Question

The angle between the two vectors $\overrightarrow{A}=\hat{i}+2\hat{j}-\hat{k}and\overrightarrow{B}=\widehat{-i}+\hat{j}-2\hat{k}$ is

• A. ${90}^o$
• B. ${30}^o$
• C. ${45}^o$
• D. ${60}^o$

Answer 1

The correct option is D ${60}^o$

$\overrightarrow{A}=\hat{i}+2\hat{j}-\hat{k}$
$\overrightarrow{B}=-\hat{i}+\hat{j}-2\hat{k}$
$\overrightarrow{A}\cdot \overrightarrow{B}=|A||B|\cos \theta$
$|A|=\sqrt{1^2+2^2+(-1{)}^2}$
$=\sqrt{6}$
$|B|=\sqrt{(-1{)}^2+(1{)}^2+(-2{)}^2}$
$|B|=\sqrt{6}$
$\therefore \cos \theta =\frac{\overrightarrow{A}\cdot \overrightarrow{B}}{|A|\times |B|}$
$=(\hat{i}+2\hat{j}-\hat{k})\cdot (-\hat{i}+\hat{j}-2\hat{k})$
$=\frac{-1+2+2}{\sqrt{6}\times \sqrt{6}}=\frac{3}{6}=\frac{1}{2}$
$\Rightarrow \cos \theta =\frac{1}{2}$
$\theta ={\cos }^{-1}\left(\frac{1}{2}\right)={60}^{\circ }$.