If I were to represent the sides of a dice as a vector formula, what would that formula be?
For example, a 6 sided die whose center is at 0,0,0 would be 1,0,0 for the center of one face, 0,1,0 for another; 0,0,1; -1,0,0; 0,-1,0 and 0,0,-1
but as you increase the number of sides in a dice the more they are like a sphere, so a 12 sided dice has equal faces and what would be the way to represent this concept of dividing a sphere into a shape with faces and basically to calculate the vectors of each of those faces.
Thanks in advance.
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$\begingroup$Wikipedia gives co-ordinates of a regular dodecahedron's $20$ vertices as $$(\pm 1, \pm 1, \pm 1), \left(0, \pm \phi, \pm \tfrac1\phi\right), \left(\pm \tfrac1\phi, 0, \pm \phi\right), \left(\pm \phi, \pm \tfrac1\phi, 0\right)$$ where $\phi = \frac{1+\sqrt{5}}{2}$. You can spot a cube in there, and there are in fact five such cubes
For an icosahedron [$12$ vertices] it gives $(0, \pm 1, \pm \phi), (\pm 1, \pm \phi, 0), (\pm \phi,0, \pm 1)$,
for an octahedron [$6$ vertices] it gives $(0, 0, \pm 1), (0,\pm 1, 0), ( \pm 1,0,0)$,
for a tetrahedron [$4$ vertices] it gives $\left(\pm1, 0, -\frac1{\sqrt{2}}\right),\left(0,\pm1,+ \frac1{\sqrt{2}}\right)$,
and for a cube [$8$ vertices] it gives $(\pm1, \pm 1, \pm1)$
There is no direct analogon for other platonic solids. This nice representation of the cube that you discovered only works because the parallel faces can be identified with the axes of the coordinate system.
If you want to compute the normal vectors of the faces of such a solid, it is probably easiest if you start by determining the angles between the faces.
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