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If I have $n$ known planes (known normal vector and a point on a plane) that intersect each other in such a way so as to form closely located 3D lines, then

(1). To get a common single 3D line to represent these intersections via least square solutions approach, minimally how many planes should be intersected?

(2). How do I get the best 3D line using least square minimization? Any formulae?

enter image description here

If i take profile view of planes, Red dots indicate the intersection lines (passing into the computer screen), so, i am looking for best line to be represented all of them.

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Mind if I ask why you added eigenvalues and svd tags? – Shuhao Cao Jun 4 '13 at 23:49
up vote 1 down vote accepted
  1. Answer should be that 2 planes suffice, provided these two planes have different normal vector. The least square approach will get you the true intersection line of two planes.

  2. When there are 3 or more planes, they may not intersect with each other at just one single line. For how to get the line when we assume there are no parallel planes (planes with the same normal), please refer to my answer in this question: least square solution for obtaining a line 3d by intersecting many planes.

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thank you.. does that link related to the finding direction vector and position on that line? – slinga Jun 5 '13 at 8:39
but, if two planes are minimum, that mean we have only one line3d, then, I feel we dont want least square... Is that 3 planes? – slinga Jun 5 '13 at 13:29
@slinga Least square is not necessary for 2 planes, for the intersection is uniquely determined, but the result least square yields will be the same with using classical methods. – Shuhao Cao Jun 5 '13 at 13:33
@slinga Yes, basically we are looking for the equation of the line $\mathbf{x}=\mathbf{x}_0+t\mathbf{b}$ where $\mathbf{x}_0$ is the point, $\mathbf{b}$ is that vector. – Shuhao Cao Jun 5 '13 at 13:35

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