Question

A line makes angles $\alpha ,\beta ,\gamma$ and $\delta$ with the diagonals of a cube, prove the $co{s}^2\alpha +co{s}^2\beta +co{s}^2\gamma +co{s}^2\delta =\frac{4}{3}$

Answer 1

Let a be the length of an edge of the cube and let one corner be at the origin.
Clearly, OP, AR, BS and CQ are the diagonals of the cube.
The direction ration of OP, AR, BS and CQ are:
a-0, a-0, a-0 i.e.a,a,a
0-a, a-0, a-0 i.e.-a,a,a
a-0, 0-a, a-0 i.e.a, -a,a
and a-0, 0-a i.e. a,a,-a
Let the direction ratios of a line be proportional to I, m, n.
Suppose this line makes angle $\alpha ,\gamma ,\gamma$ and $\delta$ with OP,AR, BS and CQ respectively.
Now $\alpha$ is the angle between OP and the line whose DR's are proportional to l,m,n.
$\therefore cos\alpha =\frac{a.l+a.m+a.n}{\sqrt{a^2+a^2+a^2.\sqrt{l^2+m^2+n^2}}}\Rightarrow cos\alpha =\frac{l+m+n}{\sqrt{3}\sqrt{l^2+m^2+n^2}}........\left(1\right)$
Now $\beta$ is the angle between AR and the line whose DR's are proportional to l,m,n.
$\therefore cos\alpha =\frac{-a.l+a.m+a.n}{\sqrt{a^2+a^2+a^2.\sqrt{l^2+m^2+n^2}}}\Rightarrow cos\alpha =\frac{-l+m+n}{\sqrt{3}\sqrt{l^2+m^2+n^2}}........\left(2\right)$
Similarly,
$\therefore cos\gamma =\frac{l-m+n}{\sqrt{3}\sqrt{l^2+m^2+n^2}}.........\left(3\right)\Rightarrow cos\delta =\frac{l+m-n}{\sqrt{3}\sqrt{l^2+m^2+n^2}}........\left(4\right)$
Therefore from eq(1), eq(2), eq(3) and eq(4):
$co{s}^2\alpha +co{s}^2\beta +co{s}^2\gamma +co{s}^2\delta =\frac{(l+m+n{)}^2}{3(l^2+m^2+n^2)}+\frac{(-l+m+n{)}^2}{3(l^2+m^2+n^2)}+\frac{(l-m+n{)}^2}{3(l^2+m^2+n^2)}+\frac{(l+m-n{)}^2}{3(l^2+m^2+n^2)}$
$=\frac{1}{3(l^2+m^2+n^2)}\left[\right(l+m+n{)}^2+(-l+m+n{)}^2+(l-m+n{)}^2+(l+m-n{)}^2]=\frac{1}{3(l^2+m^2+n^2)\left\{4\right(l^2+m^2+n^2\left)\right\}}=\frac{4}{3}$