Question

Find the angle between the vectors $\hat{i}-2\hat{j}+3\hat{k}$ and $3\hat{i}-2\hat{j}+\hat{k}$.

Answer 1

Here, let $\overrightarrow{a}=\hat{i}-2\hat{j}+3\hat{k}$ and $\overrightarrow{b}=3\hat{i}-2\hat{j}+\hat{k}$.

Then,

$\overrightarrow{a}.\overrightarrow{b}=(\hat{i}-2\hat{j}+3\hat{k}).(3\hat{i}-2\hat{j}+\hat{k})$

$=\left(1.3\right)+(-2)(-2)+\left(3.1\right)$

$=3+4+3=10$

$|\overrightarrow{a}|=\sqrt{1+4+9}=\sqrt{14}$

and, $|\overrightarrow{b}|=\sqrt{9+4+1}=\sqrt{14}$

Therefore,

$\cos \theta =\frac{10}{\sqrt{14}\sqrt{14}}=\frac{10}{14}=\frac{5}{7}$

Thus, $\theta ={\cos }^{-1}\left(\frac{5}{7}\right)$