Say I have a 3D plane in frame $A$ parametrized by $ax + by + cz = d$. Now I also have a frame $B$ and I know the $4\times4$ transformation matrix between $A$ and $B$. How do I obtain the parameters of the plane in frame $B$.
An easy way would be to take the normal of the plane and apply the $4\times4$ transformation to that normal and then solve for the new $d$. Is there a more elegant way of doing this?
In homogeneous coordinates, the implicit Cartesian equation $ax+by+cz=d$ of the plane can be written in the vector form $\mathbf\pi^T\mathbf x=0$, with $\pi=(a,b,c,-d)^T$. Suppose you have the invertible point transformation $\mathbf x'=\mathtt M\mathbf x$. Then $$\mathbf\pi^T\mathbf x = \mathbf\pi^T(\mathtt M^{-1}\mathbf x') = (\mathtt M^{-T}\mathbf\pi)^T\mathbf x',$$ i.e., a plane $\pi$ transforms as $\mathbf\pi' = \mathtt M^{-T}\mathbf\pi$.