Question

A line makes angles $\alpha ,\beta ,\gamma ,\delta$ with the four diagonals of a cube, prove that

${\cos }^2\alpha +{\cos }^2\beta +{\cos }^2\gamma +{\cos }^2\delta =\frac{4}{3}$

Answer 1

Let $a$ be the length of na edge of the cube and let one corner be at origin.

Clearly $OP,AR,BSandCQ$ are the diagonals of the cube.

The direction ratio of $OP,AR,BSandCQ$

$a-0,a-0,a-0i.ea,a,a$

$0-a,a-0,a-0i.e-a,a,a$

$a-0,0-a,a-0i.ea,-a,a$

and $a-0,a-0,0-ai.ea,a,-a$

Let the ratio of a line be proportional to $l,m,n$

Suppose this line makes angle $\alpha ,\beta ,\gamma ,\delta$ with $OP,AR,BS,CQ$ respectively

Now $\alpha$ is the angle between OP and the line whose direction ratio,s are proportional to $l,m,n$

$\therefore \cos \alpha =\frac{a\cdot l+a\cdot m+a\cdot n}{\sqrt{a^2+a^2+a^2}\cdot \sqrt{l^2+m^2+n^2}}$

$\Rightarrow \cos \alpha =\frac{l+m+n}{\sqrt{3}\cdot \sqrt{l^2+m^2+n^2}}..........\left(1\right)$

Now $\beta$ is the angle between OP and the line whose direction ratio,s are proportional to $l,m,n$

$\therefore \cos \beta =\frac{-a\cdot l+a\cdot m+a\cdot n}{\sqrt{a^2+a^2+a^2}\cdot \sqrt{l^2+m^2+n^2}}$

$\Rightarrow \cos \beta =\frac{-l+m+n}{\sqrt{3}\cdot \sqrt{l^2+m^2+n^2}}..........\left(2\right)$

Now $\gamma$ is the angle between OP and the line whose direction ratio,s are proportional to $l,m,n$

$\therefore \cos \gamma =\frac{a\cdot l-a\cdot m+a\cdot n}{\sqrt{a^2+a^2+a^2}\cdot \sqrt{l^2+m^2+n^2}}$

$\Rightarrow \cos \gamma =\frac{l-m+n}{\sqrt{3}\cdot \sqrt{l^2+m^2+n^2}}..........\left(3\right)$

Now $\delta$ is the angle between OP and the line whose direction ratio,s are proportional to $l,m,n$

$\therefore \cos \delta =\frac{a\cdot l+a\cdot m-a\cdot n}{\sqrt{a^2+a^2+a^2}\cdot \sqrt{l^2+m^2+n^2}}$

$\Rightarrow \cos \delta =\frac{l+m-n}{\sqrt{3}\cdot \sqrt{l^2+m^2+n^2}}..........\left(4\right)$

Therefore from squaring and adding (1),(2),(3) and (4)

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

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

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

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

$=\frac{4}{3}$