# Sec.12.4 - The Geometry of a Tetrahedron

Discovery Project in Sec.12.4, Calculus by Stewart

Chinese version 四面體的幾何學

A tetrahedron is a solid with four vertices, *P* ,*Q* , *R* and *S* ,and four triangular faces, as shown in the figure.

**Question 1:**

Let $v_1, v_2, v_3$ and $v_4$ be vectors with lengths equal to the areas of the faces opposite the vertices *P*, *Q*, *R* and *S*, respectively, and directions perpendicular to the respective faces and pointing outward. Show that
$$
v_1+v_2+v_3+v_4=0
$$

**Question 2:**

The volume *V* of a tetrahedron is one-third the distance from a vertex to the opposite face, times the area of the face.

(a)Find a formula for the volume of a tetrahedron in terms of the coordination of its vertices *P* ,*Q* , *R* and *S*.

(b)Find the volume of the tetrahedron whose vertices are *P*(1,1,1), *Q*(1,2,3), *R*(1,1,2), and *S*(3,-1,2).

**Answer:**

**(b)**

By (a), a=(0,1,2), b=(0,0,1), c=(2,-2,1).

So
$$
V=\frac{\left |
\begin{array}{ccc}
0 & 1 & 2 \\

0 & 0 & 1 \\

2 & -2 & 1 \\

\end{array}\right |}{6}=\frac{1}{3}
$$

**Question 3:**

Suppose the tetrahedron in the figure has a trirectangular vertex *S*. (This means that the three angles at *S* are all right angles.) Let *A*,*B*,and *C* be the areas of the three faces that meet at *S*,and let *D* be the area of the opposite face *PQR*. Using the result Question 1, or otherwise, show that
$$
D^{2}=A^{2}+B^{2}+C^{2}
$$
(This is a three-dimensional version of the Pythagorean Theorem.)