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 December 1st, 2010, 01:20 PM #1 Newbie   Joined: Dec 2010 Posts: 2 Thanks: 0 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 August 28th, 2011, 10:52 PM #2 Senior Member   Joined: Aug 2011 Posts: 334 Thanks: 8 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) August 29th, 2011, 04:27 AM #3 Math Team   Joined: Sep 2007 Posts: 2,409 Thanks: 6 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, = 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 = 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 = 0 to = 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. August 29th, 2011, 05:39 AM #4 Senior Member   Joined: Aug 2011 Posts: 334 Thanks: 8 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. September 20th, 2011, 02:03 PM #5 Newbie   Joined: Dec 2010 Posts: 2 Thanks: 0 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. Tags find, fitting, plane, points ,

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