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Subject: Re: Point Normal Form of a plane
Date: Thu, 16 Dec 2010 05:40:38 +0000 (UTC)
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"GAURAV " <gsharda@engineering.uiowa.edu> wrote in message <iec43h$k0a$1@fred.mathworks.com>...
> Hi!
> 
> I have three points. P1(x1,y1,z1), P2(x2,y2,z2), P3(x3,y3,z3).
> I need plane passing through these three points. I just need a point and a normal for this.
> Can someone help me out.
> 
> Thank You,
> Regards,
> Gaurav
- - - - - - - - - -
  You already have a point, in fact three of them.  Provided the three given points are not colinear, then cross(P2-P1,P3-P1) is an orthogonal vector to the plane.  You can also characterize the plane by the equation:

 dot((P-P1),cross(P2-P1,P3-P1)) = 0

for any point P(x,y,z) on the plane.

Roger Stafford