Find the internal angle of the triangle between $\overrightarrow a \& \overrightarrow b $. If $\overrightarrow a  = \hat j + \sqrt 3 \hat k;\overrightarrow b  =  - \hat j + \sqrt 3 \hat k;\overrightarrow c  = 2\sqrt 3 \hat k$ form a triangle then find  the internal angle of the triangle between $\overrightarrow a \& \overrightarrow b $.
My approach is as follow
Let $\overrightarrow a  + \overrightarrow b  = \overrightarrow c $
$\overrightarrow a  = \hat j + \sqrt 3 \hat k;\overrightarrow b  =  - \hat j + \sqrt 3 \hat k;\overrightarrow c  = 2\sqrt 3 \hat k$
$ \Rightarrow {\left( {\overrightarrow a  + \overrightarrow b } \right)^2} = {\left( {\overrightarrow c } \right)^2} \Rightarrow {\overrightarrow a ^2} + {\overrightarrow b ^2} + 2\overrightarrow a .\overrightarrow b  = {\overrightarrow c ^2}$
$ \Rightarrow \overrightarrow a .\overrightarrow b  = 2 \Rightarrow $
$\cos \theta  = \frac{{\overrightarrow a .\overrightarrow b }}{{\left| {\overrightarrow a } \right|.\left| {\overrightarrow b } \right|}} = \frac{1}{2} = \frac{\pi }{3}$
But official answer is $\frac{2\pi }{3}$ where I am making mistake

The best way to find the angle is tp find the lengths of the side vectors here they are $a=2,b=2, c=\sqrt{12}$, so the required angle is $C$, then by cosine law
$$\cos C=\frac{a^2+b^2-c^2}{2ab}=\frac{2+2-12}{8}=-\frac{1}{2} \implies C=\frac{2\pi}{3}.$$
If you want to it by dot product of $\vec a$ and $\vec b$, the angle between $\vec a$ and $\vec B$ is given by $$\cos \phi =\frac{\vec a. \vec b}{a b}=\frac{1}{2} \implies \phi =\frac{\pi}{3}.$$ But the internal abgke of the trangle is given by $\theta=\pi-\phi= \frac{2\pi}{3}.$
