Basic rotations in $3-D$ are represented by the matrices: $$ R^3_{xy}(\theta)= \begin{bmatrix} \cos \theta &-\sin \theta & 0\\ \sin \theta &\cos \theta & 0\\ 0& 0 & 1\\ \end{bmatrix}\quad \text{rotation in the plane}\; (xy) $$
$$ R^3_{xz}(\theta)= \begin{bmatrix} \cos \theta&0 &\sin \theta\\ 0& 1 & 0\\ -\sin \theta & 0&\cos \theta \\ \end{bmatrix}\quad \text{rotation in the plane}\; (xz) $$ $$ R^3_{yz}(\theta)= \begin{bmatrix} 1& 0 & 0\\ 0 &\cos \theta&-\sin \theta\\ 0&\sin \theta & \cos \theta \\ \end{bmatrix}\quad \text{rotation in the plane}\; (yz) $$ where te sign of the $\sin \theta$ components are determined by the right-hand rule.
In $4-D$ space the basic rotations becomes something as $$ R^4_{xy}(\theta)= \begin{bmatrix} \cos \theta &-\sin \theta & 0&0\\ \sin \theta &\cos \theta & 0&0\\ 0& 0 & 1&0\\ 0& 0 & 0&1\\ \end{bmatrix} \quad R^4_{xz}(\theta)= \begin{bmatrix} \cos \theta&0 &\sin \theta&0\\ 0& 1 & 0 &0\\ -\sin \theta & 0&\cos \theta&0 \\ 0&0&0&1 \end{bmatrix} $$
$$ R^4_{xt}(\theta)= \begin{bmatrix} \cos \theta&0&0 &-\sin \theta\\ 0& 1 & 0&0\\ 0& 0 & 1&0\\ \sin \theta&0&0 &\cos \theta \\ \end{bmatrix} \quad R^4_{yz}(\theta)= \begin{bmatrix} 1 & 0 &0 &0\\ 0&\cos \theta&\sin \theta&0\\ 0&-\sin \theta & \cos \theta &0\\ 0&0&0&1 \end{bmatrix} $$
$$ R^4_{yt}(\theta)= \begin{bmatrix} 1& 0& 0&0\\ 0&\cos \theta&0 &\sin \theta\\ 0& 0 & 1&0\\ 0&-\sin \theta&0 &\cos \theta \\ \end{bmatrix} \quad R^4_{zt}(\theta)= \begin{bmatrix} 1 & 0 &0 &0\\ 0&1&0&0\\ 0&0&\cos \theta &-\sin \theta\\ 0&0&\sin \theta &\cos \theta \\ \end{bmatrix} $$ But here (as far as i know) we don't have a simple extension of the right-hand rule, so what is the rule for the signs of the $\sin \theta$ components?