The projection of the vector $\vec{a}=2\hat{i}+3\hat{j}+2\hat{k}$ on the vector $\vec{b}=\hat{i}+2\hat{j}+\hat{k}$ is :

$\frac{11}{3}\sqrt{6}$ $\frac{1}{3}\sqrt{6}$ $\frac{10}{3}\sqrt{6}$ $\frac{5}{3}\sqrt{6}$

$\frac{5}{3}\sqrt{6}$

The correct answer is Option (4) → $\frac{5}{3}\sqrt{6}$ projection → $|\vec a|\cos θ$ $|\vec a|×\frac{(\vec a.\vec b)}{|\vec a||\vec b|}=\frac{(2+6+1)}{\sqrt{1+2^2+1}}$ $=\frac{10}{\sqrt{6}}\frac{\sqrt{6}}{\sqrt{6}}=\frac{5}{3}\sqrt{6}$