Question

A vector $\overrightarrow{OP}$ is inclined to OX at ${45}^o$ and OY at ${60}^o$. Find the angle at which $\overrightarrow{OP}$ is incilned to OZ.

Answer 1

Suppose $\overrightarrow{OP}$ is incilned at angle $\gamma$ to OZ.

Let l, m, n be the direction cosines of $\overrightarrow{OP}$. Then,

$l=\cos {45}^o=\frac{1}{\sqrt{2}}$

$m=\cos {60}^o=\frac{1}{2}$

$n=\cos \gamma$

Now,

$l^2+m^2+n^2=1$

$\frac{1}{2}+\frac{1}{4}+n^2=1$

$n^2=\frac{1}{4}$

$n=\pm \frac{1}{2}$

$\cos \gamma =\pm \frac{1}{2}$

$\gamma ={60}^o$ or ${120}^o$