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suppose I got a projective camera model. for this model I would like to back-project a ray through a point in the image plane. I know that the equation for this is the following: $$ y(\lambda) = P^+ \pmb{x} + \lambda \pmb{c} $$ where $P^+$ denotes the pseudoinverse of the camera matrix P. P has a dimensionality of 3 by 4. $x$ is the point on the image plane in homogenous coordinates. Hence it's dimensionality is 3 by 1. $c$ the center of the camera in 3-space in homogeneous coordinates. (note that this equation is taken from the book "Multiple View Geometry in Computer Vision" page 162.)

Now I don't fully get this equation. I get that $P^+ x$ results in a point on the line we are looking for. Hence we have two points that we can use for constructing a line. However I don't get the parametrization using $\lambda$. Why is the equation not in the form like: $$y(\lambda) = (1-\lambda) \pmb{a} + \lambda \pmb{b}$$

Any help in understanding the original equation of the resulting ray would be appreciated! :D

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I'm unfamiliar with this notation, child you clarify it? What is $\lambda$? How is it that you can multiply $P^+$ and $x$, when $P^+$ is a matrix and $x$ is scalar? When you say $c_0$ is the "center of the camera," what does that mean? –  Colin K Jan 26 '12 at 17:40
    
ok let me try to clearify: $x$ is in fact a point on the image plane in homogeneous coordinates. $\lambda$ is just the parameter of the ray. I think geometrically it can be interpreted as the inverse of the depth of the point. (but I'm not 100% sure about that) –  Tobias Domhan Jan 26 '12 at 17:55
    
and the center of the camera is the right null-space of $P$: $PC=0$ –  Tobias Domhan Jan 26 '12 at 18:03
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I'm sort of shocked that with a masters degree in optics, there is a notation that is not only unknown to me, but seems completely nonsensical. You keep taking about a product with $P$ either involving or producing scalars, but isn't $P$ your lens matrix? You say $\lambda$ is "the" ray parameter, but which parameter? Height? Angle? By homogenous coordinates, do you mean normalized coordinates? Is the right null space of $P$ the optical axis? –  Colin K Jan 26 '12 at 19:14
    
OH! I just figured it out. This isn't a geometrical optics question this is a computer vision thing. The "Camera Matrix" is not at all what I thought. Give me some time to figure this out now. –  Colin K Jan 26 '12 at 19:52

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