Find the position vector of a point $A$ in space such that $\overrightarrow {{\rm{OA}}}$ is inclined at ${60^\circ }$ to OX and at ${45^\circ }$ to OY and $|\overrightarrow {{\rm{OA}}} | = 10$ units.

Since, $\overrightarrow {{\rm{OA}}}$ is inclined at ${60^\circ }$ to ${\rm{OX}}$ and at ${45^\circ }$ to ${\rm{OY}}$.

Let $\overrightarrow {{\rm{OA}}}$ makes angle $\alpha$ with ${\rm{OZ}}$.

$\therefore$ ${\cos ^2}{60^\circ } + {\cos ^2}{45^\circ } + {\cos ^2}\alpha = 1$
$\Rightarrow$ ${\left( {\frac{1}{2}} \right)^2} + {\left( {\frac{1}{{\sqrt 2 }}} \right)^2} + {\cos ^2}\alpha = 1$

$\Rightarrow$ $\frac{1}{4} + \frac{1}{2} + {\cos ^2}\alpha = 1$
$\Rightarrow$ ${\cos ^2}\alpha = 1 - \left( {\frac{1}{2} + \frac{1}{4}} \right)$

$\Rightarrow$ ${\cos ^2}\alpha = 1 - \left( {\frac{6}{8}} \right)$
$\Rightarrow$ ${\cos ^2}\alpha = \frac{1}{4}$

$\Rightarrow$ $\cos \alpha = \frac{1}{2} = \cos {60^\circ }$
$\therefore$ $\alpha = {60^\circ }$

$\therefore$ $\overrightarrow {{\rm{OA}}} = |\overrightarrow {{\rm{OA}}} |\left( {\frac{1}{2}\widehat {\rm{i}} + \frac{1}{{\sqrt 2 }}\widehat {\rm{j}} + \frac{1}{2}\widehat {\rm{k}}} \right)$

$= 10\left( {\frac{1}{2}\widehat {\rm{i}} + \frac{1}{{\sqrt 2 }}\widehat {\rm{j}} + \frac{1}{2}\widehat {\rm{k}}} \right)$

$= 5\widehat {\rm{i}} + 5\sqrt 2 \widehat {\rm{j}} + 5\widehat {\rm{k}}$