Re: Sphere formula

*To*: mathgroup at smc.vnet.net*Subject*: [mg109506] Re: Sphere formula*From*: Ray Koopman <koopman at sfu.ca>*Date*: Sun, 2 May 2010 05:35:44 -0400 (EDT)*References*: <hr65rj$jq6$1@smc.vnet.net> <hrbach$hq9$1@smc.vnet.net>

On May 1, 3:51 am, "Alexander Elkins" <alexander_elk... at hotmail.com> wrote: > "S. B. Gray" <stev... at ROADRUNNER.COM> wrote in messagenews:hrbach$hq9$1 at smc.vnet.net... >> On 4/27/2010 1:05 AM, S. B. Gray wrote: >>> 1. The center of a sphere through 4 points has a very nice determinant >>> form. (http://mathworld.wolfram.com/Sphere.html) What I want is a nice >>> formula for the center of a sphere through 3 points, where the center is >>> in the plane of the three points. I have a formula but it's a horrible >>> mess of hundreds of lines, even after FullSimplify. >>> >>> 2. (Unlikely) Is there a way to get Mathematica to put a long formula >>> into a matrix/determinant form if there is a nice one? >>> >>> Any tips will be appreciated. >>> >>> Steve Gray >> >> Thanks to everyone who answered my question, but there is a simpler >> answer. I forgot the simple fact that any linear combination of two >> vectors lies in the plane of the two vectors. >> >> Let the three points be p1,p2,p3. Consider the linear function >> p=b(p2-p1)+c(p3-p1) where b,c are to be determined and p is the desired >> center. Now do >> >> Solve[{Norm(p-p1)==Norm(p-p2),Norm(p-p1)==Norm(p-p3)},{b,c}]. >> >> This gives b,c and therefore p, which will be equidistant from p1,p2, >> and p3 and lie in their plane. Very simple. (I used (p-p1).(p-p1) etc. >> instead of Norm.) >> >> Steve Gray > > Unless I misinterpreted something in this posting, the following does not > give the expected center point {1, 2, 3} using Ray Koopman' s posted values > for {p1, p2, p3} as the result: > > In[1]:=With[{p1={0.018473,-1.1359,-0.768653}, > p2={2.51514,5.25315,6.48158}, > p3={0.818313,-0.881007,-1.0825}}, > With[{p=b(p2-p1)+c(p3-p1)}, > p/.NSolve[{(p-p1).(p-p1)==(p-p2).(p-p2), > (p-p1).(p-p1)==(p-p3).(p-p3)},{b,c}]]] > > Out[1]={{0.797494,2.34596,2.76487}} > [...] The fix is simple: just change the definition of p. Block[{p1 = {0.018473,-1.1359,-0.768653}, p2 = {2.51514,5.25315,6.48158}, p3 = {0.818313,-0.881007,-1.0825}, b,c,p}, p = (1-b-c)p1 + b*p2 + c*p3; p /. Flatten@Solve[{(p-p1).(p-p1)==(p-p2).(p-p2), (p-p1).(p-p1)==(p-p3).(p-p3)},{b,c}]] {1.,2.,3.}