Math CentralQuandaries & Queries


Question from Brittany, a student:


If I am given an irregular tetrahedron with the coordinates of the 4 points how do I find the volume? For example I am asked to find the volume and only given the points A(-4,-3,5), B(2,-1,2), C(0,-5,0), and D(-2,0,0) can you shown me the working and formula to find the volume?

Much Thanks :)


First, subtract the coordinates of one point from those of the others (you can choose any of the four but if you have an option that gives you lots of zeros, take it; it will make your work below easier.)

Here, subtracting D gives: A' = (-2,-3,5), B' = ? C' = ?

What you have just done is to construct a coordinate system with the origin at one of your points.

Now stack the three differences into a 3x3 matrix

\[ \left( \begin{array}{r r r}-2 & -3 & 5\\? & ? & ?\\? & ? & ? \end{array} \right) \]

and take its determinant. The determinant of a 3x3 matrix is the sum of the three forward wraparound diagonal products minus the sum of the three backward ones:

\[ \left| \begin{array}{c c c}a & b & c\\d & e & f\\g & h & i \end{array} \right| = aei + bfg + cdh - afh -bdi - ceg \]

That gives you the volume of the parallelepiped defined by the three vectors. Divide by six to get the volume of the tetrahedron. (If it's negative, you took the three vectors in a cyclic order opposite to that of the three coordinate axes. Either swap two and recalculate, or (if you're being informal) just lose the minus sign.

Good Hunting!


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