Question

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

Answer 1

Given : $\overrightarrow{A}=\hat{i}+2\hat{j}-\hat{k}$ $\overrightarrow{B}=-\hat{i}+\hat{j}-2\hat{k}$

$\overrightarrow{A}.\overrightarrow{B}=-1+2+2=3$

$|\overrightarrow{A}|=\sqrt{1^1+2^2+1^2}=\sqrt{6}$

$|\overrightarrow{B}|=\sqrt{1^1+1^2+2^2}=\sqrt{6}$

$⟹\cos \theta =\frac{\overrightarrow{A}.\overrightarrow{B}}{|\overrightarrow{A}||\overrightarrow{B}|}$

Or $\cos \theta =\frac{3}{\sqrt{6}\times \sqrt{6}}=\frac{1}{2}$

$⟹\theta ={60}^o$