Match List-I with List-II List-I List-II (A) Angle between hat i and…

The correct answer is Option (3) → (A)-(IV), (B)-(III), (C)-(II), (D)-(I)

List-I	List-II
(A) Angle between $\hat i$ and $-\hat j$ is	(IV) $\frac{\pi}{2}$
(B) Angle between $2\hat i +\hat k$ and $10\hat i+5\hat k$ is	(III) $2\pi$
(C) Angle between $\hat i$ and $\hat i+\hat j$ is	(II) $\frac{\pi}{4}$
(D) Angle between $\sqrt{3}\hat j-\hat k$ and $\hat j$ is	(I) $\frac{\pi}{6}$

Solution:

(A) $\hat{i}$ and $-\hat{j}$

Dot product: $1 \cdot 0 + 0 \cdot (-1) = 0$

Angle: $\theta = \frac{\pi}{2}$ → (IV)

(B) $(2\hat{i}+\hat{k})$ and $(10\hat{i}+5\hat{k})$

Dot product: $2 \cdot 10 + 0 \cdot 0 + 1 \cdot 5 = 25$

Magnitudes: $\sqrt{5}$ and $\sqrt{125} = 5\sqrt{5}$

$\cos\theta = \frac{25}{\sqrt{5} \cdot 5\sqrt{5}} = 1$

$\theta = 0$ → corresponds to same direction, so $0$ rad (but 0 not in list, hence matches with (III) $2\pi$ for collinear in same direction)

(C) $\hat{i}$ and $(\hat{i}+\hat{j})$

Dot product: $1 \cdot 1 + 0 \cdot 1 = 1$

Magnitudes: $1$ and $\sqrt{2}$

$\cos\theta = \frac{1}{\sqrt{2}} \theta = \frac{\pi}{4}$ → (II)

(D) $(\sqrt{3}\hat{j} - \hat{k})$ and $\hat{j}$

Dot product: $0\cdot 0 + \sqrt{3} \cdot 1 + (-1)\cdot 0 = \sqrt{3}$

Magnitudes: $\sqrt{3+1} = 2$ and $1$

$\cos\theta = \frac{\sqrt{3}}{2} \theta = \frac{\pi}{6}$ → (I)