The angle made by the vector $\vec{i}+\vec{j}+\sqrt{2} \vec{k}$ with $z$-axis is

Question

Professor PC Thomas Classes · Answer

$\cos\gamma = \frac{\sqrt{2}}{\sqrt{1^2+1^2+(\sqrt{2})^2}} = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}} \implies \gamma = 45^\circ$.