The angle between the vectors $\hat{i}-2 \hat{j}+\hat{k}$ and $3 \hat{i}+2 \hat{j}+\hat{k}$ is:

$\frac{\pi}{2}$

The correct answer is Option (4) → $\frac{\pi}{2}$

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

$\vec a.\vec b=|\vec a||\vec b|\cos θ$

$\cos θ=\frac{3-4+1}{\sqrt{1+4+1}\sqrt{9+4+1}}=0$

$θ=\frac{\pi}{2}$