Question

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

Answer 1

given $\overrightarrow{a}=\hat{i}+\hat{j}-\hat{k},\overrightarrow{b}=\hat{i}+\hat{j}+\hat{k}$

$\cos \theta =\frac{\overrightarrow{a}.\overrightarrow{b}}{\left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|}=\frac{(\hat{i}+\hat{j}-\hat{k}).(\hat{i}+\hat{j}+\hat{k})}{|\hat{i}+\hat{j}-\hat{k}||\hat{i}+\hat{j}+\hat{k}|}$

$=\frac{1+1-1}{\sqrt{3}.\sqrt{3}}=\frac{1}{3}$

$\therefore \theta ={\cos }^{-1}\left(\frac{1}{3}\right)$