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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.

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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.

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  • $\begingroup$ Thanks for taking the time but at the end of your post you stated what the paper was claiming anyway and then you left it to me to verify it, which it was my question to begin with , I couldn't verify it. $\endgroup$ – Musaab Nov 9 '17 at 12:58
  • $\begingroup$ @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 $\mathbf\pi_w=T_{w,c}^{-T}\mathbf\pi_c$. $\endgroup$ – amd Nov 9 '17 at 19:40

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