# tetrahedron volume given rectangular parallelepiped

Let $$A$$ be a rectangular parallelepiped with edges of lengths $$15, 20, 30$$. Let $$B$$ be a tetrahedron on four non-adjacent vertices of $$A$$ (i.e no two vertices of $$B$$ share a common edge of $$A$$). Compute the volume of $$B$$.

This site gives a way to calculate but there's gotta be a closed form elegant formula for this.

Let the rectangular parallelepiped measurements be $$a,\,b,\,c$$ then the mixed product of the three vectors is $$6$$ times the volume of the desired tetrahedron $$V=\frac16\operatorname{abs} \left| \begin{array}{ccc} a&b&0\\ a&0&c\\ 0&b&c \end{array} \right|=\frac13 abc$$ Thus the desired volume is $$\frac13\cdot 15\cdot 20\cdot 30=3000$$.
• More simply (to my mind, anyway), each of the tetrahedra at the vertices $A', B, C', D$ has volume $\tfrac16abc,$ so the tetrahedron $AB'CD'$ has volume $abc - \frac23abc = \frac13abc.$ The question Volume of a tetrahedron whose 4 faces are congruent is relevant, but has more detail than is needed here. Aug 11 '20 at 21:00