Question

If the line $l$ makes angles $\alpha ,\beta ,\gamma ,\delta$ with the four diagonals of a cube, then prove that
${\cos }^2\alpha +{\cos }^2\beta +{\cos }^2\gamma +{\cos }^2\delta =\frac{4}{3}$

Answer 1

The direction of the 4 diagonal lines of a cube centered at the origin are:

$\pm (\hat{i}+\hat{j}+\hat{k})\pm (\hat{i}-\hat{j}+\hat{k})\pm (\hat{i}+\hat{j}-\hat{k})\pm (\hat{i}-\hat{j}-\hat{k})$

Let $l$ be having directions as;

$(x\hat{i}+y\hat{j}+z\hat{k})$

Hence;

$\sqrt{3}\left|l\right|\cos \alpha =\pm (x+y+z)\sqrt{3}\left|l\right|\cos \beta =\pm (x-y+z)\sqrt{3}\left|l\right|\cos \gamma =\pm (x+y-z)\sqrt{3}\left|l\right|\cos \delta =\pm (x-y-z)$

Above equations have been derived by taking dot product of the line $l$ and respective diagonals.

Now squaring,

$3{\left|l\right|}^2{\cos }^2\alpha ={(x+y+z)}^2=x^2+y^2+z^2+2(xy+yz+zx)3{\left|l\right|}^2{\cos }^2\beta ={(x-y+z)}^2=x^2+y^2+z^2+2(-xy-yz+zx)3{\left|l\right|}^2{\cos }^2\gamma ={(x+y-z)}^2=x^2+y^2+z^2+2(xy-yz-zx)3{\left|l\right|}^2{\cos }^2\delta ={(x-y-z)}^2=x^2+y^2+z^2+2(-xy+yz-zx)3{\left|l\right|}^2({\cos }^2\alpha +{\cos }^2\beta +{\cos }^2\gamma +{\cos }^2\delta )=4(x^2+y^2+z^2)+2(0+0+0)=4{\left|l\right|}^2{\cos }^2\alpha +{\cos }^2\beta +{\cos }^2\gamma +{\cos }^2\delta =\frac{4}{3}$