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Subject: Re: Triangulation using sphere intersects
Date: Sat, 22 Nov 2008 00:05:03 +0000 (UTC)
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"Roger Stafford" <ellieandrogerxyzzy@mindspring.com.invalid> wrote in message <gg7ep2\$dg\$1@fred.mathworks.com>...
> .......
>  Then solve for the plane which contains all three centers.
> .......

It occurs to me that I might have taken too large a step when I said to find the equation of the plane through the three sphere centers.  To give that step in greater detail, let C1 = [x1;y1;z1], C2 = [x2;y2;z2], and C3 = [x3;y3;z3] be the three centers.  Then the cross product, cross(C2-C1,C3-C1), must be orthogonal to the desired plane.  Hence any point on the plane P = (x,y,z) must satisfy

dot(P-C1,cross(C2-C1,C3-C1)) =
(x-x1)*((y2-y1)*(z3-z1)-(y3-y1)*(z2-z1)) +
(y-y1)*((z2-z1)*(x3-x1)-(z3-z1)*(x2-x1)) +
(z-z1)*((x2-x1)*(y3-y1)-(x3-x1)*(y2-y1)) = 0

which gives the equation of that plane.

Roger Stafford

```