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

$\frac{10}{\sqrt{6}}$

The correct answer is Option (1) → $\frac{10}{\sqrt{6}}$ ##

The projection of vector $\vec{a}$ on the vector $\vec{b}$ is given by

$\frac{1}{|\vec{b}|}(\vec{a} \cdot \vec{b}) = \frac{(2 \times 1 + 3 \times 2 + 2 \times 1)}{\sqrt{(1)^2 + (2)^2 + (1)^2}} = \frac{10}{\sqrt{6}}$