Vector JEE Questions PDF with Solutions: Download Practice Questions

Vector JEE Last 10 Years' Most Asked Topics

JEE Main Vector questions over the last 10 years have covered various topics. The table below shows the most asked JEE Main and JEE Advanced vector question topics.

JEE Main 2026 Vector Most Asked Concepts

Refer to the table given below for Vector PYQ JEE Main most asked concepts in 2026. In JEE Main, questions were asked from different topics.

Difficulty Level of Vector JEE Questions

JEE Main questions on vectors were direct, formula-based, and easy, while vector JEE Advanced questions were more conceptual and multi-concept type questions:

Also Check: JEE Main 2027 Study Guide Complete Notes, Important Concepts, Formulae and Practice Questions

Important Vector Formulas

Before you solve vector questions for JEE PDF, it’s important to learn the basic vector formulas. These formulae are used quite often in numericals.

1. For vector $\vec{A}=a \hat{i}+b \hat{j}+c \hat{k}$, the magnitude $|\vec{A}|=\sqrt{a^2+b^2+c^2}$

2. Unit Vector: $\hat{A}=\frac{\vec{A}}{|\vec{A}|}$

3. The position vector of $P(x, y, z)$, is $\overrightarrow{O P}=x \hat{i}+y \hat{j}+z \hat{k}$

4. $\vec{A}+\vec{B}=\left(a_1+b_1\right) \hat{i}+\left(a_2+b_2\right) \hat{j}+\left(a_3+b_3\right) \hat{k}$

5. $\vec{A}-\vec{B}=\left(a_1-b_1\right) \hat{i}+\left(a_2-b_2\right) \hat{j}+\left(a_3-b_3\right) \hat{k}$

6. Distance between 2 vectors

$A B=\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2+\left(z_2-z_1\right)^2}$

7. Dot products: $\vec{A} \cdot \vec{B}=|\vec{A}||\vec{B}| \cos \theta$

8. Angle between 2 vectors: $\cos \theta=\frac{\vec{A} \cdot \vec{B}}{|\vec{A}||\vec{B}|}$

9. Cross product: $|\vec{A} \times \vec{B}|=|\vec{A}||\vec{B}| \sin \theta$

10. Projection of One Vector on Another

• Scaler: $\frac{\vec{A} \cdot \vec{B}}{|\vec{B}|}$

• Vector: $\frac{\vec{A} \cdot \vec{B}}{|\vec{B}|^2} \vec{B}$

11. Vector Triple Product: $\vec{A} \times(\vec{B} \times \vec{C})=\vec{B}(\vec{A} \cdot \vec{C})-\vec{C}(\vec{A} \cdot \vec{B})$

Also Check: JEE 2027 Important Formulas

Vector JEE Questions with Solutions

The vector questions for JEE help you prepare effectively for exam. Solving vectors chapter jee questions is one of the best ways to prepare for IIT JEE.

Question 1: If $\vec{a}=\hat{i}+\hat{j}+\hat{k}, \vec{b}=\hat{j}-\hat{k}$ and $\vec{c}$ be three vectors such that $\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{c}}=\overrightarrow{\mathrm{b}}$ and $\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{c}}=3$, then $\overrightarrow{\mathrm{c}}(\overrightarrow{\mathrm{a}}-2 \overrightarrow{\mathrm{~b}})$ is equal to $\_\_\_\_$ .

(1) 3

(2) 1

(3) 2

(4) 4

Correct Answer: (1)

Solution:

Using triple product identity with $\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{c}}=\overrightarrow{\mathrm{b}}$

$\begin{aligned}
& \overrightarrow{\mathrm{a}} \times(\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{c}})=\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}} \\
& (\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{c}}) \overrightarrow{\mathrm{a}}-(\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{a}}) \overrightarrow{\mathrm{c}}=\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}
\end{aligned}$

As $\vec{a} \times \vec{b}=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ 0 & 1 & -1\end{array}\right|=-2 \hat{i}+\hat{j}+\hat{k}$
Substituting given values

$\begin{aligned}
& 3 \vec{a}-3 \vec{c}=-2 \hat{i}+j+k \\
& 3(\hat{i}+\hat{j}+\hat{k})+2 \hat{i}+\hat{j}+\hat{k}=3 \vec{c} \\
& \vec{c}=\frac{1}{3}(5 \hat{i}+2 \hat{j}+2 \hat{k})
\end{aligned}$

Now $\overrightarrow{\mathrm{c}} \cdot(\overrightarrow{\mathrm{a}}-2 \overrightarrow{\mathrm{~b}})=\frac{1}{3}(5 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+2 \hat{\mathrm{k}}) \cdot(\hat{\mathrm{i}}-\hat{\mathrm{j}}+3 \hat{\mathrm{k}})$

$=\frac{1}{3}[5-2+6]=3$

Question 2: Let $\vec{a}=2 \hat{i}+3 \hat{j}+3 \hat{k}$ and $\vec{b}=6 \hat{i}+3 \hat{j}+3 \hat{k}$. Then the square of the area of the triangle with adjacent sides determined by the vectors $(2 \vec{a}+3 \vec{b})$ and $(\overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{b}})$ is :

(1) 450

(2) 900

(3) 1800

(4) 2400

Correct Answer: (3)

Solution:

$\begin{aligned}
& 2 \vec{a}+3 \vec{b}=2(2 \hat{i}+3 \hat{j}+3 \hat{k})+3(6 \hat{i}+3 \hat{j}+3 \hat{k}) \\
& \Rightarrow 22 \hat{i}+15 \hat{j}+15 \hat{k} \\
& \vec{a}-\vec{b}=-4 \hat{i} \\
& \text { Area }=\frac{1}{2}|(2 \vec{a}+3 \vec{b}) \times(\vec{a}-\vec{b})| \\
& =\frac{1}{2}|-60 \hat{j}+60 \hat{k}|=\frac{1}{2} \sqrt{(60)^2 \times 2} \\
& A=\frac{60}{\sqrt{2}}
\end{aligned}$

Square of area is $\mathrm{A}^2=1800$

Question 3: Let $\vec{a}=4 \hat{i}-\hat{j}+3 \hat{k}, \vec{b}=10 \hat{i}+2 \hat{j}-\hat{k}$ and a vector $\vec{c}$ be such that $2(\vec{a} \times \vec{b})+3(\vec{b} \times \vec{c})=\overrightarrow{0}$. If $\vec{a} \cdot \vec{c}=15$, then $\overrightarrow{\mathrm{c}} \cdot(\hat{\mathrm{i}}+\hat{\mathrm{j}}-3 \hat{\mathrm{k}})$ is equal to :

(1) -6

(2) -5

(3) -4

(4) -3

Correct Answer: (2)

Solution:

$\begin{aligned}
& 2(\vec{a} \times \vec{b})-3(\vec{c} \times \vec{b})=\overrightarrow{0} \\
& (2 \vec{a}-3 \vec{c}) \times \vec{b}=\overrightarrow{0} \\
& \vec{b} \|(2 \vec{a}-3 \vec{c}) \\
& 2 \vec{a}-3 \vec{c}=\lambda \vec{b} \\
& \vec{c}=\frac{2 \vec{a}-\lambda \vec{b}}{3} \\
& \vec{a} \cdot \vec{c}=15 \\
& \left(\frac{2 \vec{a}-\lambda \vec{b}}{3}\right) \cdot \vec{a}=15 \\
& \frac{2|a|^2-\lambda \vec{b} \cdot \vec{a}}{3}=15 \\
& 2(26)-\lambda(40-2-3)=45 \\
& \lambda=\frac{1}{5} \\
& \Rightarrow \vec{c} \cdot(\hat{i}+\hat{j}-3 \hat{k})=\frac{\left(2(4 \hat{i}-\hat{j}+3 k)-\frac{1}{5}(10 \hat{i}+2 \hat{j}-k) \cdot(\hat{i}+\hat{j}-3 \hat{k})\right)}{3} \\
& =\frac{2(4-1-9)-\frac{1}{5}(10+2+3)}{3}=-5
\end{aligned}$

Question 4: Let O be the origin, $\overrightarrow{\mathrm{OP}}=\overrightarrow{\mathrm{a}}$ and $\overrightarrow{\mathrm{OQ}}=\overrightarrow{\mathrm{b}}$. If R is the point on $\overrightarrow{\mathrm{OP}}$ such that $\overrightarrow{\mathrm{OP}}=5 \overrightarrow{\mathrm{OR}}$, and M is the point such that $\overrightarrow{\mathrm{OQ}}=5 \overrightarrow{\mathrm{RM}}$, then $\overrightarrow{\mathrm{PM}}$ is equal to:

(1) $\frac{1}{5}(\vec{a}-4 \vec{b})$

(2) $\frac{1}{5}(\vec{b}-4 \vec{a})$

(3) $\frac{1}{5}(-\vec{a}+4 \vec{b})$

(4) $\frac{1}{5}(-\vec{b}+4 \vec{a})$

Correct Answer: (2)

Solution:

$\begin{aligned} & \overrightarrow{\mathrm{OR}}=\frac{\overrightarrow{\mathrm{OP}}}{5}=\frac{\vec{a}}{5} \\ & \overrightarrow{\mathrm{RM}}=\frac{\overrightarrow{\mathrm{OQ}}}{5}=\frac{\vec{b}}{5} \\ & \overrightarrow{\mathrm{OM}}-\overrightarrow{\mathrm{OR}}=\frac{\vec{b}}{5} \\ & \overrightarrow{\mathrm{OM}}=\frac{\vec{a}+\bar{b}}{5} \\ & \overrightarrow{\mathrm{PM}}=\overrightarrow{\mathrm{OM}}-\overrightarrow{\mathrm{OP}} \\ & \frac{\vec{a}+\vec{b}}{5}-\vec{a} \\ & =\frac{\vec{b}-4 \vec{a}}{5}\end{aligned}$

Question 5: Let $\vec{a}=\sqrt{7} \hat{i}+\hat{j}-\hat{k}$ and $\vec{b}=\hat{j}+2 \hat{k}$. If $\vec{r}$ is a vector such that $\overrightarrow{\mathrm{r}} \times \overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}=\overrightarrow{0}$ and $\overrightarrow{\mathrm{r}} \cdot \overrightarrow{\mathrm{a}}=0$, then $|3 \overrightarrow{\mathrm{r}}|^2$ is equal to :

(1) 44

(2) 54

(3) 86

(4) 132

Correct Answer: (1)

Solution:

$\begin{aligned} & \vec{r} \times \vec{a}-\vec{b} \times \vec{a}=\overrightarrow{0} \\ & (\vec{r}-\vec{b}) \times \vec{a}=\overrightarrow{0} \\ & \vec{r}-\vec{b}=\lambda \vec{a} \\ & \vec{r}=\vec{b}+\lambda \vec{a} \\ & \vec{r} . a=0 \Rightarrow \vec{a} \vec{b}+\lambda|\vec{a}|^2=0 \\ & \lambda=-\frac{\vec{a} \cdot \vec{b}}{|\vec{a}|^2}=-\frac{(1-2)}{9}=\frac{1}{9} \\ & \vec{r}=\vec{b}+\frac{\vec{a}}{9} \\ & |3 \vec{r}|^2=9|\vec{r}|^2=9\left(b^2+\frac{a^2}{81}+\frac{2(\vec{a} \cdot \vec{b})}{9}\right)=44\end{aligned}$

Question 6: Let $\hat{\mathrm{u}}$ and $\hat{\mathrm{v}}$ be unit vectors inclined at an acute angle such that $|\hat{\mathrm{u}} \times \hat{\mathrm{v}}|=\frac{\sqrt{3}}{2}$. If

$\overrightarrow{\mathrm{A}}=\lambda \hat{\mathrm{u}}+\hat{\mathrm{v}}+(\hat{\mathrm{u}} \times \hat{\mathrm{v}})$. Then $\lambda$ is equal to :

(1) $\frac{4}{3}(\vec{A} \cdot \hat{u})-\frac{2}{3}(\vec{A} \cdot \hat{v})$

(2) $\frac{2}{3}(\overrightarrow{\mathrm{~A}} \cdot \hat{\mathrm{u}})-\frac{1}{3}(\overrightarrow{\mathrm{~A}} \cdot \hat{\mathrm{v}})$

(3) $\frac{4}{3}(\vec{A} \cdot \hat{u})+\frac{2}{3}(\vec{A} \cdot \hat{v})$

(4) $(\vec{A} \cdot \hat{u})-\frac{1}{2}(\vec{A} \cdot \hat{v})$

Correct Answer: (1)

Solution:

$\begin{aligned} & |\hat{\mathrm{u}} \times \hat{\mathrm{v}}|=\frac{\sqrt{3}}{2} \\ & |\hat{\mathrm{u}}||\hat{\mathrm{v}}| \sin \theta=\frac{\sqrt{3}}{2} \quad \therefore \theta=\frac{\pi}{3} \\ & \text { and } \hat{\mathrm{u}} \cdot \hat{\mathrm{v}}=|\hat{\mathrm{u}}||\hat{\mathrm{v}}| \cos \frac{\pi}{3}=\frac{1}{2}\end{aligned}$

$\overrightarrow{\mathrm{A}}=\lambda \hat{\mathrm{u}}+\hat{\mathrm{v}}+\hat{\mathrm{u}} \times \hat{\mathrm{v}}$

Dot with û

$\begin{aligned}
& \overrightarrow{\mathrm{A}} \cdot \hat{\mathrm{u}}=\lambda(1)+\hat{\mathrm{u}} \cdot \hat{\mathrm{v}}+\hat{\mathrm{u}} \cdot(\hat{\mathrm{u}} \times \hat{\mathrm{v}}) \\
& \overrightarrow{\mathrm{A}} \cdot \hat{\mathrm{u}}=\lambda+\frac{1}{2} \\
& \Rightarrow 2 \overrightarrow{\mathrm{~A}} \cdot \hat{\mathrm{u}}=2 \lambda+1
\end{aligned}$

Dot equation (1) with $\hat{\mathrm{v}}$

$\begin{aligned}
& \overrightarrow{\mathrm{A}} \cdot \hat{\mathrm{v}}=\lambda(\hat{\mathrm{u}} \cdot \hat{\mathrm{v}})+\hat{\mathrm{v}} \cdot \hat{\mathrm{v}}+\hat{\mathrm{v}} \cdot(\hat{\mathrm{u}} \times \hat{\mathrm{v}}) \\
& \overrightarrow{\mathrm{A}} \cdot \hat{\mathrm{v}}=\frac{\lambda}{2}+1 \\
& \overrightarrow{\mathrm{~A}} \cdot \hat{\mathrm{v}}-\frac{\lambda}{2}=1
\end{aligned}$

From (2) and (3)

$\begin{aligned}
& 2 \overrightarrow{\mathrm{~A}} \cdot \hat{\mathrm{u}}-2 \lambda=\overrightarrow{\mathrm{A}} \cdot \hat{\mathrm{v}}-\frac{\lambda}{2} \\
& \therefore \lambda=\frac{4}{3} \overrightarrow{\mathrm{~A}} \cdot \hat{\mathrm{u}}-\frac{2}{3} \overrightarrow{\mathrm{~A}} \cdot \hat{\mathrm{v}}
\end{aligned}$

Question 7: Let $\overrightarrow{\mathrm{a}}_{\mathrm{k}}=\left(\tan \theta_{\mathrm{k}}\right) \hat{\mathrm{i}}+\hat{\mathrm{j}}$ and $\overrightarrow{\mathrm{b}}_{\mathrm{k}}=\hat{\mathrm{i}}-\left(\cos \theta_{\mathrm{k}}\right) \hat{\mathrm{j}}$, where $\theta_k=\frac{2^{k-1} \pi}{2^n+1}$, for some $n \in N, n>5$. Then the value of $\frac{\sum_{k=1}^n\left|\vec{a}_k\right|^2}{\sum_{k=1}^n\left|\vec{b}_k\right|^2}$ is $\_\_\_\_$ .

(1) 3

(2) 4

(3) 2

(4) 1

Correct Answer: (1)

Solution:

$\begin{aligned} & \left|a_k\right|^2=1+\tan ^2 \theta_k=\sec ^2 \theta_k \\ & \left|b_k\right|^2=1+\cot ^2 \theta_k=\operatorname{cosec}^2 \theta_k \\ & \text { Since, } \cot \theta-\tan \theta=2 \cot 2 \theta \\ & \therefore-\operatorname{cosec}^2 \theta-\sec ^2 \theta=-4 \operatorname{cosec}^2 2 \theta \\ & \sec ^2 \theta=4 \operatorname{cosec}^2 2 \theta-\operatorname{cosec}^2 \theta \\ & \Sigma \sec ^2 \theta_k=4 \Sigma \operatorname{cosec}^2 2 \theta_k-\Sigma \operatorname{cosec}^2 \theta_k \\ & \text { Now } \\ & \Sigma \operatorname{cosec} 2 \theta_k=\Sigma \operatorname{cosec}^2 \theta_k \\ & \text { since } \\ & \sum_{k=1}^n \operatorname{cosec}^2 \frac{2^k \pi}{2^n+1}=\sum_{k=1}^n \operatorname{cosec}^2 \frac{2^{k-1} \pi}{2^n+1} \\ & \text { because } \operatorname{cosec}^n \frac{2^n \pi}{2^n+1}=\operatorname{cosec} \frac{\pi}{2^n+1} \\ & \therefore \Sigma \sec ^2 \theta_k=3 \Sigma \operatorname{cosec}^2 \theta_k \\ & \frac{\sum \sec ^2 \theta_k}{\sum \operatorname{cosec}^2 \theta_k}=3\end{aligned}$

Question 8: Two adjacent sides of a parallelogram PQRS are given by $\overrightarrow{\mathrm{PQ}}=\hat{\mathrm{j}}+\hat{\mathrm{k}}$ and $\overrightarrow{\mathrm{PS}}=\hat{\mathrm{i}}-\hat{\mathrm{j}}$. If the side PS is rotated about the point P by an acute angle $\alpha$ in the plane of the parallelogram so that it becomes perpendicular to the side PQ , then $\sin ^2\left(\frac{5 \alpha}{2}\right)-\sin ^2\left(\frac{\alpha}{2}\right)$ is equal to :

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

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

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

(4) $\frac{2 \sqrt{3}}{5}$

Correct Answer: (2)

Solution:

Let angle between $\overrightarrow{\mathrm{PQ}} \& \overrightarrow{\mathrm{PS}}$ is $\theta$

$\begin{aligned}
& \cos \theta=\frac{\overrightarrow{\mathrm{PQ}} \cdot \overrightarrow{\mathrm{PS}}}{|\overrightarrow{\mathrm{PQ}}||\overrightarrow{\mathrm{PS}}|} \\
& \cos \theta=\frac{0-1+0}{\sqrt{2} \sqrt{2}} \\
& \cos \theta=\frac{-1}{2} \\
& \theta=\frac{2 \pi}{3} \\
& \alpha=\frac{2 \pi}{3}-\frac{\pi}{2} \text { So } \alpha=\frac{\pi}{6} \\
& \sin ^2 \frac{5 \alpha}{2}-\sin ^2 \frac{\alpha}{2}=\sin \left(\frac{5 \alpha}{2}+\frac{\alpha}{2}\right) \sin \left(\frac{5 \alpha}{2}-\frac{\alpha}{2}\right) \\
& =\sin \frac{6 \alpha}{2} \times \sin \frac{4 \alpha}{2} \\
& =\sin 3 \alpha \times \sin 2 \alpha \\
& =\sin \frac{\pi}{2} \times \sin \frac{\pi}{3} \\
& =1 \times \frac{\sqrt{3}}{2}=\frac{\sqrt{3}}{2}
\end{aligned}$

Question 9: Let the vectors $\vec{a}=-\hat{i}+\hat{j}+3 \hat{k}$ and $\vec{b}=\hat{i}+3 \hat{j}+\hat{k}$. For some $\lambda, \mu \in \mathbf{R}$, let $\overrightarrow{\mathrm{c}}=\lambda \overrightarrow{\mathrm{a}}+\mu \overrightarrow{\mathrm{b}}$. If $\vec{c} \cdot(3 \hat{i}-6 \hat{j}+2 \hat{k})=10$ and $\vec{c} \cdot(\hat{i}+\hat{j}+\hat{k})=-2$, then $|\overrightarrow{\mathrm{c}}|^2$ is equal to :

(1) 8

(2) 12

(3) 14

(4) 15

Correct Answer: (2)

Solution:

$\begin{aligned}
& \overrightarrow{\mathrm{c}}=\lambda \overrightarrow{\mathrm{a}}+\mu \overrightarrow{\mathrm{b}} \\
& \overrightarrow{\mathrm{c}}=(\mu-\lambda) \hat{\mathrm{i}}+(\lambda+3 \mu) \hat{\mathrm{j}}+(3 \lambda+\mu) \hat{\mathrm{k}} \\
& \overrightarrow{\mathrm{c}} \cdot(3 \hat{\mathrm{i}}-6 \hat{\mathrm{j}}+2 \hat{\mathrm{k}})=10 \\
& 3(\mu-\lambda)-6(\lambda+3 \mu)+2(3 \lambda+\mu)=10 \\
& -3 \lambda-13 \mu=10 \\
& \overrightarrow{\mathrm{c}} \cdot(\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}})=-2 \\
& 5 \mu+3 \lambda=-2
\end{aligned}$

Solving (1) & (2)

$\begin{aligned}
& \lambda=1, \mu=-1 \\
& \overrightarrow{\mathrm{c}}=-2 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}+2 \hat{\mathrm{k}} \\
& |\overrightarrow{\mathrm{c}}|^2=12
\end{aligned}$

Question 10: If $\vec{a}$ and $\vec{b}$ are two vectors such that $|\vec{a}|=2$ and $|\overrightarrow{\mathrm{b}}|=3$, then the maximum value of $3|(3 \vec{a}+2 \vec{b})|+4|(3 \vec{a}-2 \vec{b})|$ is :

(1) 30

(2) 36

(3) 60

(4) 72

Correct Answer: (3)

Solution:

$\begin{aligned} & \mathrm{E}=3 \sqrt{9 \mathrm{a}^2+4 \mathrm{~b}^2+12 \mathrm{a} \overrightarrow{\mathrm{b}}}+4 \sqrt{9 \mathrm{a}^2+4 \mathrm{~b}^2-12 \overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}} \\ & =3 \sqrt{36+36+12 \times 6 \cos \theta}+4 \sqrt{36+36-72 \cos \theta} \\ & =3 \sqrt{72+72 \cos \theta}+4 \sqrt{72-72 \cos \theta} \\ & =18 \sqrt{2} \sqrt{1+\cos \theta}+24 \sqrt{2} \sqrt{1-\cos \theta} \\ & =18 \sqrt{2} \sqrt{2} \cos \frac{\theta}{2}+24 \sqrt{2}\left(\sqrt{2} \sin \frac{\theta}{2}\right) \\ & =36\left(\cos \frac{\theta}{2}\right)+48 \sin \left(\frac{\theta}{2}\right) \\ & \mathrm{E}_{\max }=\sqrt{(36)^2+(48)^2}=\sqrt{6^4+6^2 \times 8^2}=60\end{aligned}$

Question 11: Let P be a point in the plane vectors $A B=3 \hat{i}+\hat{j}-\hat{k}$ and $A C=\hat{i}-\hat{j}+3 \hat{k}$ such that P is equidistant from the lines AB and AC . If $|A P|=\frac{\sqrt{5}}{2}$, then the area of the triangle ABP is :

(1) $\frac{\sqrt{26}}{4}$

(2) 2

(3) $\frac{\sqrt{30}}{4}$

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

Correct Answer: (3)

Solution:

$\begin{aligned} & \cos 2 \theta=\frac{-1}{11} \Rightarrow \operatorname{Sin} \theta=\sqrt{\frac{6}{11}} \\ & \text { Area }=\frac{1}{2} \sqrt{11} \times \sqrt{5} / 2 \times \sqrt{\frac{6}{11}}=\frac{\sqrt{30}}{4}\end{aligned}$

Question 12: For three unit vectors $a, b, c$ satisfying $|a-b|^2+|b-c|^2+|c-a|^2=9$ and $|2 a+k b+k c|=3$, the positive value of $k$ is

(1) 3

(2) 5

(3) 4

(4) 6

Correct Answer: (2)

Solution:

$6-2(\bar{a} \cdot \bar{b}+\bar{b} \cdot \bar{c}+\bar{c} \cdot \bar{a})=9$

$\bar{a} \cdot \bar{b}+\bar{b} \cdot \bar{c}+\bar{c} \cdot \bar{a}=\frac{-3}{2}$

$|2 \bar{a}+k \bar{b}+k \bar{c}|=3$

$|2 \bar{a}+k(-\bar{a})|=3$

$|(2-k)||(\bar{a})|=3$

$2-k= \pm 3 \Rightarrow k=5,-1$

Question 13: Let $\vec{a}$ and $\vec{b}$ be two unit vectors such that the angle between them is $\frac{\pi}{3}$. If $\lambda \vec{a}+2 \vec{b}$ and $3 \vec{a}-\lambda \vec{b}$ are perpendicular to each other, then the number of values of in [-1,3] is:

(1) 3

(2) 2

(3) 1

(4) 0

Correct Answer: (4)

Solution:

We are given that $ \vec{a} $ and $ \vec{b} $ are unit vectors and the angle between them is $ \frac{\pi}{3} $, so:

$
\vec{a} \cdot \vec{b} = \cos\left(\frac{\pi}{3}\right) = \frac{1}{2}
$

We are also given that the vectors $ \lambda \vec{a} + 2 \vec{b} $ and $ 3 \vec{a} - \lambda \vec{b} $ are perpendicular. Therefore:

$
(\lambda \vec{a} + 2 \vec{b}) \cdot (3 \vec{a} - \lambda \vec{b}) = 0
$

Expanding the dot product:

$
= \lambda \cdot 3 (\vec{a} \cdot \vec{a}) - \lambda^2 (\vec{a} \cdot \vec{b}) + 6 (\vec{a} \cdot \vec{b}) - 2\lambda (\vec{b} \cdot \vec{b})
$

Since $ \vec{a} \cdot \vec{a} = \vec{b} \cdot \vec{b} = 1 $ and $ \vec{a} \cdot \vec{b} = \frac{1}{2} $, we get:

$
3\lambda - \lambda^2 \cdot \frac{1}{2} + 6 \cdot \frac{1}{2} - 2\lambda = 0
$

$
3\lambda - \frac{\lambda^2}{2} + 3 - 2\lambda = 0
$

$
\lambda - \frac{\lambda^2}{2} + 3 = 0
\Rightarrow -\frac{\lambda^2}{2} + \lambda + 3 = 0
$

Multiply through by 2 to eliminate the denominator:

$
-\lambda^2 + 2\lambda + 6 = 0
\Rightarrow \lambda^2 - 2\lambda - 6 = 0
$

Solving this quadratic equation:

$
\lambda = \frac{2 \pm \sqrt{(-2)^2 + 4 \cdot 1 \cdot 6}}{2} = \frac{2 \pm \sqrt{28}}{2} = \frac{2 \pm 2\sqrt{7}}{2} = 1 \pm \sqrt{7}
$

We now check whether these values lie within the interval $ [-1, 3] $:

$
1 + \sqrt{7} \approx 1 + 2.645 \approx 3.645 \notin [-1, 3]
$
$
1 - \sqrt{7} \approx 1 - 2.645 \approx -1.645 \notin [-1, 3]
$

None of the values lie in the interval $ [-1, 3] $. So, the number of values of $ \lambda $ in the interval is: 0

Hence, the answer is option (4).

Question 14: Let $\overline{\mathrm{a}}=2 \hat{\mathrm{i}}+\hat{\mathrm{j}}-2 \hat{\mathrm{k}}, \overline{\mathrm{b}}=\hat{\mathrm{i}}+\hat{\mathrm{j}}$ and $\overline{\mathrm{c}}=\overline{\mathrm{a}} \times \overline{\mathrm{b}}$. Let $\overline{\mathrm{d}}$ be a vector such that $|\overline{\mathrm{d}}-\overline{\mathrm{a}}|=\sqrt{11},|\overline{\mathrm{c}} \times \overline{\mathrm{d}}|=3$ and angle between $\overline{\mathrm{c}}$ and $\overline{\mathrm{d}}$ is $\frac{\pi}{4}$. Then $\overline{\mathrm{a}} . \overline{\mathrm{d}}$ is equal to:

(1) 11

(2) 0

(3) 1

(4) 3

Correct Answer: (2)

Solution:

$\begin{aligned} & \varepsilon=\left|\begin{array}{ccc}\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ 2 & 1 & -2 \\ 1 & 1 & 0\end{array}\right|=\hat{\mathrm{i}}(2)-\hat{\mathrm{j}}(2)+\hat{\mathrm{k}}(1) \\ & |\overline{\mathrm{c}} \times \overline{\mathrm{d}}|=3 \Rightarrow|\overline{\mathrm{c}}||\overline{\mathrm{d}}| \sin (\overline{\mathrm{c}}, \overline{\mathrm{d}})=3 \Rightarrow 3|\overline{\mathrm{~d}}| \frac{1}{\sqrt{2}}=3 \\ & |\mathrm{~d}|=\sqrt{2} \\ & |\overline{\mathrm{~d}}-\overline{\mathrm{a}}|=\sqrt{11} \Rightarrow \overline{\mathrm{~d}}^2+\overline{\mathrm{a}}^2-2 \overline{\mathrm{a}} \cdot \overline{\mathrm{d}}=1 \overline{1} \Rightarrow \overline{\mathrm{~d}}^2+9-2 \overline{\mathrm{a}} \cdot \overline{\mathrm{d}}=11 \Rightarrow \overline{\mathrm{a}} \cdot \overline{\mathrm{d}}=0\end{aligned}$

Question 15: 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) $\frac{1}{\sqrt{2}}(-\hat{\mathrm{i}}+\hat{\mathrm{k}})$

Correct Answer: (4)

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).

Also check: JEE Mains 2027: Maths Set of 5 Sample Paper with Solution

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