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A tetrahedron $ABCD$ has vertices $A$, $B$, $C$ and $D$, as shown below:
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Show that the edges $AD$ and $BC$ of the tetrahedron are mutually perpendicular if and only if $AB^2+CD^2 = AC^2+BD^2$.
If the position vector of $A$ is ${\bf a}$ (and similar for the other vertices), can you use the scalar product to find an expression for $AB^2$?
It might be helpful to note that $AB^2=|\overrightarrow{AB}|^2$.
Sometimes it is easier to try and show $p-q=0$ rather than trying to show $p=q$.