Best landing plane to minimize orthogonal distances

Question

I have a set of points (in the form x1, y1, z1 ... xn, yn, zn) obtained from a surface mesh. I want to find a suitable three-dimensional plane for these points, minimizing orthogonal distances. x, y, z are independent, that is, I want to get the coefficient A, B, C, D for the plane equation Ax + By + Cz + D = 0.

What will be the algorithm for obtaining A, B, C, D?

Note: in the previous post we discussed the plane of best fit in the sense of least squares, considering the z coordinate as a linear function of x, y. However, this is not my business.

Answer 1

From memory, this turns into a task of its own vector. The distance from the point to your plane is proportional to Ax + By + Cz + D - one of the ways to see this is to note that it is normal to the plane (A, B, C). Constant D is a pain in the neck, but I think you can get rid of it by redefining your variables to transfer it to a constant, so that everything has a value of 0. In this case, I think that the best suitable plane will go through the origin.

Then you find that you want to minimize SUM_i (X_i. A) ^ 2, where A is a 3-vector. Of course, you can make it arbitrarily small by multiplying all the components of A by some small scalar, so you want to minimize this topic by limiting it, for example, || A || ^ 2 = 1, which makes sense of proportionality, making A a unit vector. (X_i. A) ^ 2 = A '(X_i' X) A, so you want to minimize A '(SUM_i (X_i'X_i)) A So, I think you need a minimum eigenvector SUM_i X_i'X_i

One of the reasons why this is often not used in statistics is that your answer will change if you scale the units of any of the coordinate vectors without increasing the same unit sizes in other directions.

Think about it, you can see that it all worked out right at http://en.wikipedia.org/wiki/Total_least_squares

Answer 2

Least Square Data Filtering , Section 2: "Linear fit of nD points using orthogonal regression."

As mentioned in mcdowella, you need to solve a 3x3 system.