Find the internal angle of the triangle between $\overrightarrow a \& \overrightarrow b $.

Question

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

Answer 1

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, $$\cos\theta=\cos\measuredangle ABC=\cos\left(180^{\circ}-\measuredangle(\vec{a},\vec{b})\right)=-\frac{1}{2}.$$$$

Answer 2

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 $\theta=\pi-\phi= \frac{2\pi}{3}.$

Answer 3

You have calculated the external angle. Draw vector signs on your diagram, it would be clear to you.