Vectors JEE Mains Questions - Practice MCQs with Solutions

Mathematics: Vector PYQs of JEE Main

By practicing previous year questions, you get a clear understanding of the question pattern, the techniques to solve them, and the scoring strategies that can help you perform better in the exam. Below, we have provided a set of Vectors JEE Main previous year questions for your practice.

Question 1: Let $\vec{a}=2 \hat{i}-3 \hat{j}+ \hat k, \vec{b}=3 \hat{i}+2 \hat{j}+5 \hat k$ and a vector $\vec{c}$ be such that $(\vec{a}-\vec{c}) \times \vec{b}=-18 \hat{i}-3 \hat{j}+12 \hat k$ and $\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{c}}=3$. If $\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}}=\overrightarrow{\mathrm{d}}$, then $|\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{d}}|$ is equal to :

1) 18

2) 12

3) 9

4) 15

Solution:

Given:
$\vec{a} = 2\hat{i} - 3\hat{j} + \hat{k}, \quad \vec{b} = 3\hat{i} + 2\hat{j} + 5\hat{k}$

Calculate $\vec{a} \times \vec{b}$:
$\vec{a} \times \vec{b} = \begin{vmatrix}
\hat{i} & \hat{j} & \hat{k} \\
2 & -3 & 1 \\
3 & 2 & 5
\end{vmatrix} = -17\hat{i} - 7\hat{j} + 13\hat{k}$

Given:
$(\vec{a} - \vec{c}) \times \vec{b} = -18\hat{i} - 3\hat{j} + 12\hat{k}$

Then:
$(\vec{a} - \vec{c}) \times \vec{b} = \vec{a} \times \vec{b} - \vec{c} \times \vec{b}$

So:
$\vec{b} \times \vec{c} = -18\hat{i} - 3\hat{j} + 12\hat{k} - (-17\hat{i} - 7\hat{j} + 13\hat{k})$

$\vec{b} \times \vec{c} = -\hat{i} + 4\hat{j} - \hat{k}$

Let $\vec{d} = \vec{b} \times \vec{c}$

Now:
$\vec{a} \cdot \vec{d} = (2\hat{i} - 3\hat{j} + \hat{k}) \cdot (-\hat{i} + 4\hat{j} - \hat{k})$

$= -2 - 12 - 1 = -15$

Therefore:
$|\vec{a} \cdot \vec{d}| = 15$

Hence, the answer is option 4.

Question 2: Let $\vec{a}$ and $\vec{b}$ be the vectors of the same magnitude such that $\frac{|\vec{a}+\vec{b}|+|\vec{a}-\vec{b}|}{|\vec{a}+\vec{b}|-|\vec{a}-\vec{b}|}=\sqrt{2}+1$. Then $\frac{|\vec{a}+\vec{b}|^2}{|\vec{a}|^2}$ is :

1) $2+\sqrt{2}$

2) $2+4 \sqrt{2}$

3) $4+2 \sqrt{2}$

4) $1+\sqrt{2}$

Solution:

$\frac{|\bar{a}+\bar{b}|+|\bar{a}-\bar{b}|}{|\bar{a}+\bar{b}|-|\bar{a}-\bar{b}|}=\sqrt{2}+1$

Apply componendo and dividendo

$\begin{aligned} & \Rightarrow \frac{2|\bar{a}+\bar{b}|}{2|\bar{a}-\bar{b}|}=\frac{\sqrt{2}+2}{\sqrt{2}} \\ & \Rightarrow|\overline{\mathrm{a}}+\overline{\mathrm{b}}|=(1+\sqrt{2})|\overline{\mathrm{a}}-\overline{\mathrm{b}}| \\ & \Rightarrow|\overline{\mathrm{a}}+\overline{\mathrm{b}}|^2=(3+2 \sqrt{2})|\overline{\mathrm{a}}-\overline{\mathrm{b}}|^2 \\ & \Rightarrow 2|\overline{\mathrm{a}}|^2+2 \overline{\mathrm{a}} \cdot \overline{\mathrm{b}}=(3+2 \sqrt{2})\left(2|\overline{\mathrm{a}}|^2-2 \overline{\mathrm{a}} \cdot \overline{\mathrm{b}}\right) \\ & \Rightarrow 2|\overline{\mathrm{a}}|^2(2+2 \sqrt{2})=2 \overline{\mathrm{a}} \cdot \overline{\mathrm{b}}(4+2 \sqrt{2}) \\ & \Rightarrow \frac{\overline{\mathrm{a}} \cdot \overline{\mathrm{b}}}{|\overline{\mathrm{a}}|^2}=\frac{2+2 \sqrt{2}}{4+2 \sqrt{2}}=\frac{1}{\sqrt{2}}\end{aligned}$

Now

$\begin{aligned}
& \frac{|\overline{\mathrm{a}}+\overline{\mathrm{b}}|^2}{|\overline{\mathrm{a}}|^2}=1+\frac{|\overline{\mathrm{b}}|^2}{|\overline{\mathrm{a}}|^2}+\frac{2 \overline{\mathrm{a}} \cdot \overline{\mathrm{~b}}}{|\overline{\mathrm{a}}|^2} \\
& =1+1+2\left(\frac{1}{\sqrt{2}}\right)=2+\sqrt{2}
\end{aligned}$

Hence, the correct answer is option 1

Question 3: Consider two vectors $\vec{u}=3 \hat{i}-\hat{j}$ and $\vec{v}=2 \hat{i}+\hat{j}-\lambda \hat{k}, \lambda>0$. The angle between them is given by $\cos ^{-1}\left(\frac{\sqrt{5}}{2 \sqrt{7}}\right)$. Let $\vec{v}=\vec{v}_1+\vec{v}_2$, where $\vec{v}_1$ is parallel to $\vec{u}$ and $\overrightarrow{v_2}$ is perpendicular to $\vec{u}$. Then the value $\left|\overrightarrow{v_1}\right|^2+\left|\overrightarrow{v_2}\right|^2$ is equal to

1) $\frac{23}{2}$

2) $\frac{25}{2}$

3) 10

4) 14

Solution:

$\begin{aligned} & \vec{u} \cdot \vec{v}=|u| \cdot|v| \cdot \cos \theta \\ & \Rightarrow 6-1=\sqrt{10} \cdot \sqrt{5+\lambda^2} \cdot \frac{\sqrt{5}}{2 \sqrt{7}} \\ & \Rightarrow 1=\sqrt{2} \cdot \sqrt{5+\lambda^2} \cdot \frac{1}{2 \sqrt{7}} \\ & \Rightarrow 14=5+\lambda^2 \\ & \Rightarrow \lambda^2=9 \\ & \Rightarrow \lambda=3\end{aligned}$

$\begin{aligned} & v_1=k \vec{u} \\ & \vec{v}=\vec{v}_1+\vec{v}_2 \\ & \Rightarrow \vec{v}=k \vec{u}+\vec{v}_2 \\ & \vec{v} \cdot \vec{u}=k \cdot|\vec{u}|^2\end{aligned}$

$\begin{aligned} & \Rightarrow 5=k \cdot 10 \Rightarrow k=\frac{1}{2} \\ & \therefore \quad \vec{v}_1=\frac{\vec{u}}{2}=\frac{3 \hat{i}}{2}-\frac{\hat{j}}{2}\end{aligned}$

$\begin{aligned} & \left|\vec{v}_1\right|^2=\frac{10}{4} \\ & \vec{v}_2=\vec{v}-\vec{v}_1\end{aligned}$

$=\frac{1}{2} \hat{i}+\frac{3 \hat{j}}{2}-3 \hat{k}$

$\begin{aligned} & \left|\vec{v}_2\right|^2=\frac{10}{4}+9 \\ & \left|\vec{v}_1\right|^2+\left|\vec{v}_2\right|^2=\frac{10}{4}+\frac{10}{4}+9=14\end{aligned}$

Hence, the correct answer is option 4

Question 4: Let the angle $\theta, 0<\theta<\frac{\pi}{2}$ between two unit vectors $\hat{a}$ and $\hat{b}$ be $\sin ^{-1}\left(\frac{\sqrt{65}}{9}\right)$. If the vector $\vec{c}=3 \hat{a}+6 \hat{b}+9(\hat{a} \times \hat{b})$, then the value of $9(\vec{c} \cdot \hat{a})-3(\vec{c} \cdot \hat{b})$ is

1) 31

2) 29

3) 24

4) 27

Solution:

To solve the problem, we begin with

$\vec{c}=3 \hat{a}+6 \hat{b}+9(\hat{a} \times \hat{b})$

Let's calculate:

$\begin{aligned}
& \vec{c} \cdot \hat{a}=3+6(\hat{a} \cdot \hat{b}) \\
& \vec{c} \cdot \hat{b}=3(\hat{a} \cdot \hat{b})+6
\end{aligned}$

We need to find the value of $9(\vec{c} \cdot \hat{a})-3(\vec{c} \cdot \hat{b})$.

Substituting the expressions, we get:

$\begin{aligned} & 9(\vec{c} \cdot \hat{a})=9(3+6 \hat{a} \cdot \hat{b})=27+54(\hat{a} \cdot \hat{b}) \\ & 3(\vec{c} \cdot \hat{b})=3(3 \hat{a} \cdot \hat{b}+6)=9 \hat{a} \cdot \hat{b}+18\end{aligned}$

Therefore,

$9(\vec{c} \cdot \hat{a})-3(\vec{c} \cdot \hat{b})=(27+54(\hat{a} \cdot \hat{b}))-(9 \hat{a} \cdot \hat{b}+18)$

Simplifying,

$\begin{aligned}
& =27-18+54(\hat{a} \cdot \hat{b})-9(\hat{a} \cdot \hat{b}) \\
& =9+45(\hat{a} \cdot \hat{b})
\end{aligned}$

Given $\sin \theta=\frac{\sqrt{65}}{9}$, and knowing that for unit vectors $\cos \theta=\hat{a} \cdot \hat{b}$, we use the identity $(\cos \theta)^2=1-(\sin \theta)^2$ :

$\begin{aligned}
& (\hat{a} \cdot \hat{b})^2=1-\left(\frac{\sqrt{65}}{9}\right)^2 \\
& =1-\frac{65}{81} \\
& =\frac{16}{81}
\end{aligned}$

Thus, $\hat{a} \cdot \hat{b}=\frac{4}{9}$.
Substitute back into the equation:

$9+45 \times \frac{4}{9}=9+20=29$

Hence, the correct answer is option 2

Question 5: If $\overrightarrow{\mathrm{a}}$ is a nonzero vector such that its projections on the vectors $2 \hat{i}-\hat{j}+2 \hat{k}, \hat{i}+2 \hat{j}-2 \hat{k}$ and $\hat{k}$ are equal, then a unit vector along $\overrightarrow{\mathrm{a}}$ is

1) $\frac{1}{\sqrt{155}}(-7 \hat{i}+9 \hat{j}+5 \hat{k})$

2) $\frac{1}{\sqrt{155}}(-7 \hat{i}+9 \hat{j}-5 \hat{k})$

3) $\frac{1}{\sqrt{155}}(7 \hat{i}+9 \hat{j}-5 \hat{k})$

4) $\frac{1}{\sqrt{155}}(7 \hat{i}+9 \hat{j}+5 \hat{k})$

Solution:

Projection of $\vec{a}$ on $\vec{v}$

$=\frac{\vec{a} \cdot \vec{v}}{|\vec{v}|}-$

$\begin{aligned} & \Rightarrow \frac{\vec{a} \cdot(2 \hat{i}-\hat{j}+2 \hat{k})}{3}=\frac{\vec{a} \cdot \hat{k}}{1}=\frac{\vec{a} \cdot(\hat{i}+2 \hat{j}-2 \hat{k})}{3} \\ & \Rightarrow \vec{a} \cdot(2 \hat{i}-\hat{j}-\hat{k})=0 \text { and } \vec{a} \cdot(\hat{i}+2 \hat{j}-5 \hat{k})=0 \\ & \Rightarrow \vec{a} \perp(2 \hat{i}-\hat{j}-\hat{k}) \text { and }(\hat{i}+2 \hat{j}-5 \hat{k}) \\ & \Rightarrow \vec{a} \|(2 \hat{i}-\hat{j}-\hat{k}) \times(\hat{i}+2 \hat{j}-5 \hat{k})\end{aligned}$

$\begin{aligned} & \Rightarrow \vec{a}= \pm k\left|\begin{array}{ccc}\hat{i} & -\hat{j} & \hat{k} \\ 2 & -1 & -1 \\ 1 & 2 & -5\end{array}\right|= \pm k(7 \hat{i}+9 \hat{j}-5 \hat{k}) \\ & \Rightarrow \text { Unit vector will be } \frac{1}{\sqrt{155}}(7 \hat{i}+9 \hat{j}+5 \hat{k})\end{aligned}$

Hence, the correct answer is option 4

Question 6: Let the area of the triangle formed by the lines $x+2=y-1=z, \frac{x-3}{5}=\frac{y}{-1}=\frac{z-1}{1}$ and $\frac{x}{-3}=\frac{y-3}{3}=\frac{z-2}{1}$ be $A$. Then $A^2$ is equal to ______

Solution:

$L_1=\frac{x+2}{1}=\frac{y-1}{1}=\frac{z}{1}=\lambda$, any point on it $(\lambda-2, \lambda$ $+1, \lambda)$

$L_2=\frac{x-3}{5}=\frac{y}{-1}=\frac{z-1}{1}=\mu$, any point on it $(5 \mu+3$, $-\mu, \mu+1)$

$L_3=\frac{x}{-3}=\frac{y-3}{3}=\frac{z-2}{1}=k$, any point on it $(-3 k$, $3 k+3, k+2)$

$P \equiv$ point of intersection of $L_1$ and $L_2=(-2,1,0)$
$Q=$ point of intersection of $L_1$ and $L_3=(0,3,2)$
$R \equiv$ point of intersection of $L_2$ and $L_3=(3,0,1)$

$\begin{aligned} & \overline{P Q}=2 \hat{i}+2 \hat{j}+2 \hat{k} \\ & \overline{P R}=5 \hat{i}-\hat{j}+\hat{k}\end{aligned}$

$A=\frac{1}{2}|\overline{P Q} \times \overline{P R}|=\sqrt{56}$

$A^2=56$
Hence, the answer is (56).

Question 7:Let $\overrightarrow{\mathrm{a}}=\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\hat{\mathrm{k}}$ and $\overrightarrow{\mathrm{b}}=2 \hat{\mathrm{i}}+\hat{\mathrm{j}}-\hat{\mathrm{k}}$. Let $\hat{\mathrm{c}}$ be a unit vector in the plane of the vectors $\vec{a}$ and $\vec{b}$ and be perpendicular to $\vec{a}$. Then such a vector $\hat{\mathbf{c}}$ is :

1) $\frac{1}{\sqrt{5}}(\hat{\mathrm{j}}-2 \hat{\mathrm{k}})$

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

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

4) $\frac{1}{\sqrt{2}}(-\hat{\mathrm{i}}+\hat{\mathrm{k}})$

Solution:

$\begin{aligned} & \vec{c}=x \vec{a}+y \vec{b} \\ & \vec{c}=x(\hat{i}+2 \hat{j}+\hat{k})+y(2 \hat{i}+\hat{j}-\hat{k}) \\ & \vec{a} \cdot \vec{c}=(\hat{i}+2 \hat{j}+\hat{k}) \cdot(x(\hat{i}+2 \hat{j}+\hat{k})+y(2 \hat{i}+\hat{j}-\hat{k})) \\ & (\hat{i}+2 \hat{j}+\hat{k}) \cdot(x \hat{i}+2 x \hat{j}+x \hat{k})+2 y \hat{i}+y \hat{j}-y \hat{k}=0 \\ & \Rightarrow \quad(x+2 y)+2(x+9)+(x-y)=0 \\ & \Rightarrow \quad y=-2 x \\ & \therefore \quad \vec{c}=x(-3 \hat{i}+3 \hat{k}) \\ & |\vec{c}|=|x| \sqrt{9+9}=3|x| \sqrt{2} \\ & \therefore|\vec{c}|=1 \\ & 3|x| \sqrt{2}=1 \\ & |x|=\frac{1}{3 \sqrt{2}} \\ & \text { Let } x=\frac{1}{3 \sqrt{2}} \\ & \vec{c}=\frac{1}{3 \sqrt{2}}(-3 \hat{i}+3 \hat{k}) \\ & \text { or } \vec{c}=\frac{1}{\sqrt{2}}(-\hat{i}+\hat{k})\end{aligned}$

Hence, the answer is option (4).

Question 8: Let the three sides of a triangle $A B C$ be given by the vectors $2 \hat{i}-\hat{j}+\hat{k}, \hat{i}-3 \hat{j}-5 \hat{k}$ and $3 \hat{i}-4 \hat{j}-4 \hat{k}$. Let G be the centroid of the triangle ABC . Then $6\left(|\overrightarrow{\mathrm{AG}}|^2+|\overrightarrow{\mathrm{BG}}|^2+|\overrightarrow{\mathrm{CG}}|^2\right)$ is equal to ______

Solution:

The position vector of a point is usually taken from the origin. If a point $ A $ is at the origin, then $ \vec{A} = \vec{0} $.

For two points $ A $ and $ B $, the vector $ \vec{AB} $ is given by:
$
\vec{AB} = \vec{B} - \vec{A}
$

The centroid $ G $ of a triangle with vertices having position vectors $ \vec{A}, \vec{B}, \vec{C} $ is:
$
\vec{G} = \frac{\vec{A} + \vec{B} + \vec{C}}{3}
$

The vector from point $ A $ to centroid $ G $ is:
$
\vec{AG} = \vec{G} - \vec{A}
$

The square of the magnitude (length) of a vector $ \vec{v} = a\hat{i} + b\hat{j} + c\hat{k} $ is:
$
|\vec{v}|^2 = a^2 + b^2 + c^2
$

Key Benefits of Studying Vectors for JEE Main

Simplifies 3D Geometry Problems – Vectors are useful to make it simple to calculate distances, angles, and intersections of points, lines, and planes. Suppose the shortest distance between skew lines can be easily and effortlessly found by utilizing the cross product.

Boosts Visualization Skills – As we know vectors deal with both magnitude and direction which helps students to think in three dimensions, which is crucial for solving 3D questions and helps in solving geometry-based problems.

Useful for Maximum/Minimum Problems – When you optimize questions like the maximum area of a triangle or the volume of a parallelepiped, it can be solved directly with the use of dot and cross product formulas, which saves time and effort.

Bridges Algebra and Geometry – Dot products and cross products connect algebraic calculations with geometric interpretation, which helps to explain angles, perpendicularity, and areas easily.

Continuously Tested Topic – Vectors come repeatedly, which makes them a high-scoring topic. When you master it, it ensures better accuracy and confidence in 3D geometry and coordinate geometry problems.

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