The angle which a vector $\hat{i}-\hat{j}+\sqrt{2}\hat{k}$ makes with x-axis is

The correct option is B ${120}^o$
Let $\overrightarrow{A}=\hat{i}-\hat{j}+\sqrt{2}\hat{k}$ and vector representing y - axis be $\overrightarrow{B}=\hat{j}$.
We know that $\cos \beta =\frac{\overrightarrow{A}\cdot \overrightarrow{B}}{|\overrightarrow{A}||\overrightarrow{B}|}=\frac{-1}{\sqrt{(1{)}^2+(-1{)}^2+(\sqrt{2}{)}^2}}=-\frac{1}{2}$
$\therefore \beta ={120}^o$