Question

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

Answer 1

Given that

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

We know that $\Rightarrow \overrightarrow{a}.\overrightarrow{b}=|a||b|cos\theta$ .....$\left(1\right)$

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

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

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

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

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

$\Rightarrow \overrightarrow{a}.\overrightarrow{b}=(3\times 1)+(-2\times -2)+(1\times -3)$

$\Rightarrow \overrightarrow{a}.\overrightarrow{b}=3+4-3=4$

$\Rightarrow \overrightarrow{a}.\overrightarrow{b}=4$

Substituting these values in equation $\left(1\right)$ we get

$\Rightarrow 4=\sqrt{14}\times \sqrt{14}\cos \theta$

$\Rightarrow 4=14\cos \theta$

$\Rightarrow \cos \theta =\frac{4}{14}=\frac{2}{7}$

$\Rightarrow \theta =co{s}^{-1}\frac{2}{7}$