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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.}
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