Question

Find the cosine of the angle between the vectors:
$\overline{a}=2\hat{i}-\hat{j}+2\hat{k},\overline{b}=\hat{i}+\hat{j}+\hat{k}$

Answer 1

Let the angle between them be $\theta$

Given $\overline{a}=2\hat{i}-\hat{j}+2\hat{k},\overline{b}=\hat{i}+\hat{j}+\hat{k}$

As we know that $\overrightarrow{a}\cdot \overrightarrow{b}=|\overrightarrow{a}||\overrightarrow{b}|\cos \theta$

$\cos \theta =\frac{\overrightarrow{a}.\overrightarrow{b}}{|\overrightarrow{a}||\overrightarrow{b}|}=\frac{(2\hat{i}-\hat{j}+2\hat{k}).(\hat{i}+\hat{j}+\hat{k})}{\sqrt{2^2+1^2+2^2}\sqrt{1^2+1^2+1^2}}=\frac{2-1+2}{3\sqrt{3}}=\frac{1}{\sqrt{3}}$