Standard matrix of rotation of unit cube A unit cube is centered at $(0,0,0)$ with vertices $(\pm1, \pm1, \pm1)$. What is the standard matrix for a 120 degree rotation about the line joining the points $(-1,-1,-1)$ and $(1,1,1)$?
Know: To get the standard matrix for the rotation, I need to map the original points $(0,0,1), (0,1,0), (1,0,0)$ to their post-rotation position in space. The new coordinates will comprise the rotation matrix. The new points/vectors can be calculated using the Rodrigues Rotation Formula involving cross-products. 
However, is there a more organic method that does not involve cross-products?
 A: Concentrating on just the first octant, we can see that a rotation by $\frac{2\pi}{3}$ (counterclockwise of course, with respect to the plane through the tips of $i,j,k$ (this is normal to $(1,1,1)$) sends $i$ to $j$, $j$ to $k$ and $k$ to $i$, where $i,j,k$ are my names for the standard basis. This plane cuts the first octant at an equilateral triangle, and the rotation is just permuting the vertices.
Writing this as a matrix on the left of column vectors, we would have then
$$T=\begin{bmatrix}0&0&1 \\1&0&0\\0&1&0\end{bmatrix}$$
As you can see:
$$T\begin{bmatrix}1\\0\\0\end{bmatrix}=\begin{bmatrix}0\\1\\0\end{bmatrix}$$
$$T\begin{bmatrix}0\\1\\0\end{bmatrix}=\begin{bmatrix}0\\0\\1\end{bmatrix}$$
$$T\begin{bmatrix}0\\0\\1\end{bmatrix}=\begin{bmatrix}1\\0\\0\end{bmatrix}$$
So it is indeed the transformation that permutes $i,j,k$ cyclically.
A: I don't know whether you'll find this approach to be more "organic", but here's one that does not involve cross products:
The matrix for a $120^\circ$ rotation about the $z$-axis is given by
$$
R = \frac 12 \pmatrix{-1 & \sqrt{3}&0\\-\sqrt 3 & -1&0\\0&0&2}
$$
One way to get the desired matrix is to take this matrix and apply a "change of basis".  That is, select an $x,y,z$-axis corresponding to this new rotation.
Of course, our new $z$-axis will be the unit vector in the $(1,1,1)$ direction.  For $x,y$ axes, it suffices to take any two orthogonal directions in the normal plane.  In particular, we can take the unit vectors in the $(1,-1,0)$ and $(1,1,-2)$ directions.  
Let $U$ be the matrix whose columns form this orthonormal basis (in x,y,z order).  That is,
$$
U = 
\pmatrix{
1/\sqrt 2  & 1/\sqrt 6 & 1/\sqrt 3\\
-1/\sqrt 2 & 1/\sqrt 6 & 1/\sqrt 3\\
0          & -2/\sqrt 6 & 1/\sqrt 3
}
$$
The matrix of the desired rotation will be given by
$$
A = URU^T = \\
\frac 12 
\pmatrix{
1/\sqrt 2  & 1/\sqrt 6 & 1/\sqrt 3\\
-1/\sqrt 2 & 1/\sqrt 6 & 1/\sqrt 3\\
0          & -2/\sqrt 6 & 1/\sqrt 3
}
\pmatrix{-1 & \sqrt{3}&0\\-\sqrt 3 & -1&0\\0&0&2}
\pmatrix{
1/\sqrt 2  & 1/\sqrt 6 & 1/\sqrt 3\\
-1/\sqrt 2 & 1/\sqrt 6 & 1/\sqrt 3\\
0          & -2/\sqrt 6 & 1/\sqrt 3
}^T
$$
where $T$ denotes the matrix transpose.

Note: there are two $120^\circ$ rotations, each corresponding to a "clockwise" or "counterclockwise" rotation.  To get the opposite rotation, begin with 
$$
R = \frac 12 \pmatrix{-1 & -\sqrt{3}&0\\ \sqrt 3 & -1&0\\0&0&2}
$$
