Match List-I with List-II

Consider two vectors $\vec{a} = \hat{i} + 2\hat{j} - \hat{k}$ and $\vec{b} = -3\hat{i} - 6\hat{j} + 3\hat{k}$, then

List-I	List-II
(A) Angle between $\vec{a}$ and $\vec{b}$ is	(I) $\cos^{-1}\left(\frac{1}{\sqrt{6}}\right)$
(B) Angle between $\vec{a}$ and $x$-axis is	(II) $\cos^{-1}\left(\frac{2}{\sqrt{6}}\right)$
(C) Angle between $\vec{b}$ and $x$-axis is	(III) $\pi$
(D) Angle between $\vec{a}$ and $y$-axis is	(IV) $\cos^{-1}\left(-\frac{1}{\sqrt{6}}\right)$

Choose the correct answer from the options given below:

Given $\vec{a} = \hat{i} + 2\hat{j} - \hat{k}$ and $\vec{b} = -3\hat{i} - 6\hat{j} + 3\hat{k}$

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(A) Angle between $\vec{a}$ and $\vec{b}$

Comparing components:

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

$\vec{b} = (-3, -6, 3)$

$\vec{b} = -3(1, 2, -1)$

$\vec{b} = -3\vec{a}$

When one vector is a negative scalar multiple of another, they point in exactly opposite directions.

Angle between $\vec{a}$ and $\vec{b}$ = $\pi$

(A) → (III)

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(B) Angle between $\vec{a}$ and $x$-axis

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

$= \sqrt{1 + 4 + 1}$

$= \sqrt{6}$

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

$= 1$

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

$= \frac{1}{\sqrt{6}}$

$\theta = \cos^{-1}\left(\frac{1}{\sqrt{6}}\right)$

(B) → (I)

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(C) Angle between $\vec{b}$ and $x$-axis

$|\vec{b}| = \sqrt{(-3)^2 + (-6)^2 + (3)^2}$

$= \sqrt{9 + 36 + 9}$

$= \sqrt{54}$

$= 3\sqrt{6}$

$\vec{b} \cdot \hat{i} = (-3)(1) + (-6)(0) + (3)(0)$

$= -3$

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

$= \frac{-3}{3\sqrt{6}}$

$= \frac{-1}{\sqrt{6}}$

$\theta = \cos^{-1}\left(\frac{-1}{\sqrt{6}}\right)$

(C) → (IV)

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(D) Angle between $\vec{a}$ and $y$-axis

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

$= 2$

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

$= \frac{2}{\sqrt{6}}$

$\theta = \cos^{-1}\left(\frac{2}{\sqrt{6}}\right)$

(D) → (II)

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Final Matching:

(A) Angle between $\vec{a}$ and $\vec{b}$ → (III) $\pi$

(B) Angle between $\vec{a}$ and $x$-axis → (I) $\cos^{-1}\left(\frac{1}{\sqrt{6}}\right)$

(C) Angle between $\vec{b}$ and $x$-axis → (IV) $\cos^{-1}\left(\frac{-1}{\sqrt{6}}\right)$

(D) Angle between $\vec{a}$ and $y$-axis → (II) $\cos^{-1}\left(\frac{2}{\sqrt{6}}\right)$