Coordinates of a rhombohedron

Question

I have found this question (Coordinates for vertices of the "silver" rhombohedron.) which asks:

"The "silver" rhombohedron (a.k.a the trigonal trapezohedron) is a three-dimensional object with six faces composed of congruent rhombi. You can see it visualised here.

I am interested in replicating the visualisation linked to above in MATLAB, but to do that I need to know the coordinates of the rhombohedron's eight vertices (I am using the patch function).

How do I calculate the coordinates of the vertices of the silver rhombohedron?"

The question has this answer:

"Use vectors e1=(1,0,0), e2=(cosα,sinα,0) and e3=(cosα,0,sinα) as basis. Then vertices are set of all points with each coordinate 0 or 1: (0,0,0), (0,0,1), ..., (1,1,1). Or 0, e1, e2, e3, e1+e2, ..., e1+e2+e3.

Multiply coordinates by constant if needed."

Using this methodology I find that two of the faces of the resulting rhombohedron are not congruent and do not have angle "α".

What vectors would I use for all faces of the rhombohedron to be congruent?

Answer 1

One construction is to choose the basis $$e_1 = (a,b,b), \quad e_2 = (b,a,b), \quad e_3 = (b,b,a),$$ with $a > 0$ and $b \not \in \{-a/2, a\}$. This symmetry guarantees the congruence. All that is needed now is to normalize the edge length, which is to say, divide by $$\sqrt{a^2 + 2b^2}.$$ To obtain the interior angle $\alpha$, it is a straightforward calculation of the dot product: $$e_1 \cdot e_2 = 2ab + b^2 = |a||b| \cos \alpha = (a^2 + 2b^2) \cos \alpha,$$ consequently $$\alpha = \cos^{-1} \frac{2ab + b^2}{a^2 + 2b^2}.$$ We can express $b$ in terms of $a$, then after normalization, obtain a basis solely in terms of $\alpha$.