If $\vec{a} $ and $\vec{b}$ are unit vectors, then the angle between $\vec{a}$ and $\vec{b}$ for $\vec{a}+\sqrt{3}\vec{b}$ to be a unit vector is given by :

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

The correct answer is Option (2) → $\frac{5\pi }{6}$

$|\vec a+\sqrt{3}\vec b|$

$⇒(\vec a+\sqrt{3}\vec b).(\vec a+\sqrt{3}\vec b)=1$

So $|\vec a|^2+3|\vec b|^2+2\sqrt{3}\vec a.\vec b=1$

$2\sqrt{3}\vec a.\vec b=-3$

$|\vec a||\vec b|\cos θ=-\frac{\sqrt{3}}{2}$

$⇒θ=π-\frac{π}{6}=\frac{5π}{6}$