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### Volume of a Tetrahedron

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Date: 01/23/2002 at 22:17:19
From: Andrew
Subject: Geometry of a Tetrahedron

The volume of a tetrahedron is one-third the distance from a vertex
to the opposite face, times the area of that face. Find a formula for
the volume of a tetrahedron in terms of the coordinates of its
vertices P, Q, R, and S.

I'm not even sure where to begin. I think it may have something to do
with cross product multiplication of vectors.
```

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Date: 01/24/2002 at 22:24:44
From: Doctor Pete
Subject: Re: Geometry of a Tetrahedron

Hi,

Thanks for writing to Dr Math. Don't despair - you do know where to
begin, because you mentioned vectors. So I'll begin by setting up some
vectors, which will be denoted by capital letters, in terms of their
coordinates, which will be lowercase letters.  Suppose you have

P = (x1, y1, z1),
Q = (x2, y2, z2),
R = (x3, y3, z3),
S = (x4, y4, z4).

Furthermore, we will write

A = Q - P,
B = R - P,
C = S - P.

In essence, we translated vector P to the origin, and moved Q, R, S
accordingly, to obtain A, B, and C; this will simplify our work.

Now recall the dot product of two vectors

M = (m1, m2, m3),  N = (n1, n2, n3)

satisfy the following properties:

[D1]     M . N = m1*n1 + m2*n2 + m3*n3,
[D2]     M . N = |M||N|Cos[t].

Here |M| signifies the magnitude (length) of M, and t is the angle
between vectors M and N.  As for the cross product, we have

|  i  j  k |
[C1]     M x N = | m1 m2 m3 |,
| n1 n2 n3 |

[C2]     M x N = |M||N|Sin[t]*U,

where in [C1], i, j, k are the unit x-, y-, and z-vectors, and in
[C2], U is the unit vector that is orthogonal to M and N and points in
the direction as specified by the right-hand rule.  (In particular, we
have i x j = k, j x k = i, k x i = j.) A proof of these facts is given
in all textbooks dealing with linear algebra.

The second property about cross products is the main connection to the
geometry of the problem, because geometrically it says that the cross
product of two vectors is a third vector orthogonal to the other two,
with magnitude equivalent to the area of the parallelogram defined by
the two vectors. Perhaps a picture will show this:

N_______________
/|             /
/ |h           /
/  |_          /
/___|_|________/
0               M

In the above diagram we are looking at the plane containing the
vectors M and N. The height h of the parallelogram is simply
|N|Sin[t], where t is the angle M0N, the angle between M and N.
Thus the area of the parallelogram is |M||N|Sin[t]. The vector M x N
is pointing in a direction perpendicular to this plane (straight at
you), by the right-hand rule.

The curious thing about the cross product, then, is that the area of
the triangle determined by points M, N, and 0, is simply half the
magnitude of the cross product, because the parallelogram consists of
two congruent copies of triangle M0N. Thus, in the case of our vectors
A, B, C, we may choose any two of these to show that the area of the
triangular face determined by, say, vectors B and C, is simply

|B x C|/2.

But wait - there's more. We observe that the vector B x C is parallel
to the height from the vertex at A to the opposite face. If we draw
another picture in the plane that contains the vectors B x C, A, and
the length from A to the plane containing B and C, as follows,

B x C
|
|     A
|    /|
|   / |
|  /  |d
| /   |
|/____|_____B___C
0      G

we see that B and C are now projected onto this plane and appear as a
single line. The important thing to realize is that in this picture,
vector A is in the same plane as B x C and the line segment AG. If we
let s = angle 0AG, then we may write the distance d of AG as simply

d = |A|Cos[s].

Therefore, the volume of the tetrahedron is

V = (1/3)d|B x C|/2
= (1/6)|A||B x C|Cos[s].

But s is also the angle between A and B x C, and if we recall the
formula [D2] for the dot product of two vectors, we find that

V = (1/6)|A . (B x C)|.

The product A . (B x C) is more commonly called the (scalar) triple
product, because (with some slight details omitted) the symmetry of
our argument reveals that

A . (B x C) = B . (C x A) = C . (A x B).

Now we may write A . (B x C) in terms of the coordinates using the
formulas [D1, C1].  We have

|  i     j     k  |
B x C = |x3-x1 y3-y1 z3-z1|,
|x4-x1 y4-y1 z4-z1|

and since A = (x2-x1, y2-y1, z2-z1), we immediately see that

|x2-x1, y2-y1, z2-z1|
A . (B x C) = |x3-x1, y3-y1, z3-z1|.
|x4-x1, y4-y1, z4-z1|

Thus we have a formula for V in terms of the coordinates of P, Q, R,
and S. But shouldn't this formula be symmetric in the coordinates? It
is - it's just that it isn't obvious from looking at it. I leave it to
you to show that the above determinant is equivalent to

|x1  y1  z1  1|
|x2  y2  z2  1|
|x3  y3  z3  1|
|x4  y4  z4  1| .

- Doctor Pete, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
College Linear Algebra
College Polyhedra
High School Linear Algebra
High School Polyhedra

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