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

Let a vector be equally inclined to the axes $OX, OY$,and $OZ$ at an angle $\alpha.$
Then,the direction cosines of the vector are $l = \cos \alpha, m = \cos \alpha,$ and $n = \cos \alpha.$
We know that for any vector,the sum of the squares of its direction cosines is $1.$
Therefore,$\cos^2 \alpha + \cos^2 \alpha + \cos^2 \alpha = 1.$
This simplifies to $3 \cos^2 \alpha = 1.$
Solving for $\cos \alpha,$ we get $\cos^2 \alpha = \frac{1}{3},$ which implies $\cos \alpha = \pm \frac{1}{\sqrt{3}}.$
Assuming the vector is in the first octant,the direction cosines are $\left(\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right).$