# Does the dimension of R have anything to do with proportion of inscribed triangles to their paralellepipeds?

I had a hard time deciding on how to title this question. I hope it gives people some idea so they can give me their two cents. We all know that the formula for a triangle is $$(1/2)b*h$$

In $\mathbf{R}^2$ of course because that's where a triangle can exist. I know it can exist in higher dimension but it can always be represented in two dimensions. Why is this the formula? It's just obvious, I've never seen anyone write a proof for it but I'm sure it would be about 4 lines and involve some simple vector math.

Now, in Linear class we've come across another, still obvious but slightly less so, formula that the volume of a tetrahedron described by $v_{1}=(a,b,c), v_{2}=(d,e,f) , v_{3}=(g,h,i)$ is

$$(1/6)*volume(Parallelepiped)$$

Where Parallelepiped is a parallelepiped described by those same vectors $v_{1}, v_{2},v_{3}$.

Again just drawing these things out will show why this is the case. But I was wondering, do these proportions have anything to do with the lowest dimension in which the guys can be represented in? Like in 2 dimensions, it's $\frac{1}{2!} = \frac{1}{2}$, then in 3 it jumps to $\frac{1}{3!} = \frac{1}{6}$.

I asked my math teacher (obvious first move) and he said he never thought about it that way and didn't have an answer. So I thought I'd ask somewhere a little more crowdsourced and see if this formula will continue all the way through these triangular-esque inscriptions of parallelepipeds in $\mathbf{R}^{n}$.

• – Macavity Apr 2 '14 at 18:41
• Well... that was easy. Thank you – Alex Chumbley Apr 2 '14 at 18:43