If $|\vec{a} \times \vec{b}| = \sqrt{3}$ and $\vec{a} \cdot \vec{b} = -3$, then angle between $\vec{a}$ and $\vec{b}$ is

$\frac{5\pi}{6}$

The correct answer is Option (4s) → $\frac{5\pi}{6}$ ##

Given that, $|\vec{a} \times \vec{b}| = \sqrt{3}$ and $\vec{a} \cdot \vec{b} = -3$.

This implies

$|\vec{a}||\vec{b}| \sin \theta = \sqrt{3} \quad ...{i}$

and $|\vec{a}||\vec{b}| \cos \theta = -3 \quad ...{ii}$

Divide equation (i) by (ii):

$\tan \theta = \frac{\sqrt{3}}{-3} = -\frac{1}{\sqrt{3}}$

$\theta = \frac{5\pi}{6}$