Body系下空间平面如何转到World系下

题目

已知:传感器坐标系（Body系）下有一平面P方程为Ax+By+Cz+D=0，简写为$$\left[\begin{array}{cc}n& D\end{array}\right]\ast \left[\begin{array}{c}x\\ y\\ z\\ 1\end{array}\right]=0$$,其中n=[A B C];
另外传感器在世界坐标系（W系）下的位姿为$$T=\left[\begin{array}{cc}R& t\\ 0& 1\end{array}\right]$$
求：平面P在W系下的方程

解法

$$\left[\begin{array}{cc}n& D\end{array}\right]\ast \left[\begin{array}{c}{T}^{-1}\end{array}\right]\ast \left[\begin{array}{c}T\end{array}\right]\left[\begin{array}{c}x\\ y\\ z\\ 1\end{array}\right]=0$$

$$\left[\begin{array}{cc}n& D\end{array}\right]\ast \left[\begin{array}{cc}{R}^T& -R^{T}t\\ 0& 1\end{array}\right]\ast \left[\begin{array}{c}T\end{array}\right]\left[\begin{array}{c}x\\ y\\ z\\ 1\end{array}\right]=0$$

$$\left[\begin{array}{cc}n{R}^T& -n{R}^{T}t+D\end{array}\right]\ast \left[\begin{array}{c}T\end{array}\right]\left[\begin{array}{c}x\\ y\\ z\\ 1\end{array}\right]=0$$

$$\left[\begin{array}{cc}n{R}^T& -n{R}^{T}t+D\end{array}\right]\ast \left[\begin{array}{c}{x}^{{}^{\prime }}\\ y^{{}^{\prime }}\\ z^{{}^{\prime }}\\ 1\end{array}\right]=0$$

其中$$\left[\begin{array}{c}{x}^{{}^{\prime }}\\ y^{{}^{\prime }}\\ z^{{}^{\prime }}\\ 1\end{array}\right]$$
为W系的基，因此W系下P的方程为
$$\left[\begin{array}{cc}n{R}^T& -n{R}^{T}t+D\end{array}\right]\ast \left[\begin{array}{c}{x}^{{}^{\prime }}\\ y^{{}^{\prime }}\\ z^{{}^{\prime }}\\ 1\end{array}\right]=0$$

推导完毕。