Construct orthogonal projection for plane (matrix form)

Question

I am currently trying to get up-to-date with my university level math (i kinda slacked off a little), using some homework that our professor provied for us. Now, one task is like this (original Language is German, so translation might be inaccurate):

There is a cube in a $R^3$ space with the following coordinates for its corners: $$ \begin{pmatrix}0\\0\\0 \end{pmatrix} \qquad \begin{pmatrix}1\\0\\0 \end{pmatrix}\qquad \begin{pmatrix}0\\1\\0 \end{pmatrix}\qquad \begin{pmatrix}0\\0\\1 \end{pmatrix} $$ $$ \begin{pmatrix}1\\1\\0 \end{pmatrix} \qquad \begin{pmatrix}1\\0\\1 \end{pmatrix}\qquad \begin{pmatrix}0\\1\\1 \end{pmatrix}\qquad \begin{pmatrix}1\\1\\1 \end{pmatrix} $$ Now the task is divided in two subpoints: 1) Construct a plane $E$, so that the orthogonal projection of the cube onto $E$ creates a uniform hexagon (again, don't really know how to translate this). 2) Construct the matrix for this orthogonal projection, with appropriate coordinates.

Now, im not asking anyone to solve my homework for me. I'm just kinda stuck here - can anybody give me tips/outline how i should approach this problem? I know how i have to rotate the cube to get a uniform hexagon from a certain viewpoint, i just don't know how to apply this properly.

Update:

Is there any way to get the plane formula for $E$ mathematically, that is not using concepts like "get a physical cube and try it out"?

Answer 1

Hint: There's no substitute for actually getting yourself a physical cube and trying to figure out how to cut it to see a regular hexagon (this is the word, not uniform hexagon).

Answer 2

Hint:

Consider a plane $\alpha$ passing through the origin $O=(0,0,0)$. If this plane is orthogonal to the vector of the main diagonal of the cube $\vec D=(1,1,1)^T$ than the orthogonal projection of the vertex $D$ is coincident with $O$ and, by symmetry, the projections of all other vertices are equidistant from $O$ and form a regular hexagon.

You can prove this noting that the projection of a vertex $\vec X$ is given by: $$ \vec X'=\vec X- \frac{(\vec X, \vec D)}{|\vec D|^2} \vec D $$ so the matrix that represents the projection is: $$ \frac{1}{3} \begin{pmatrix} 2&-1&1\\ -1&2&-1\\ -1&-1&2 \end{pmatrix} $$

Obviously, by symmetry, any plane orthogonal to a main diagonal can work.