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

Choose the correct answer from the options given below:

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

The correct answer is Option (2) → (A)-(III), (B)-(I), (C)-(IV), (D)-(II)

$\vec a=(1,2,-1),\;\vec b=(-3,-6,3)$

Since $\vec b=-3\vec a$, the vectors are opposite ⇒ angle = $\pi$.

$|\vec a|=\sqrt{6},\;|\vec b|=3\sqrt{6}$

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

$\vec{a} \cdot \vec{b} = (1)(-3) + (2)(-6) + (-1)(3) = -3 - 12 - 3 = -18$

$|\vec{a}| = \sqrt{1^2 + 2^2 + (-1)^2} = \sqrt{6}, \quad |\vec{b}| = \sqrt{9 + 36 + 9} = \sqrt{54}$

$\cos \theta = \frac{-18}{\sqrt{6} \cdot \sqrt{54}} = \frac{-18}{\sqrt{324}} = \frac{-18}{18} = -1$

$\theta = \pi \Rightarrow (III)$

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

$\cos \alpha = \frac{a_x}{|\vec{a}|} = \frac{1}{\sqrt{6}} \Rightarrow \cos^{-1} \left( \frac{1}{\sqrt{6}} \right) = (I)$

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

$\cos \alpha = \frac{-3}{\sqrt{54}} = \frac{-3}{3\sqrt{6}} = \frac{-1}{\sqrt{6}} \Rightarrow (IV)$

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

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