Question

If the edges of a rectangular parallelpiped are a,b,c the angles between four diagonals are given by

Answer 1

Take O as corner of the rectangular parallelpiped as origin and $OA,OB,OC$ the edges through axes.

( diagram )

Let $OA=a,OB=b,OC=c$ then the coordinate of $O,A,B,C$ are $(0,0,0),(a,0,0),(0,b,0),(0,0,c)$ respectively.

The coordinates of other points are shown in figure.

The four diagonals are $OP,AL,BM,CN$

Direction cosines of OP are $a-0,b-0,c-0$ i.e, $a,b,c$ respectively.

Direction cosines of AL are $0-a,b-0,c-0$ i.e, $-a,b,c$ respectively.

Direction cosines of BM are $a-0,0-b,c-0$ i.e, $a,-b,c$ respectively.

Let $Q$ be the angle $OP$ and $AL$

$\therefore \cos \alpha =\frac{\left(a\right)(-a)+\left(b\right)\left(b\right)+\left(c\right)\left(c\right)}{\sqrt{a^2+b^2+c^2}\sqrt{a^2+b^2+c^2}}$

$=\frac{-a^2+b^2+c^2}{a^2+b^2+c^2}$

$\therefore \alpha ={\cos }^{-1}\left(\frac{-a^2+b^2+c^2}{a^2+b^2+c^2}\right)$

$\therefore$ angle between $OP$ and $AL$ is $={\cos }^{-1}\left(\frac{-a^2+b^2+c^2}{a^2+b^2+c^2}\right)$

similarly angle between OP and BM $={\cos }^{-1}\left(\frac{a^2+b^2-c^2}{a^2+b^2+c^2}\right)$

Proceeding this way we see that the angle between the diagonals are given by ${\cos }^{-1}\left(\frac{a^2\pm b^2\pm c^2}{a^2+b^2+c^2}\right)$

Hence, solved.