Finding volume of pyramid in $\mathbb{R}^3$

Question

Assume we have a pyramid (triangular base) sitting in $\mathbb{R}^3$, i.e. we have the coordinates for the vertices $A,B,C$ and $D$. How can we find the volume of the pyramid without using either its height or its base area and no advanced tools like integration?

Answer 1

Hint:

Project the vertices onto the horizontal plane. The projected faces of the pyramid will form an arrangement of four triangles.

Consider the right truncated prism formed by a face and its projection. The volume of this prism is the same as that of a prism with two horizontal faces passing through the centroids of the original faces, i.e. the area of the (triangular) base times the average altitudes of the vertices of the oblique base.

Now the volume of the pyramid is that of the volume of the prisms defined by the up-facing faces minus the volume of the prisms defined by the down-facing faces.

This is a 3D generalization of the shoelace formula.

Answer 2

Hint. Take the vectors $v_1 = B - A$, $v_2 = C - A$ and $v_3 = D- A$. The determinant of the matrix whose lines are $v_1$, $v_2$ and $v_3$ gives the volume of the parallelepiped with edges given by these vectors.