Reading at Mathworld, I came across the subject of tetrahedrons. Particularly calculating the volume with four known vertices. There's a formula which uses the triple product to calculate the volume of a parallelepiped. I'm very aware of why the triple product represents the volume of a parallelepiped.

However, I don't understand how to derive the relation that the volume of a tetrahedrons is one-sixth of a parallelepiped. I can accept it, sure. And graphically, I'm sure that it is true.

Is there any way to prove it using linear algebra?

Wikipedia provides me with one proof. However, I don't understand how they went from pyramid volume being $1/3\,A_0h$ to tetrahedrons being $1/6\,A_0h$. It must mean that the volume of a tetrahedron is half that of a pyramid, must it not?


In the unit cube, consider the vertices with $x \le y \le z$; these form a tetrahedron. Indeed, for any of the six possible orderings of the variables, you get a tetrahedron, and the interiors of these tets are disjoint, and every point of the unit cube lies in one of the tets. In fact, all the tets have the same shape (the long edge is the main diagonal from $(0,0,0)$ to $(1,1,1)$, etc.) So the volume of each is 1/6 the volume of the cube.

That's only for a cube, but other cases follow by applying linear transformations.


Volume of tetrahedron = 1/3 (base area) (height) Volume of parallelopiped = (base area) (height) They have same heights, but the base area of the tetrahedron is half of that of the parallelopiped. So,

Volume of paralellopiped= 6 times volume of tetrahedron.


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