Path: utzoo!utgpu!watserv1!watmath!att!att!bu.edu!rpi!zaphod.mps.ohio-state.edu!usc!apple!well!levine
From: levine@well.sf.ca.us (Ron Levine)
Newsgroups: comp.graphics
Subject: Re: How to determine 4 points in a 3D plane??
Summary: elementary vector algebra
Message-ID: <21401@well.sf.ca.us>
Date: 29 Oct 90 19:54:44 GMT
References: <31644@netnews.upenn.edu>
Lines: 35

In article <31644@netnews.upenn.edu>, sichen@grasp.cis.upenn.edu (the S T R A N G E R ???!!!) writes:
> 
>   I am looking for the algorithm (pseudocode is fine) to determine any 4 given
> abitrary points p(x,y,z) that lie in the same plane...e.g. The algorithm should
> work for the following points ..
> 
> 	a)	p1=[1,1,0]
> 		p2=[0,2,1]
> 		p3=[1,0,2]
> 		p4=[3,-1,2]
> 
> 	b)	p1=[4, 2, 1]
> 		p2=[0, -3, 4]
> 		p3=[2, 0, 3]
> 		p4=[1, 4, 1]


Form the 3 vectors v1 = p1 - p4, v2 = p2 - p4, and v3 = p3 - p4.

Then the four points are coplanar if and only if the "box 
product"  v1 dot ( v2 cross v3 ) = 0.

Another way of expressing the same formula is that the 
determinant
                | v1x v1y v1z |
                | v2x v2y v2z | = 0
                | v3x v3y v3z |

If you don't understand "dot", "cross", or "determinant", see any 
book on elementary vector algebra.  

Applying this test shows that neither of the point sets a) nor b) 
is coplanar.

        --Ron