Question

If a vector $\overrightarrow{P}$ makes angles $\alpha ,\beta$ and $\gamma$ with X, Y and Z axes respectively. Then find $si{n}^2\alpha +si{n}^2\beta +si{n}^2\gamma$

Answer 1

$cos\alpha =\frac{p_x}{p}$
$cos\beta =\frac{p_y}{p}$
$cos\gamma =\frac{P_z}{p}$
$\therefore co{s}^2\alpha +co{s}^2\beta +co{s}^2\gamma =\frac{p_x^2+p_y^2+p_z^2}{p^2}=1$


And,
$si{n}^2\alpha +si{n}^2\beta +si{n}^2\gamma$
$=(1-co{s}^2\alpha )+(1-co{s}^2\beta )+(1-co{s}^2\gamma )$
$=3-(co{s}^2\alpha +co{s}^2\beta +co{s}^2\gamma )$
$=3-1$
$=2$