Suppose, i have been given different planes which orients to different direction (i.e. i know only the plane parameter of those planes). If i am able to find out planes (probably more than 3 planes) which should intersect (or meet) at a certian point (this is the corner point where those planes make), then how can i find that point. this point would be the one which is closest to all those seleted planes. if any one can explain the way to go further then it is gratefull as i have to use least square to get a more accurate values.i found the CramersRule's but i am not sure whether it fit for this problem and i am not sure how to implement it as i am poor in this geometric and matrices things. thank you in advance.
Computing the point which is closest to many Planar surfaces
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geometry
matrices
algorithms
computational-geometry
solid-geometry
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0@Suresh Venkat: thanks for commenting. but, for me, i can not imagine eaven what Joseph O'Rourke has meant. OK, I will decribe my problem in another way. if i know intersection line of a certain pair of planes, then, i think i can calculate the intersection point of all the lines. and, there will not be a point where all the lines are passing. so, i have to find the point which is closest to all the lines. SO, MY INTENTION IS TO AVOIDING OF WORKING WITH LINES. I want to use planes. So, as i said i need the point which is closest to all the planes, which minimises sum of the squre distances. – 2011-12-29
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I just realized that since you're working with sum of squares, there might be an easy answer. Suppose your plane $h$ is $\sum a_i x_i +a_0 = 0$. Then the distance of the point $p = (p_1, \ldots p_d)$ from the plane is $d(p,h) = \frac{\sum a_i p_i + a_0}{\sqrt{\sum_i a_i^2}}$
which is a linear expression in $p_i$. Squaring gives you a quadratic in the $p_i$. Now you want to minimize the sum of these squared distances, and that's just a quadratic expression in the $p_i$. Moreover, this expression is convex and therefore has a unique minimum, so taking the partial derivatives and setting them to zero will yield a linear system that you can solve to get the answer.
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0Sure, that's not a problem. just add a comment here. – 2012-01-03