Question

Let $\overrightarrow{a}$ and $\overrightarrow{b}$ be two given vectors such that $|\overrightarrow{a}|=\sqrt{3},|\overrightarrow{b}|=2$ and $\overrightarrow{a}\cdot \overrightarrow{b}=\sqrt{6}.$ Find the angle between $\overrightarrow{a}$ and $\overrightarrow{b}.$

Answer 1

$\begin{array}{c}Let\theta betheanglebetween\overrightarrow{a}and\overrightarrow{b}.\\ Then,\\ \overrightarrow{a}.\overrightarrow{b}=\sqrt{6}\\ \Rightarrow \left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|\cos \theta =\sqrt{6}\\ \Rightarrow \left(\sqrt{3}\right)\left(2\right)\cos \theta =\sqrt{6}\\ \Rightarrow \cos \theta =\frac{\sqrt{6}}{2\sqrt{3}}\\ \Rightarrow \cos \theta =\frac{1}{\sqrt{2}}\\ \Rightarrow \theta ={\cos }^{-1}\left(\frac{1}{\sqrt{2}}\right)=\frac{\pi }{4}\\ Hence,therequiredangleis\frac{\pi }{4}.\end{array}$

Hence, this is the answer.