Compute the volume of the tetrahedron with vertices at the origin, (1, 2, 0), (2, 1, 3) and (0, 1, 1), and give the general formula for the volume of the n dimensional simplex spanned by n edge vectors.

Compute the volume of the tetrahedron with vertices at the origin, (1, 2, 0), (2, 1, 3) and (0, 1, 1), and give the general formula for the volume of the n dimensional simplex spanned by n edge vectors.

Approach: Place the three edge vectors from the origin into a matrix, take the absolute value of its determinant for the parallelepiped volume, then divide by the factorial that converts a box into a simplex.

1. The three edge vectors from the origin are (1, 2, 0), (2, 1, 3) and (0, 1, 1). Their determinant, which is the scalar triple product, is 1*(1*1 - 3*1) - 2*(2*1 - 3*0) + 0*(2*1 - 1*0) = -2 - 4 + 0 = -6, so the parallelepiped they span has volume 6. The simplex volume is that divided by 3! = 6, giving 1. In n dimensions the simplex spanned by edge vectors v_1, ..., v_n from a common vertex has volume |det[v_1 ... v_n]| / n!, because the unit cube splits into exactly n! congruent simplices sorted by the ordering of the coordinates.

Follow-up: What fraction of the unit cube in R^10 lies in the simplex x_1 >= x_2 >= ... >= x_10, and what does that ratio say about sampling by sorting?

Key concepts: simplex volume, determinant, parallelepiped, scalar triple product.