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December 1st, 2010, 01:20 PM   #1
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Find a fitting plane to given 3d points

Hi All,

This is my first post so pardon me if it's in the wrong place.

I am developing a solution where I need to find a plane that is fitted to given points in 3d space. I tried looking at orthogonal distance regression, but I found it a bit difficult to understand. Here is one link (http://mathforum.org/library/drmath/view/63765.html) that I found quite helpful, however I am stuck when the matrix form begins. Can someone help me with either understanding that link or give some alternative solutions?

I appreciate any help. Thanks.

-Maulik
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August 28th, 2011, 10:52 PM   #2
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Re: Find a fitting plane to given 3d points

Hello !

This is a link to a paper dealing with 3-D. Linear Regression and 3-D. Plane Regression :
http://www.scribd.com/people/documents/ ... jjacquelin
Then select "Regressions et trajectoires en 3D".
A non-recursive algorithm for orthogonal mean squares fitting is ginen page 21 (Updated edition July 11, 2011)
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August 29th, 2011, 04:27 AM   #3
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Re: Find a fitting plane to given 3d points

Any (non-vertical) plane can be written as z= Ax+ By+ C. Given points ,, ..., we can think of this as solving the system of equations , , ... , .

And we can cast that into a matrix problem:


Well, of course, if n> 3, this is problematical. We can think of this as "Ax= y" where the matrix, A, maps all of into a 3 dimensional subspace of . If y does not happen to be in that subspace there is no such x. But we can ask for the x such that Ax is closest to y. Of course, in that case, the difference, Ax- y, must be perpendicular to the subspace. That is, <Au, Ax- y>= 0 for any u in .

Now, a little theory. If A is a linear transformation from one inner product space, U, to another, V, then its adjoint, A*, is the linear transformation from V back to U such that <Au, v>= <u, A*v> for all u in U, all v in V. Notice that the inner product on the left is in U, the inner product on the right is in V. It is easy to show that if A is from to , written as a matrix, its adjoint is just its transpose.

Here, we can go from <Au, Ax- y>= 0 to <u, A*(Ax- y)>= 0. The important difference is that now the innerproduct is in and u could be any vector in that space. The only vector that has inner product 0 with all vectors (including itself) is 0. We must have A*(Ax- y)= A*Ax- A*y= 0. Note that A is from to , so a "3 by n" matrix, while A* is from to , written as an "n by 3" matrix so that A*A is a 3 by 3 matrix. There are circumstances in which even A*A does not have an inverse but typically it does. Assuming that, we have, from A*Ax= A*y, x= (A*A)^{-1}A*y.

Here, as before, so that and .

That last is a symmetric 3 by 3 matrix and fairly easily invertible so it is not at all difficult to fine where x, of course, is the vector of coefficients of the desired linear function.
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August 29th, 2011, 05:39 AM   #4
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Re: Find a fitting plane to given 3d points

Hi , HallsofIvy ,
as far as I can understand, maulik13 asked for orthogonal distance regression. The method that you propose is not an orthogonal one.
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September 20th, 2011, 02:03 PM   #5
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Re: Find a fitting plane to given 3d points

Hi Guys! Thank you for you replies. I will have to study your replies and I will post in a bit more detail. And yes I was looking for orthogonal regression analysis.
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