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