Cross product is usually defined such that $$\lVert\bf{a} \times \bf{b} \rVert =\lVert \mathbf{a}\rVert \cdot\lVert \mathbf{b} \rVert \sin\theta$$ where the result vector is perpendicular to both $a$ and $b$, with a direction given by the right-hand rule.
However, I am looking for an explanation of the following equivalence of the cross product, which states ( in Computer Vision Algorithms and Applications, p.310) that:
The cross-product operator projects a vector onto a set of orthogonal basis vectors that include $\hat{t}$, zeros out the $\hat{t}$ component, and rotates the other two by 90◦ $$[\hat{\bf{t}}]_\times = \bf{SZR_{90\circ}S}^T = \begin{bmatrix} \bf{s_0} & \bf{s_1} & \bf{\hat{\bf{t}}}\\ \end{bmatrix} \begin{bmatrix} 1&& \\ &1& \\ &&0 \end{bmatrix} \begin{bmatrix} 0&-1& \\ 1&0& \\ &&1 \end{bmatrix} \begin{bmatrix} \bf{s}^T_0 \\ \bf{s}^T_1 \\ \hat{\bf{t}}^T\\ \end{bmatrix} $$ where $\hat{\bf{t}} = \bf{s_0}\times\bf{s_1}$
Why is this the case?
Note:
$[\hat{\bf{t}}]_\times$ is the matrix form of the cross product operator with $\hat{\bf{t}} = (\hat{t}_x, \hat{t}_y, \hat{t}_z),$ $$[\hat{\bf{t}}]_\times = \begin{bmatrix} 0 & -\hat{t}_z & \hat{t}_y \\ \hat{t}_z & 0 & -\hat{t}_x \\ -\hat{t}_y & \hat{t}_x & 0 \end{bmatrix} $$