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Volume of a Tetrahedron

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.

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


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:

           /|             /
          / |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
      | /   |
     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   
Associated Topics:
College Linear Algebra
College Polyhedra
High School Linear Algebra
High School Polyhedra

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