Question

Find the angle between the vectors $\overrightarrow{a}$ and $\overrightarrow{b}$ with magnitude $\sqrt{3}$ and $2$ respectively and $\overrightarrow{a}\cdot \overrightarrow{b}=\sqrt{6}$.

Answer 1

$\left|\overrightarrow{a}\right|=\sqrt{3}\Rightarrow \left|\overrightarrow{b}\right|=2\Rightarrow \overrightarrow{a}.\overrightarrow{b}=\sqrt{6}Letanglebetween\overrightarrow{a}and\overrightarrow{b}is\theta \Rightarrow \frac{\overrightarrow{a}.\overrightarrow{b}}{\left|a\right|\left|b\right|}=\cos \theta \Rightarrow \frac{\sqrt{6}}{2\times \sqrt{3}}=\cos \theta \Rightarrow \cos \theta =\frac{1}{\sqrt{2}}\therefore \theta ={45}^{\circ }$