Match List-I with List-II.

Choose the correct answer from the options given below :

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

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

for two vector $\vec a, \vec b$

angle θ between them

$⇒\vec a.\vec b=|\vec a||\vec b|\cos θ⇒θ=\cos^{-1}(\frac{\vec a.\vec b}{|\vec a||\vec b|})$

(A) $\vec a=\hat i-2\hat j+3\hat k,\vec b=2\hat i+\hat j$

$\vec a.\vec b=2-2=0⇒θ=90°$ (III)

(B) $\vec a=\hat{i}+\hat{j}+2\hat{k},\vec b=2\hat{i} + 2\hat{j} +4\hat{k}$

$\vec a=2\vec b⇒θ=0°$ (II)

(C) $\vec a=2\hat{i}-\hat{j}+\hat{k},\vec b=\hat{i} + \hat{j} +\hat{k}$

$\vec a.\vec b=2-1+1=2,|\vec a|=\sqrt{6},|\vec b|=\sqrt{3}$

so $θ=\cos^{-1}(\frac{2}{\sqrt{18}})$ (I)

(D) $\vec a=\hat i+\hat j-\hat k,\vec b=\hat i+\hat j+\hat k$

$\vec a.\vec b=1+1-1=1,|\vec a|=\sqrt{3}=|\vec b|$

$θ=\cos^{-1}(\frac{1}{3})$ (IV)