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MathGroup Archive 1995

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Re: Equation for Circle in 3-D surrounding Equilateral Triangle

  • To: mathgroup at smc.vnet.net
  • Subject: [mg2356] Re: [mg2321] Equation for Circle in 3-D surrounding Equilateral Triangle
  • From: Preston Nichols <nichols at godel.math.cmu.edu>
  • Date: Fri, 27 Oct 1995 01:49:42 -0400

John Duggan, duggan at ecs.umass.edu,Tue, 24 Oct 1995, wrote:

"The question is:

"Given three point at the intersection of the sides of an  
Equilateral triangle in 3D what is and how does one code Mathematica  
for the equation of the circle in 3D that contains all 3 points. I  
felt that the intersection of a sphere and a plane would solve the  
problem. But my mathematica coding appears to fall short for this."

-----------------------------------------------------------

I'm not sure exactly what you are trying to get as a result, but  
the following might help.  The steps below work for any triangle in  
3-space, not only equilateral ones.  (However, very thin triangles  
might lead to numerical trouble.)
-----------------------------------------------------------

(* First, we gather some tools and set the initial data *)

Needs["LinearAlgebra`Orthogonalization`"];
Needs["LinearAlgebra`CrossProduct`"];

norm[v_] := Sqrt[v.v];

p1 = { -0.256, -0.256, 3.423 };
p2 = {-3.206, 0.474, -2.690 };
p3 = {-6.532, 2.766, 2.769 };

(* Now get some vectors parallel to the sides, *)

s1 := p2 - p1;
s2 := p3 - p2;
s3 := p1 - p3;

(* and find the midpoints of the sides. *)

m1 := p1 + s1/2;
m2 := p2 + s2/2;
m3 := p3 + s3/2;

(* Acutally, s3 and m3 will not be needed.  Note the use of := ,  
which means that the si and mi will be automatically recomputed  
(when called) if the values of the pi are changed (i = 1,2,3). *)

(* A useful fact from Euclid:  in any triangle, the perpendicular  
bisectors (in the plane of the triangle) of the three sides all meet  
at a single point, and this point is equidistant from the three  
vertices.  This point is called the circumcenter of the triangle,  
and it is the center of the circle we are seeking.

If the triangle is `floating' in three dimensions, the  
perpendicular bisecting lines in the plane of the triangle are not  
as easy to get hold of as the orthogonal planes which bisect the  
sides.  These three planes will intersect each other in a single  
line, orthogonal to the plane of the triangle, through the  
circumcenter.  The equations of these planes are

({x,y,z} -m1).s1 == 0; ({x,y,z} -m2).s2 == 0; ({x,y,z} -m3).s3 == 0;

because we can use s1, s2, and s3 as the normal vectors to the  
planes.  To find the circumcenter we need to find the intersection  
of any two of these planes with the plane of the triangle, whose  
normal vector is: *)

trianglenormal := Cross[s1,s2];

(* If we let

triangleplane := ({x,y,z} - p1). trianglenormal == 0;
midplane1 := ({x,y,z} -m1).s1 == 0;
midplane2 := ({x,y,z} -m2).s2 == 0;

we could use

Solve[{triangleplane, midplane1, midplane2}, {x,y,z}]

but it is more graceful and more efficient to use the following,  
which combines the equations for the three planes into a single  
system of linear equations: *)

circumcenter := LinearSolve[{s1,s2, trianglenormal},
		{m1.s1,m2.s2,p1. trianglenormal}];

(*  So now we have the center of the desired circumscribed circle,  
and here's the radius: *)

radius := norm[circumcenter-p1];

(* One of the best ways to represent curves such as a circle in  
3-space is by a parametric equation.    This will be easy if we can  
get a pair of orthonormal vectors parallel to the plane of the  
triangle.  Fortunately, there's a neat trick in the  
LinearAlgebra`Orthogonalization` package: *)

{v1,v2} = GramSchmidt[{s1,s2}];

(* The GramSchmidt process returns an orthonormal set of vectors  
with the same span as the vectors you give it.  Thus {v1,v2} are a  
basis for the plane through the origin which is parallel to the  
plane of the triangle.  Now we can parametrically plot the circle.  
*)

Off[ParametricPlot3D::ppcom];

thecircle = ParametricPlot3D[
		radius Cos[t] v1 + radius Sin[t] v2 + circumcenter,
			{t,0,2 Pi}];

(* To see why this works, think about what happens when v1 =  
{1,0,0}, v2 = {0,1,0}, circumcenter = {0,0,0}, and radius =  
anything. *)

(* Finally, to check the result: *)

thetriangle = Graphics3D[Polygon[{p1,p2,p3}]];
Show[{thetriangle, thecircle}];

-----------------------------------------------------------
Preston Nichols
Department of Mathematics
Carnegie Mellon University


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