Question

A line makes angles $\alpha ,\beta ,\gamma$ and $\delta$ with the diagonals of a cube. Then ${\cos }^2\alpha +{\cos }^2\beta +{\cos }^2\gamma +{\cos }^2\delta =\frac{x}{3}$. Find $x$.

Answer 1

$\cos \alpha =\frac{l+m+n}{\sqrt{3}}$

$\cos \beta =\frac{l+m-n}{\sqrt{3}}$

$\cos \gamma =\frac{l-m+n}{\sqrt{3}}$

$\cos \delta =\frac{-l+m+n}{\sqrt{3}}$ $[L^2+M^2+N^2-1]$

${\cos }^2\alpha +{\cos }^2\beta +{\cos }^2\gamma +{\cos }^2\gamma$

$=\frac{1}{3}+(l+m+n{)}^2+(l+m-n{)}^2+(l-m+n{)}^2+(-l+m-n)$

$=\frac{1}{3}\left(4\right(l^2+m^2+n^2\left)\right)=\frac{x}{3}$

$\frac{4}{3}=\frac{x}{3}(x=4)$