If $\vec a =\hat j +\sqrt{3}\hat k, \vec b =−\hat j +\sqrt{3}\hat k$ and $\vec c=2\sqrt{3}\hat k$ form a triangle, then the internal angle of the triangle between $\vec a$ and $\vec b$ is

$\frac{2π}{3}$

Clearly, $\vec a + \vec b = \vec c$ and $|\vec a|=|\vec b|=2$

So, $\vec a, \vec b, \vec c$ form an isosceles triangle as shown below.

Let θ be the internal angle between $\vec a$ and $\vec b$. Then,

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

$⇒-\cos θ=\frac{-1+3}{2×2}=\frac{1}{2}⇒\cos θ=-\frac{1}{2}⇒θ=\frac{2π}{3}$