# relationship between two planes (or their (N,d) vectors)

I'm reading this article in computer vision and I just can't get my head around eqution(6). The scenario is as follows: We have the pose of a camera in world coordinate system as $T_{w,c}$ so that a 3D point defined in local camera coordinate system $p_c$ could be transformed to the world coordinate system using $p_w=T_{w,c}p_c$. Then a plane in camera coordinate system, represented by a 4D vector of its normal and diatnce to the origin $\pi_c=[n^T,d]$. could be transformed to world coordinates using $\pi_w=T_{c,w}^{-T}\pi_c$

I have no idea how we got to this final equation for planes from the info given above. Any help on this will be appreciated.

A general equation of a plane is $Ax+By+Cz+D=0$, which can be written in vector form as $[A,B,C]\cdot[x,y,z]=-D$. From this equation we can find that the point $$\mathbf x_0 = {-D\over A^2+B^2+C^2}[A,B,C]$$ lies on the plane. Now, for any point $\mathbf x$ on the plane, we have $[A,B,C]\cdot(\mathbf x-\mathbf x_0)=-D-(-D)=0$, but $\mathbf x-\mathbf x_0$ is parallel to the plane, so $[A,B,C]$ is normal to it. This also means that $$d=\|\mathbf x_0\| = \left\| {-D\over A^2+B^2+C^2}[A,B,C] \right\| = {|D| \over \sqrt{A^2+B^2+C^2}}$$ is the distance of the plane from the origin. Without the absolute value, you have $d=-D/\sqrt{A^2+B^2+C^2}$ as the signed distance from the origin, i.e., it is the distance you need to move along $[A,B,C]$ to reach the plane, which is negative if this vector points away from the plane.
If we take $\mathbf n$ to be a unit normal to the plane, we then have $\mathbf n\cdot\mathbf x = d$ as the vector equation of the plane, which we can rewrite as $[\mathbf n,-d]\cdot[\mathbf x,1]=0$. Switching to column vectors, we get $\mathbf\pi=[\mathbf n^T,-d]^T$ (note the change of sign from what you have in your question). This vector is covariant, which means that if points transform as $\mathbf p_w=T_{w,c}\mathbf p_c$, it transforms as $\mathbf\pi_w = T_{w,c}^{-T}\mathbf\pi_c$. I’ll leave that to you to verify.
• @Musaab It was not at all clear to me from the way you phrased your question that it was this just final step that you didn’t understand. It’s almost trivial: $\mathbf\pi_c^T\mathbf p_c=0 \Leftrightarrow \mathbf\pi_c^TT_{w,c}^{-1}T_{w,c}\mathbf p_c=0 \Leftrightarrow (T_{w,c}^{-T}\mathbf\pi_c)^T(T_{w,c}\mathbf p_c)=0 \Leftrightarrow (T_{w,c}^{-T}\mathbf\pi_c)^T\mathbf p_w=0$, so $\mathbf\pi_w=T_{w,c}^{-T}\mathbf\pi_c$. – amd Nov 9 '17 at 19:40