Question

Find the cosine of the angle between the vectors:
$\overline{a}=3\hat{i}+2\hat{j},\overline{b}=\hat{i}+3\hat{j}+\hat{k}$

Answer 1

Let the angle between them be $\theta$

Given $\overline{a}=3\hat{i}+2\hat{j},\overline{b}=\hat{i}+3\hat{j}+\hat{k}$

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

$\cos \theta =\frac{\overrightarrow{a}.\overrightarrow{b}}{|\overrightarrow{a}||\overrightarrow{b}|}=\frac{(3\hat{i}+2\hat{j}).(\hat{i}+3\hat{j}+\hat{k})}{\sqrt{3^2+2^2}\sqrt{1^2+3^2+1^2}}=\frac{3+6+0}{\sqrt{143}}=\frac{9}{\sqrt{143}}$