= Orthogonal opposite edges of a tetrahedron
Place one vertex of a <tetrahedron> at the origin and denote the other vertex vectors by $a,b,c$. Orthogonality of two pairs of <opposite edges of a tetrahedron> forces $a\cdot b=a\cdot c=b\cdot c$. It follows that the third pair is orthogonal too. The three sums of squared lengths of opposite edges all equal $|a|^2+|b|^2+|c|^2-2a\cdot b$. This gives a <dot product> test for the special geometry without solving for the face planes.
Back to article page
