Q. $If \overrightarrow{a}=\hat{j}+\sqrt{3}\hat{k}, \overrightarrow{b}=-\hat{j}+\sqrt{3}\hat{k} and \overrightarrow{c}=2\sqrt{3}\hat{k}$ form a  triangle, then find out the internal angle of the triangle between $\overrightarrow{a} and \overrightarrow{b}$?

Dear Student,

$$\begin{gathered} \overrightarrow{a} = \hat{j} + \sqrt{3}\hat{k}, \overrightarrow{b} = -\hat{j} + \sqrt{3}\hat{k} and \overrightarrow{c} = 2\sqrt{3}\hat{k} form a triangle. \\ \overrightarrow{a}.\overrightarrow{b} = \left(1\right)\left(-1\right) + \left(\sqrt{3}\right)\left(\sqrt{3}\right) = -1 + 3 = 2 \\ and, \left|\overrightarrow{a}\right| = \sqrt{{\left(1\right)}^2 + {\left(\sqrt{3}\right)}^2} = \sqrt{4} = 2 and \left|\overrightarrow{b}\right| = \sqrt{{\left(-1\right)}^2 + {\left(\sqrt{3}\right)}^2} = \sqrt{4} = 2 \\ Let \theta be the angle between \overrightarrow{a} and \overrightarrow{b}. Since \theta is an internal angle of a triangle, 0 < \theta < \mathrm{\pi } \\ Now, \overrightarrow{a}.\overrightarrow{b} = \left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|\cos \theta \\ \Rightarrow 2 = \left(2 \times 2\right)\cos \theta \\ \Rightarrow \frac{2}{4} = \cos \theta \\ \Rightarrow \cos \theta = \frac{1}{2} = \cos \frac{\mathrm{\pi }}{3} \\ \Rightarrow \boxed{\mathbf{\theta }\mathbf{ }\mathbf{=}\mathbf{ }\frac{\mathbf{\pi }}{\mathbf{3}}\mathbf{ }\mathbf{=}\mathbf{ }\mathbf{a}\mathbf{n}\mathbf{g}\mathbf{l}\mathbf{e}\mathbf{ }\mathbf{b}\mathbf{e}\mathbf{t}\mathbf{w}\mathbf{e}\mathbf{e}\mathbf{n}\mathbf{ }\overrightarrow{\mathbf{a}}\mathbf{ }\mathbf{a}\mathbf{n}\mathbf{d}\mathbf{ }\overrightarrow{\mathbf{b}}} \end{gathered}$$

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