JEE Main Motion in Plane PYQ

6. (JEE Main 2023 (Online) 11th April Evening Shift )

When vector $\vec{A} = 2 \hat{i} + 3 \hat{j} + 2 \hat{k}$ is subtracted from vector $\vec{B}$ , it gives a vector equal to $2 \hat{j}$ . Then the magnitude of vector $\vec{B}$ will be :

(A) 3

(B) $\sqrt{33}$

(C) $\sqrt{6}$

(D) $\sqrt{5}$

Correct answer is (B)

Given that when vector $\vec{A} = 2 \hat{i} + 3 \hat{j} + 2 \hat{k}$ is subtracted from vector $\vec{B}$ , it gives a vector equal to $2 \hat{j}$ .

We can write this as:

$\vec{B} - \vec{A} = 2 \hat{j}$

Now, let's express the vector $\vec{B}$ in terms of its components:

$\vec{B} = B_{x} \hat{i} + B_{y} \hat{j} + B_{z} \hat{k}$

Subtract vector $\vec{A}$ from vector $\vec{B}$ :

$(B_{x} \hat{i} + B_{y} \hat{j} + B_{z} \hat{k}) - (2 \hat{i} + 3 \hat{j} + 2 \hat{k}) = 2 \hat{j}$

Comparing the components, we get:

$B_{x} - 2 = 0 B_{y} - 3 = 2 B_{z} - 2 = 0$

Solving these equations, we find the components of vector $\vec{B}$ :

$B_{x} = 2 B_{y} = 5 B_{z} = 2$

Now, we can find the magnitude of vector $\vec{B}$ :

$| \vec{B} | = \sqrt{B_{x}^{2} + B_{y}^{2} + B_{z}^{2}} = \sqrt{2^{2} + 5^{2} + 2^{2}} = \sqrt{4 + 25 + 4} = \sqrt{33}$

Therefore, the magnitude of vector $\vec{B}$ is $\sqrt{33}$

7. (JEE Main 2023 (Online) 8th April Morning Shift )

Two forces having magnitude $A$ and $\frac{A}{2}$ are perpendicular to each other. The magnitude of their resultant is:

(A) $\frac{5 A}{2}$

(B) $\frac{\sqrt{5} A}{4}$

(C) $\frac{\sqrt{5} A}{2}$

(D) $\frac{\sqrt{5} A^{2}}{2}$

Correct answer is (C)

The resultant of two perpendicular vectors is given by the Pythagorean theorem.

If we have two vectors of magnitudes $A$ and $\frac{A}{2}$ , the resultant $R$ is:

$R = \sqrt{A^{2} + {(\frac{A}{2})}^{2}} = \sqrt{A^{2} + \frac{A^{2}}{4}} = \sqrt{\frac{5 A^{2}}{4}} = \frac{\sqrt{5} A}{2}$ .

8. (JEE Main 2023 (Online) 24th January Evening Shift )

If two vectors $\vec{P} = \hat{i} + 2 m \hat{j} + m \hat{k}$ and $\vec{Q} = 4 \hat{i} - 2 \hat{j} + m \hat{k}$ are perpendicular to each other. Then, the value of m will be :

(A) $- 1$

(B) 3

(C) 1

(D) 2

Correct answer is (D)

$\vec{P} & \vec{Q}$ are perpendicular

$\begin{aligned} \vec{P} \cdot \vec{Q} = 0 \\ (\hat{i} + 2 m \hat{j} + m \hat{k}) \cdot (4 \hat{i} - 2 \hat{j} + m \hat{k}) = 0 \\ \Rightarrow 4 - 4 m + m^{2} = 0 \\ \Rightarrow (m - 2)^{2} = 0 \Rightarrow m = 2 \end{aligned}$

9. (JEE Main 2022 (Online) 29th June Morning Shift )

Two vectors $\vec{A}$ and $\vec{B}$ have equal magnitudes. If magnitude of $\vec{A}$ + $\vec{B}$ is equal to two times the magnitude of $\vec{A}$ $-$ $\vec{B}$ , then the angle between $\vec{A}$ and $\vec{B}$ will be :

(A) $\sin^{- 1} (\frac{3}{5})$

(B) $\sin^{- 1} (\frac{1}{3})$

(C) $\cos^{- 1} (\frac{3}{5})$

(D) $\cos^{- 1} (\frac{1}{3})$

Correct answer is (C)

$\sqrt{A^{2} + A^{2} + 2 A^{2} \cos θ} = 2 \sqrt{A^{2} + A^{2} + 2 A^{2} (- \cos θ)}$

$\Rightarrow 2 A^{2} + 2 A^{2} \cos θ = 8 A^{2} + 8 A^{2} - (- \cos θ)$

$\Rightarrow 5 \cos θ = 3$

$\Rightarrow θ = \cos^{- 1} (\frac{3}{5})$

10. (JEE Main 2022 (Online) 25th June Morning Shift )

$\vec{A}$ is a vector quantity such that $| \vec{A} |$ = non-zero constant. Which of the following expression is true for $\vec{A}$ ?

(A) $\vec{A} . \vec{A} = 0$

(B) $\vec{A} \times \vec{A} < 0$

(C) $\vec{A} \times \vec{A} = 0$

(D) $\vec{A} \times \vec{A} > 0$

Correct answer is (C)

$\vec{A} \times \vec{A} = A \times A \times \sin 0^{\circ}$

$= 0$