Volume of a tetrahedron
I have this problem which I am stuck on!
Find the volume of the tetrahedron with vertices at
The problem also has the note: the volume of the tetrahedron is one sixth of the volume of the
parallelepiped spanned by the vectors .
Any help is much appreciated,
Once you have found the vectors then all you need is a way of computing the volume of the parallelepiped spanned by these vectors. I suggest that you take a look at the geometrical relationship between the determinant of a matrix and the shape spanned by its column vectors.
I didn't want to give too much away! You can find the answer in most good books on Linear Algebra, or even the Wikipedia page about determinants.
This assumes, of course, that the hint you were given is true. I don't actually know how to prove that, so if anyone knows I would like to know!
It's treated in complete detail very early on in any good text on computational geometry (O'Rourke and de Berg et al. are a couple).
Originally Posted by Dranalion
If you know the determinant formulation for the area of a triangle (i.e., a 3x3 determinant which gives (2!)A(T), where A(T) denotes
the area of the triangle), then the volume of a tet is a natural generalization of that formula (which is now a 4x4 determinant which
gives (3!)V(T), where V(T) denotes the volume of the tet).
(This generalization continues into higher dimensions, say d, giving the volume of the d-dimensional simplex, and is why the determinant formulation is so important. Now the constant you'll be stating at will be d!.)
Finally note: 3! = 6 -- the "magic" six (or one sixth as they've chosen to express it) that shows up in the hint.
One can also write the determinant referred to, with , , and as rows (or columns) as .