Question

Find the angle between two vectors $(\overrightarrow{a}=2\hat{i}+\hat{j}-3\hat{k})$ and $(\overrightarrow{b}=3\hat{i}-2\hat{j}-\hat{k})$.

Answer 1

Given: $\overrightarrow{a}=2\hat{i}+\hat{j}-3\hat{k}$ and $\overrightarrow{b}=3\hat{i}-2\hat{j}-\hat{k}$
$\overrightarrow{a}.\overrightarrow{b}=(2\hat{i}+\hat{j}-3\hat{k})(3\hat{i}-2\hat{j}-\hat{k})$
$=6-2+3=\mathrm{7....}\left(1\right)$
$\left|\overrightarrow{a}\right|=\sqrt{2^2+1^2+{(-3)}^2}=\sqrt{14}....\left(2\right)$
$\left|\overrightarrow{b}\right|=\sqrt{3^2+{(-2)}^2+{(-1)}^2}=\sqrt{14}....\left(3\right)$
$\cos \theta =\frac{\overrightarrow{a}.\overrightarrow{b}}{\left|\overrightarrow{a}\right|.\left|\overrightarrow{b}\right|}=\frac{7}{\sqrt{14}\sqrt{14}}=\frac{1}{2}$
$\cos \theta =\cos {60}^o$
$\theta ={60}^o$
Hence the required angle is ${60}^o$