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 :
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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
$
