MHB Symmetry Operations of a Cube: Geometric Descriptions and Matrix Representations

[mathmari]

The vector $v$ is fine.
The problem is that we have 2 possible choices for $v'$ and we need to pick the right one.
That is, the one such that $v\times v'$ is in the same direction as $u$.

Taking $v'=\left (0, 1, 1\right )$ we have the following:

We have the vector $w=u\times v=(1,1,-1)\times (-1,0,-1)=(-1,2,1)$ and we get $\displaystyle{w'=\|w\|\frac{u\times v'}{\|u\times v'\|}=\sqrt{6}\frac{(1,1,-1)\times (0, 1, 1)}{\|(1,1,-1)\times (0, 1, 1)\|}=\sqrt{6}\frac{(2,-1,1)}{\sqrt{6}}=(2,-1,1)}$.

The condition is now satisfied, isn't it? We have $w\cdot v'=3$.

So we get the matrix \begin{equation*}\begin{pmatrix}0 & 1 & 0 \\ 0 & 0 & -1 \\ -1 & 0 & 0\end{pmatrix}\end{equation*}

:unsure:

 

[I like Serena]

The condition is now satisfied, isn't it? We have $w\cdot v'=3$.

So we get the matrix \begin{equation*}\begin{pmatrix}0 & 1 & 0 \\ 0 & 0 & -1 \\ -1 & 0 & 0\end{pmatrix}\end{equation*}

Yep. All correct. (Nod)