Match List-I with List-II

List-I	List-II
(A) Angle between $\vec{i} - \vec{j}$ and $\vec{j} + \vec{k}$	(I) 0
(B) Angle between $2\vec{j} - \vec{k}$ and $\vec{j} + 2\vec{k}$	(II) $\frac{2\pi}{3}$
(C) Angle between $\vec{i} + 2\vec{j}$ and $5\vec{i} + 10\vec{j}$	(III) $\frac{\pi}{6}$
(D) Angle between $\sqrt{3}\vec{i} + \vec{j}$ and $\vec{i}$	(IV) $\frac{\pi}{2}$

Choose the correct answer from the options given below:

To find the angle between two vectors, the dot product formula is used:

$\cos \theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}| |\vec{b}|}$

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For $\vec{i} - \vec{j}$ and $\vec{j} + \vec{k}$:

$\vec{a} = (1, -1, 0)$

$\vec{b} = (0, 1, 1)$

$\vec{a} \cdot \vec{b} = (1)(0) + (-1)(1) + (0)(1)$

$= 0 - 1 + 0$

$= -1$

$|\vec{a}| = \sqrt{1^2 + (-1)^2 + 0^2} = \sqrt{2}$

$|\vec{b}| = \sqrt{0^2 + 1^2 + 1^2} = \sqrt{2}$

$\cos \theta = \frac{-1}{\sqrt{2} \times \sqrt{2}}$

$= \frac{-1}{2}$

$\theta = \cos^{-1}\left(-\frac{1}{2}\right)$

$= \frac{2\pi}{3}$

(A) matches with (II)

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For $2\vec{j} - \vec{k}$ and $\vec{j} + 2\vec{k}$:

$\vec{a} = (0, 2, -1)$

$\vec{b} = (0, 1, 2)$

$\vec{a} \cdot \vec{b} = (0)(0) + (2)(1) + (-1)(2)$

$= 0 + 2 - 2$

$= 0$

When the dot product equals 0, the vectors are perpendicular.

$\theta = \frac{\pi}{2}$

(B) matches with (IV)

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For $\vec{i} + 2\vec{j}$ and $5\vec{i} + 10\vec{j}$:

$\vec{a} = (1, 2, 0)$

$\vec{b} = (5, 10, 0)$

$\vec{b} = 5\vec{a}$

The vectors are parallel and point in the same direction.

$\theta = 0$

(C) matches with (I)

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For $\sqrt{3}\vec{i} + \vec{j}$ and $\vec{i}$:

$\vec{a} = (\sqrt{3}, 1, 0)$

$\vec{b} = (1, 0, 0)$

$\vec{a} \cdot \vec{b} = (\sqrt{3})(1) + (1)(0) + (0)(0)$

$= \sqrt{3}$

$|\vec{a}| = \sqrt{(\sqrt{3})^2 + 1^2 + 0^2}$

$= \sqrt{3 + 1}$

$= 2$

$|\vec{b}| = \sqrt{1^2 + 0^2 + 0^2} = 1$

$\cos \theta = \frac{\sqrt{3}}{2 \times 1}$

$= \frac{\sqrt{3}}{2}$

$\theta = \cos^{-1}\left(\frac{\sqrt{3}}{2}\right)$

$= \frac{\pi}{6}$

(D) matches with (III)

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(A) - (II), (B) - (IV), (C) - (I), (D) - (III)