Show that the direction cosines of a vector equally inclined to the axes OX, OY and OZ are $\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}$.

Solution: Let $\alpha$ is the angle of the vector equally inclined to the axes OX, OY and OZ.
Then, direction cosines will be $\left(\cos \alpha ,\cos \alpha ,\cos \alpha \right)$.
We know, sum of squares of direction cosines is always 1.
${\cos }^2\alpha +{\cos }^2\alpha +{\cos }^2\alpha =1$
$\Rightarrow 3{\cos }^2\alpha =1$
$\Rightarrow \cos \alpha =\frac{1}{\sqrt{3}}$
So, direction cosines are $\left(\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}\right)$.