Vector Algebra – JEE Main Previous Year Question with Solutions
JEE Main Previous Year Question of Math with Solutions are available at eSaral. Practicing JEE Main Previous Year Papers Questions of mathematics will help the JEE aspirants in realizing the question pattern as well as help in analyzing weak & strong areas.eSaral helps the students in clearing and understanding each topic in a better way. eSaral is providing complete chapter-wise notes of Class 11th and 12th both for all subjects.Besides this, eSaral also offers NCERT Solutions, Previous year questions for JEE Main and Advance, Practice questions, Test Series for JEE Main, JEE Advanced and NEET, Important questions of Physics, Chemistry, Math, and Biology and many more.Download eSaral app for free study material and video tutorials.
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]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

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]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

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] Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... 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] Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... 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] Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... 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] Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... 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]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

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]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (3)$\overrightarrow{\mathrm{AD}}=\frac{\overrightarrow{\mathrm{AB}}+\overrightarrow{\mathrm{AC}}}{2}=4 \hat{\mathrm{j}}-\hat{\mathrm{j}}+4 \hat{\mathrm{k}}$$|\overrightarrow{\mathrm{AD}}|=\sqrt{33} Q. Let \overrightarrow{\mathrm{a}}=2 \hat{\mathrm{i}}-\hat{\mathrm{j}}+\hat{\mathrm{k}}, \overrightarrow{\mathrm{b}}=\hat{\mathrm{i}}+2 \hat{\mathrm{j}}-\hat{\mathrm{k}} and \overrightarrow{\mathrm{c}}=\hat{\mathrm{i}}+\hat{\mathrm{j}}-2 \hat{\mathrm{k}} be three vectors. A vectors. A vectors of the type \overrightarrow{\mathrm{b}}+\lambda \overrightarrow{\mathrm{c}} for some scalar \lambda, whose projection on \overrightarrow{\mathrm{a}} is of magnitude \sqrt{\frac{2}{3}}, is :(1) 2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}-3 \hat{\mathrm{k}}(2) 2 \hat{\mathrm{i}}+\hat{\mathrm{j}}+5 \hat{\mathrm{k}}(3) 2 \hat{\mathrm{i}}-\hat{\mathrm{j}}+5 \hat{\mathrm{k}}(4) 2 \hat{i}+3 \hat{j}+3 \hat{k} [JEE-MAINS Online 2013] Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. (1)\frac{|(\overline{\mathrm{b}}+\lambda \overline{\mathrm{c}}) \cdot \overline{\mathrm{a}}|}{|\overline{\mathrm{a}}|}=\sqrt{\frac{2}{3}}$$\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]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

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]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

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}))$$=(\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}) \cdot((\overrightarrow{\mathrm{b}} \overrightarrow{\mathrm{c}} \cdot \overrightarrow{\mathrm{a}}) \overrightarrow{\mathrm{c}}-[\overrightarrow{\mathrm{b}} \overrightarrow{\mathrm{c}} \cdot \overrightarrow{\mathrm{c}}] \overrightarrow{\mathrm{a}})=[\overrightarrow{\mathrm{a}} \overrightarrow{\mathrm{b}} \overrightarrow{\mathrm{c}}]^{2} Q. Let \vec{a}, \vec{b} and \vec{c} be three non-zero vectors such that no two of them are collinear and (\vec{a}, \times \vec{b}) \times \vec{c}=\frac{1}{3}|\vec{b}||\vec{c}| \vec{a} \cdot If \theta is the angle between vectors \vec{b} and \vec{c}, then a value of \sin \theta is : [JEE(Main)-2015] Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. (3)(\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{c}}) \overrightarrow{\mathrm{b}}-(\overrightarrow{\mathrm{b}} \cdot \overrightarrow{\mathrm{c}}) \overrightarrow{\mathrm{a}}=\frac{1}{3}|\overrightarrow{\mathrm{b}}||\overrightarrow{\mathrm{c}}| \overrightarrow{\mathrm{a}}$$\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.$$\Rightarrow \cos \theta=-\frac{1}{3} \Rightarrow \sin \theta=\frac{2 \sqrt{2}}{3} Q. Let \overrightarrow{\mathrm{a}}, \overrightarrow{\mathrm{b}} and \overrightarrow{\mathrm{c}} be three unit vectors such that \overrightarrow{\mathrm{a}} \times(\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}})=\frac{\sqrt{3}}{2}(\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}}) . If \overrightarrow{\mathrm{b}} is not parallel to \overrightarrow{\mathrm{c}} then the angle between a and \overrightarrow{\mathrm{b}} is :-(1) \frac{5 \pi}{6} (2) \frac{3 \pi}{4} (3) \frac{\pi}{2} (4) \frac{2 \pi}{3} [JEE(Main)-2016] Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. (1)\left(\overrightarrow{\mathrm{a} .} \overrightarrow{\mathrm{c}}-\frac{\sqrt{3}}{2}\right) \overrightarrow{\mathrm{b}}-\left(\overrightarrow{\mathrm{a} .} \overrightarrow{\mathrm{b}}+\frac{\sqrt{3}}{2}\right) \overrightarrow{\mathrm{c}}=0$$\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]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

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}$$\therefore|\overrightarrow{\mathrm{c}}|=2Now :|\overrightarrow{\mathrm{c}}-\overrightarrow{\mathrm{a}}|=3$$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]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

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]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

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]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

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]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

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]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (4)

• March 16, 2022 at 3:07 pm

dash kahike

• April 10, 2022 at 11:49 pm

Why Not latest questions?
19 20 and 21?

• December 3, 2021 at 5:19 pm

It is very helpful for jeee

• August 11, 2021 at 3:35 pm

Useful

• July 11, 2021 at 12:50 pm

Sol. Plz

• March 13, 2021 at 12:03 pm

• February 14, 2021 at 5:55 am

👌

• October 11, 2020 at 7:54 am

nice probles but easy to solve

• December 23, 2020 at 1:15 am

go for jee advanced ques then….those ques might be challenging for you.. ^_^

• September 30, 2020 at 4:16 pm

Last ques soln pls

• September 21, 2020 at 1:06 pm

Lalalllalllalallalalllla

• September 10, 2020 at 6:16 pm

Awesome 😀😇

• September 3, 2020 at 12:20 pm

Good

• August 30, 2020 at 6:38 pm

Last question solution pls

• August 28, 2020 at 7:05 am

Helped a lot

• August 1, 2020 at 12:05 pm