Let $S \subset \mathbb{R^3}$ be a regular simplex (tetrahedron) with boundary $\partial S$ and edge length $1$.

How to compute

1) the angle between two different edges?

2) the Lebesgue measure $\lambda_3(S)$?

For 1) I know that on a regular simplex, the edges on each face is forming a equilateral triangle. For the angle I know it is $60°$ or $70.5°$ but what is the difference?

I tried to verify this value with the formula for the dot product (I chose the vertices $v_1=(1,1,1),v_2=(1,-1,-1)$ and $v_3=(-1,1,-1)$), since the edge length is $1$, but as a result I get $109°$. How the $60°$ can be computed?

For 2) the volume of an n-simplex in n-dimensional space with vertices $(v_0, ..., v_n)$ is

$|\frac{1}{n!}det(v_1-v_0, v_2-v_0, …, v_n-v_0)|$

So in this case I used

$|\frac{1}{3!}$det$ \begin{pmatrix} 1 & 1 & -1 \\ 1 & -1 & 1 \\ 1 & -1 & -1 \end{pmatrix}|=\frac{2}{3}$

Is this solution right for the volume/Lebesgue measure or is there another way to compute this?


You state that the edge length ought be 1, but then set up vertex vectors $v_i$ differently.

Okay, taking your vertex vectors the edge vectors then would be $$w_2=v_2-v_1=\begin{pmatrix}0\\-2\\-2\end{pmatrix}, \ \ \ \ w_3=v_3-v_1=\begin{pmatrix}-2\\0\\-2\end{pmatrix}$$ Thus wrt. your question 1: $$w_s\cdot w_3=\begin{pmatrix}0\\-2\\-2\end{pmatrix}\cdot\begin{pmatrix}-2\\0\\-2\end{pmatrix}=0\cdot(-2)+(-2)\cdot0+(-2)\cdot)(-2)=4$$ OTOH $$||w_2||^2=||w_3||^2=\begin{pmatrix}-2\\0\\-2\end{pmatrix}\cdot\begin{pmatrix}-2\\0\\-2\end{pmatrix}=(-2)^2+0^2+(-2)^2=8$$ And so $$\cos(\phi)=\frac{w_2\cdot w_3}{||w_2||\cdot||w_3||}=\frac48=\frac12$$ or $$\phi=60°$$ Wrt. your question 2 you likewise should use the edge vectors: $$|\frac1{3!}\det\begin{pmatrix}0&-2&-2\\-2&0&-2\\-2&-2&0\end{pmatrix}|=\\=|\frac16\left[0\cdot0\cdot0+(-2)(-2)(-2)+(-2)(-2)(-2)-0(-2)(-2)-(-2)0(-2)-(-2)(-2)0\right]|=\\=\frac{16}6=\frac83$$ But be aware, $||w_2||^2=||w_3||^2=||w_4||^2=8$ and $\lambda_3(S)$ scales according to the cube of the edge size. Thus you get $$\lambda_3(S)=\frac{8/3}{(\sqrt{8})^3}=\frac1{6\sqrt{2}}$$ --- rk

  • $\begingroup$ The vertex vectors $v_1=(1,1,1),v_2=(1,-1,-1)$ and $v_3=(-1,1,-1)$ weren't given, just the edge length $1$, I just picked those vectors $v_1, v_2, v_3$ randomly. Does it matter for the angle and Lebesgue measure or is this solution general? $\endgroup$
    – Tartulop
    Nov 5 '18 at 21:39
  • $\begingroup$ Those $v_i$ are vertex vectors of a regular tetrahedron which is the alternation of an axes parallel cube with edge length 2. Thus $||w_i||^2=8$. $\endgroup$ Nov 6 '18 at 12:55
  • $\begingroup$ If $\mathcal{S}=\{{\mathbf{x} }\in \mathbb{R}^3: \, x_j \in [0,1], 1\leq j \leq 3; \, (x_1+x_2+x_3)=1\}$, then $\mathcal{S}$ is a lower dimensional subspace of $\mathbb{R}^3$. Shouldn't it be $\lambda_3(\mathcal{S})=0$? $\endgroup$ Apr 15 '20 at 8:44
  • $\begingroup$ Sure it is, @JackLondon, but your $S$ then has nothing to do with the $S$ of the question. $\endgroup$ Apr 17 '20 at 7:35

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