+(-1-2 \lambda) \hat{\mathrm{k}}] \cdot \frac{2 \hat{\mathrm{i}}-\hat{\mathrm{j}}+\hat{\mathrm{k}}}{\sqrt{6}}\right|=\sqrt{\frac{2}{3}}$

$|(2+2 \lambda-2-\lambda-1-2 \lambda)|=2$

$|\lambda+1|=2$

$\lambda+1=\pm 2$

$\lambda=1,-3$

Vect $=2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}-3 \hat{\mathrm{k}}$ or $-2 \hat{\mathrm{i}}-\hat{\mathrm{j}}+5 \hat{\mathrm{k}}$


Q. Let $\vec{a}=2 \hat{i}+\hat{j}-2 \hat{k}, \vec{b}=\hat{i}+\hat{j} .$ If $\vec{c}$ is a vector such that $\vec{a} \bullet \vec{c}=|\vec{c}|,|\vec{c}-\vec{a}|=2 \sqrt{2}$ and the angle between $\vec{a} \times \vec{b}$ and $\vec{c}$ is $30^{\circ},$ then $|(\vec{a} \times \vec{b}) \times \vec{c}|$ equals :

( 1)$\frac{3}{2}$

(2) 3

( 3)$\frac{1}{2}$

(4) $\frac{3 \sqrt{3}}{2}$

[JEE-MAINS Online 2013]

Sol. (1)

$|\bar{c}|^{2}+|\bar{a}|^{2}-2 \bar{a} \bar{c}=8$

$|\bar{c}|^{2}+9-2|\bar{c}|=8$

$|\overline{\mathrm{c}}|=1$

$|(\overline{\mathrm{a}} \times \overline{\mathrm{b}}) \times \overline{\mathrm{c}}|=|\overline{\mathrm{a}} \times \overline{\mathrm{b}}||\overline{\mathrm{c}}| \sin 30^{\circ}$

$=\frac{|\overline{\mathrm{a}} \times \overline{\mathrm{b}}|}{2}$

$|\overline{\mathrm{a}} \times \overline{\mathrm{b}}|=\left|\begin{array}{ccc}{\hat{\mathrm{i}}} & {\hat{\mathrm{j}}} & {\hat{\mathrm{k}}} \\ {2} & {1} & {-2} \\ {1} & {1} & {0}\end{array}\right|$

$=|2 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}+\hat{\mathbf{k}}|=3$


Q. If $[\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}} \overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{a}}]=\lambda[\overrightarrow{\mathrm{a}} \overrightarrow{\mathrm{b}} \overrightarrow{\mathrm{c}}]^{2}$ then $\lambda$ is equal to :

(1) 2 (2) 3 (3) 0 (4) 1

[JEE(Main)-2014]

Sol. (4)

$[\vec{a} \times \vec{b} \quad \vec{b} \times \vec{c} \quad \vec{c} \times \vec{a}]=(\vec{a} \times \vec{b}) \cdot((\vec{b} \times \vec{c}) \times(\vec{c} \times \vec{a}))$

$=(\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}) \cdot((\overrightarrow{\mathrm{b}} \overrightarrow{\mathrm{c}} \cdot \overrightarrow{\mathrm{a}}) \overrightarrow{\mathrm{c}}-[\overrightarrow{\mathrm{b}} \overrightarrow{\mathrm{c}} \cdot \overrightarrow{\mathrm{c}}] \overrightarrow{\mathrm{a}})=[\overrightarrow{\mathrm{a}} \overrightarrow{\mathrm{b}} \overrightarrow{\mathrm{c}}]^{2}$


Q. Let $\vec{a}, \vec{b}$ and $\vec{c}$ be three non-zero vectors such that no two of them are collinear and $(\vec{a}, \times \vec{b}) \times \vec{c}=\frac{1}{3}|\vec{b}||\vec{c}| \vec{a} \cdot$ If $\theta$ is the angle between vectors $\vec{b}$ and $\vec{c},$ then a value of $\sin \theta$ is :

[JEE(Main)-2015]

Sol. (3)

$(\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{c}}) \overrightarrow{\mathrm{b}}-(\overrightarrow{\mathrm{b}} \cdot \overrightarrow{\mathrm{c}}) \overrightarrow{\mathrm{a}}=\frac{1}{3}|\overrightarrow{\mathrm{b}}||\overrightarrow{\mathrm{c}}| \overrightarrow{\mathrm{a}}$

$\Rightarrow\left[\begin{array}{c}{\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{c}}=0} \\ {-\overrightarrow{\mathrm{b}} \cdot \overrightarrow{\mathrm{c}}=\frac{1}{3}|\overrightarrow{\mathrm{b}} \| \overrightarrow{\mathrm{c}}|}\end{array}\right.$

$\Rightarrow \cos \theta=-\frac{1}{3} \Rightarrow \sin \theta=\frac{2 \sqrt{2}}{3}$


Q. Let $\overrightarrow{\mathrm{a}}, \overrightarrow{\mathrm{b}}$ and $\overrightarrow{\mathrm{c}}$ be three unit vectors such that $\overrightarrow{\mathrm{a}} \times(\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}})=\frac{\sqrt{3}}{2}(\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}}) .$ If $\overrightarrow{\mathrm{b}}$ is not parallel to $\overrightarrow{\mathrm{c}}$ then the angle between a and $\overrightarrow{\mathrm{b}}$ is :-

(1) $\frac{5 \pi}{6}$                 (2) $\frac{3 \pi}{4}$              (3) $\frac{\pi}{2}$              (4) $\frac{2 \pi}{3}$

[JEE(Main)-2016]

Sol. (1)

$\left(\overrightarrow{\mathrm{a} .} \overrightarrow{\mathrm{c}}-\frac{\sqrt{3}}{2}\right) \overrightarrow{\mathrm{b}}-\left(\overrightarrow{\mathrm{a} .} \overrightarrow{\mathrm{b}}+\frac{\sqrt{3}}{2}\right) \overrightarrow{\mathrm{c}}=0$

$\Rightarrow \overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}=\cos \theta=-\sqrt{3} / 2 \Rightarrow \theta=5 \pi / 6$


Q. Let $\overrightarrow{\mathrm{a}}=2 \hat{\mathrm{i}}+\hat{\mathrm{j}}-2 \hat{\mathrm{k}}$ and $\overrightarrow{\mathrm{b}}=\hat{\mathrm{i}}+\hat{\mathrm{j}} .$ Let $\overrightarrow{\mathrm{c}}$ be a vector such that $|\overrightarrow{\mathrm{c}}-\overrightarrow{\mathrm{a}}|=3,|(\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}) \times \overrightarrow{\mathrm{c}}|=3$ and the angle between $\overrightarrow{\mathrm{c}}$ and $\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}$ be $30^{\circ} .$ Then $\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{c}}$ is equal to :

( 1)$\frac{1}{8}$                     (2) $\frac{25}{8}$                (3) 2                    (4) 5

[JEE(Main)-2018]

Sol. (3)

$\overrightarrow{\mathrm{a}}=2 \hat{\mathrm{i}}+\hat{\mathrm{j}}-2 \hat{\mathrm{k}}, \quad \overrightarrow{\mathrm{b}}=\hat{\mathrm{i}}+\hat{\mathrm{j}}$ and $|\overrightarrow{\mathrm{a}}|=3$

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

$|\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}|=3$

Now $:(\vec{a} \times \vec{b}) \times \vec{c}=|\vec{a} \times \vec{b}||\vec{c}| \sin 30 \hat{n}$

$|(\vec{a} \times \vec{b}) \times \vec{c}|=3 \cdot|\vec{c}| \cdot \frac{1}{2}$

$3=3|\overrightarrow{\mathrm{c}}| \cdot \frac{1}{2}$

$\therefore|\overrightarrow{\mathrm{c}}|=2$

Now $:|\overrightarrow{\mathrm{c}}-\overrightarrow{\mathrm{a}}|=3$

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

$4+9-2 \overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{c}}=9$

$\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{c}}=2$


Q. If the position vectors of the vertices $\mathrm{A}, \mathrm{B}$ and $\mathrm{C}$ of a $\Delta \mathrm{ABC}$ are respectively $4 \hat{i}+7 \hat{j}+8 \hat{k}, 2 \hat{i}+3 \hat{j}+4 \hat{k}$ and $2 \hat{i}+5 \hat{j}+7 \hat{k},$ then the position vector of the point, where the bisector of $\angle \mathrm{A}$ meets $\mathrm{BC}$ is :

(1) $\frac{1}{2}(4 \hat{i}+8 \hat{j}+11 \hat{k})$

(2) $\frac{1}{3}(6 \hat{i}+13 \hat{j}+18 \hat{k})$

(3) $\frac{1}{4}(8 \hat{\mathrm{i}}+14 \hat{\mathrm{j}}+19 \hat{\mathrm{k}})$

(4) $\frac{1}{3}(6 \hat{i}+11 \hat{j}+15 \hat{k})$

[JEE(Main)-2018]

Sol. (2)


Q. If $\overrightarrow{\mathrm{a}}, \overrightarrow{\mathrm{b}},$ and $\overrightarrow{\mathrm{c}}$ are unit vectors such that $\overrightarrow{\mathrm{a}}+2 \overrightarrow{\mathrm{b}}+2 \overrightarrow{\mathrm{c}}=\overrightarrow{0},$ then $|\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{c}}|$ is equal to :

[JEE(Main)-2018]

Sol. (4)


Q. If the angle between the lines, $\frac{x}{2}=\frac{y}{2}=\frac{z}{1}$ and $\frac{5-x}{-2}=\frac{7 y-14}{p}=\frac{z-3}{4}$ is $\cos ^{-1}\left(\frac{2}{3}\right),$ then $p$ is equal to:

[JEE(Main)-2018]

Sol. (2)


Q. Let $\overrightarrow{\mathrm{a}}=\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}, \overrightarrow{\mathrm{c}}=\hat{\mathrm{j}}-\hat{\mathrm{k}}$ and a vector $\overrightarrow{\mathrm{b}}$ be such that $\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}=\overrightarrow{\mathrm{c}}$ and $\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}=3 .$ Then $|\overrightarrow{\mathrm{b}}|$ equals :

(1) $\sqrt{\frac{11}{3}}$

(2) $\frac{11}{\sqrt{3}}$

(2) $\frac{11}{\sqrt{3}}$

( 4)$\frac{11}{3}$

[JEE(Main)-2018]

Sol. (1)


Q. Let $\vec{u}$ be a vector coplanar with the vectors $\vec{a}=2 \hat{i}+3 \hat{j}-\hat{k}$ and $\vec{b}=\hat{j}+\hat{k} .$ If $\vec{u}$ is perpendicular to $\overrightarrow{\mathrm{a}}$ and $\overrightarrow{\mathrm{u} . \overrightarrow{\mathrm{b}}}=24,$ then $|\overrightarrow{\mathrm{u}}|^{2}$ is equal to-

(1) 315               (2) 256                  (3) 84                 (4) 336

[JEE(Main)-2018]

Sol. (4)


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