What is or how do you get the rotational matrix of 4-D vector onto the xyz-space?

Question

which would make the 4-D component 0.

To be honest I'm not really sure how 4-D rotations work. I know about the simple rotations but not the mechanism in how it rotates, and I'm not sure whether to use a simple or a double rotation in this case. Other than that I can do rotations in less than 4 dimensions.

If someone is aware of an additional piece of information, that would help solve my whole problem, would be the the transformation of a unit vector along one of the axes into the long diagonal of a unit hypercube. The scaling is easy (multiply by $\sqrt{d}$) for any dimension $d$ but I can't get the rotation down.

Thanks in advance.

Answer 1

If I understand your question, you want to find a 4-D rotation matrix $R$ such that $$R\begin{bmatrix}0\\ 0\\ 0\\ 1\end{bmatrix} = \frac 12\begin{bmatrix}1\\ 1\\ 1\\ 1\end{bmatrix}.$$ To calculate $R$, see here and here.