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

By your work $$\cos\theta=-\frac{2}{2\cdot2}-\frac{1}{2}.$$
Let $\vec{a}=\vec{AB},$ $\vec{b}=\vec{BC}$ and $\vec{c}=\vec{AC}.$
Thus, $$\cos\theta=\cos\measuredangle ABC=\cos\left(180^{\circ}-\measuredangle(\vec{a},\vec{b})\right)=-\frac{1}{2}.$$$$
