A vector makes equal angles α with x and y axes and 90 ° with z - axis. Then α =

We know, the sum of squares of direction cosines of a line is $1$

${\cos }^2\alpha +{\cos }^2\beta +{\cos }^2\gamma =1$

Here given that $\alpha =\beta$ and $\gamma =90°$

$\Rightarrow {\cos }^2\alpha +{\cos }^2\alpha +{\cos }^{2}90°=1$

$\Rightarrow 2{\cos }^2\alpha =1$  $\left(\because \cos 90°=0\right)$

$\Rightarrow {\cos }^2\alpha =\frac{1}{2}$

$\Rightarrow \cos \alpha =\pm \frac{1}{\sqrt{2}}$

$\Rightarrow \alpha =45°$ or $135°$