JEE Main Motion in Plane PYQ

1. (JEE Main 2024 (Online) 5th April Morning Shift)

The angle between vector $\vec{Q}$ and the resultant of $(2 \vec{Q} + 2 \vec{P})$ and $(2 \vec{Q} - 2 \vec{P})$ is :

A. $\tan^{- 1} (P / Q)$

B. $\tan^{- 1} \frac{(2 \vec{Q} - 2 \vec{P})}{2 \vec{Q} + 2 \vec{P}}$

C. $\tan^{- 1} \frac{(2 \vec{Q} - 2 \vec{P})}{2 \vec{Q} + 2 \vec{P}}$

D. $\tan^{- 1} (2 Q / P)$

Correct answer option is (B)

To find the angle between the vector $\vec{Q}$ and the resultant of $(2 \vec{Q} + 2 \vec{P})$ and $(2 \vec{Q} - 2 \vec{P})$ , we first find the resultant vector of $(2 \vec{Q} + 2 \vec{P})$ and $(2 \vec{Q} - 2 \vec{P})$ .

The resultant vector of $(2 \vec{Q} + 2 \vec{P})$ and $(2 \vec{Q} - 2 \vec{P})$ can be simply found by adding these two vectors:

Resultant = (2 \vec{Q} + 2 \vec{P}) + (2 \vec{Q} - 2 \vec{P}) = 4 \vec{Q}

Now, we need to find the angle between the vector $\vec{Q}$ and this resultant vector $4 \vec{Q}$ . Since the resultant vector is just a scaled version of $\vec{Q}$ , they are in the same direction. The angle between any vector and another vector that is a scaled version of the first vector is always $0^{\circ}$ , because they are parallel to each other.

Therefore, the angle between $\vec{Q}$ and the resultant of $(2 \vec{Q} + 2 \vec{P})$ and $(2 \vec{Q} - 2 \vec{P})$ is $0^{\circ}$ .

So, the correct option is:

Option B: $0^{\circ}$

2. (JEE Main 2024 (Online) 31st January Evening Shift)

If two vectors $\vec{A}$ and $\vec{B}$ having equal magnitude $R$ are inclined at angle $θ$ , then

A. $| \vec{A} + \vec{B} | = 2 R \cos (\frac{θ}{2})$

B. $| \vec{A} - \vec{B} | = 2 R \cos (\frac{θ}{2})$

C. $| \vec{A} - \vec{B} | = \sqrt{2} R \sin (\frac{θ}{2})$

D. $| \vec{A} + \vec{B} | = 2 R \sin (\frac{θ}{2})$

Correct answer option is (A)

The magnitude of resultant vector

$R^{'} = \sqrt{a^{2} + b^{2} + 2 a b \cos θ}$

Here $a = b = R$

Then $R^{'} = \sqrt{R^{2} + R^{2} + 2 R^{2} \cos θ}$

$\begin{aligned} = R \sqrt{2} \sqrt{1 + \cos θ} \\ = \sqrt{2} R \sqrt{2 \cos^{2} \frac{θ}{2}} \\ = 2 R \cos \frac{θ}{2} \end{aligned}$

3. (JEE Main 2023 (Online) 25th January Morning Shift )

If $\vec{P} = 3 \hat{i} + \sqrt{3} \hat{j} + 2 \hat{k}$ and $\vec{Q} = 4 \hat{i} + \sqrt{3} \hat{j} + 2.5 \hat{k}$ then, the unit vector in the direction of $\vec{P} \times \vec{Q}$ is $\frac{1}{x} (\sqrt{3} \hat{i} + \hat{j} - 2 \sqrt{3} \hat{k})$ . The value of $x$ is _________.

Correct answer is (4)

$\vec{P} = 3 \hat{i} + \sqrt{3} \hat{j} + 2 \hat{k}$

$\vec{Q} = 4 \hat{i} + \sqrt{3} \hat{j} + 2.5 \hat{k}$

$\vec{P} \times \vec{Q} = | \begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 3 & \sqrt{3} & 2 \\ 4 & \sqrt{3} & 2.5 \end{array} |$

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

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

$| \vec{P} \times \vec{Q} | = \sqrt{\frac{3}{4} + \frac{1}{4} + 3} = 2$

Unit vector along $\vec{P} \times \vec{Q} = \frac{1}{4} (\sqrt{3} \hat{i} + \hat{j} - 2 \sqrt{3} \hat{k})$

$x = 4$

4. (JEE Main 2023 (Online) 24th January Morning Shift )

Vectors $a \hat{i} + b \hat{j} + \hat{k}$ and $2 \hat{i} - 3 \hat{j} + 4 \hat{k}$ are perpendicular to each other when $3 a + 2 b = 7$ , the ratio of $a$ to $b$ is $\frac{x}{2}$ . The value of $x$ is ____________.

Correct answer is (1)

For two perpendicular vectors

$\begin{aligned} (a \hat{i} + b \hat{j} + \hat{k}) \cdot (2 \hat{i} - 3 \hat{j} + 4 \hat{k}) = 0 \\ 2 a - 3 b + 4 = 0 \end{aligned}$

On solving, $2 a - 3 b = - 4$

Also given

$3 a + 2 b = 7$

We get $a = 1, b = 2$

$\begin{aligned} \frac{a}{b} = \frac{x}{2} \Rightarrow x = \frac{2 a}{b} = \frac{2 \times 1}{2} \\ \Rightarrow x = 1 \end{aligned}$

5. (JEE Main 2023 (Online) 15th April Morning Shift )

A vector in $x - y$ plane makes an angle of $30^{\circ}$ with $y$ -axis. The magnitude of $y$ -component of vector is $2 \sqrt{3}$ . The magnitude of $x$ -component of the vector will be :

(A) $\sqrt{3}$

(B) 2

(C) 6

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

Correct answer is (B)

JEE Main 2023 (Online) 15th April Morning Shift Physics - Vector Algebra Question 3 English Explanation
$\begin{aligned} A_{y} = A \cos 30^{\circ} = 2 \sqrt{3} \\ \Rightarrow A \frac{\sqrt{3}}{2} = 2 \sqrt{3} \\ \Rightarrow A = 4 \end{aligned}$

Now $A_{x} = A \sin 30^{\circ} = 4 \times \frac{1}{2} = 2$