# N-dimensional trianglehedron?

1. Jan 20, 2005

### T@P

if you take two unit orthogonal vectors in R2, the triangle they form has area 1/2. take 3 in R3, the volume is 1/6. i claim (and half proved) that the voluarea of an n dimensional bunch of orthognal vectors would give you 1/n!. can anyone prove it fully?

(go easy on the linear algebra, im infantile when it comes to that)

2. Jan 20, 2005

### NateTG

I take it you mean the n-volume of the convex hull formed by the origin and the usual unit vectors in $\Re^n$ under the usual metric.

Let's instead of looking at just the unit length version, look at $v_n(r)$ the 'trianglehedron' with legs of length $r$.

Now, let's prove by induction that
$$v_n(r)=\frac{1}{n!} r^n$$
Then it's easy to see that
$$v_1(r)=\frac{1}{n!} r^1$$
And that for the n+1 case, we can integrate by slicing to get:
$$v_{n+1}(r)=\int_0^rv_n(x)dx$$
but from induction we have:
$$v_n(x)=\frac{1}{n!}x^n$$
so
$$v_{n+1}(r)=\int_0^r\frac{1}{n!}x^n dx$$
which readily works out to
$$v_{n+1}(r)=\frac{1}{n!} \times \frac{1}{n+1} r^{n+1}=\frac{1}{(n+1)!}r^{n+1}$$

Sepecifically,
$$v_n(1)=\frac{1}{n!}$$