Question

If $\overrightarrow{A}=2\hat{i}+7\hat{j}+3\hat{k}$ and $\overrightarrow{B}=3\hat{i}+2\hat{j}+5\hat{k}$, then find the projection of $\overrightarrow{a}$ on $\overrightarrow{b}$.

Answer 1

Unit vector along b is $\hat{b}=\frac{\overrightarrow{b}}{|\overrightarrow{b}|}=\frac{2\hat{i}+6\hat{j}+3\hat{k}}{\sqrt{2^2+6^2+3^2}}=\frac{2\hat{i}+6\hat{j}+3\hat{k}}{7}$

So, the length of projection of $\overrightarrow{a}$ on $\overrightarrow{b}$ is $\overrightarrow{a}.\hat{b}=(7\hat{i}+\hat{j}-4\hat{k}).\frac{2\hat{i}+6\hat{j}+3\hat{k}}{7}=\frac{8}{7}$

Therefore the vector projection of $\overrightarrow{a}$ on $\overrightarrow{b}$ is:

$(\overrightarrow{a}.\hat{b})\hat{b}=\frac{8}{7}\left(\frac{2\hat{i}+6\hat{j}+3\hat{k}}{7}\right)=\frac{16}{49}\hat{i}+\frac{48}{49}\hat{j}+\frac{24}{49}\hat{k}.$