If $\vec a, \vec b$ are two unit vectors such that $|\vec a+\vec b|=2\sqrt{3}$ and $|\vec a-\vec b|=6$, then the angle between $\vec a$ and $\vec b$, is

$\frac{2π}{3}$

Let θ be the angle between unit vectors $\vec a$ and $\vec b$.

Since $\vec a$ and $\vec b$ are unit vectors.

$∴\tan\frac{θ}{2}=\frac{|\vec a-\vec b|}{|\vec a+\vec b|}=\frac{6}{2\sqrt{3}}=\sqrt{3}⇒\frac{θ}{2}=\frac{π}{3}⇒θ=\frac{2π}{3}$