Question

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

Answer 1

Given, $\overrightarrow{a}=3i-2j+k$

$\left|\overrightarrow{a}\right|=\sqrt{3^2+{(-2)}^2+1^2}=\sqrt{14}$

$\overrightarrow{b}=i-2j-3k$

$\left|\overrightarrow{b}\right|=\sqrt{1^2+{(-2)}^2+{(-3)}^2}=\sqrt{14}$

$\overrightarrow{a}\cdot \overrightarrow{b}=(3i-2j+k)\cdot (i-2j-3k)=3+4-3=4$

As we know that,

$\overrightarrow{a}\cdot \overrightarrow{b}=\left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|\cos \theta$

Here $\theta$ is the angle between the vectors $\overrightarrow{a}$ and $\overrightarrow{b}$

Therefore,

$\cos \theta =\frac{4}{\sqrt{14}\sqrt{14}}$

$\Rightarrow \theta ={\cos }^{-1}\left(\frac{2}{7}\right)$

Hence, the angle between the vectors $\overrightarrow{a}$ and $\overrightarrow{b}$ is ${\cos }^{-1}\left(\frac{2}{7}\right)$