Intersection of Three Spheres
Date: 06/05/2003 at 08:48:12 From: Rob Subject: Intersection of three spheres How do you find the intersection of three spheres? (x-x1)^2 + (y-y1)^2 + (z-z1)^2 = r1^2 (x-x2)^2 + (y-y2)^2 + (z-z2)^2 = r2^2 (x-x3)^2 + (y-y3)^2 + (z-z3)^2 = r3^2 Point 1 (x1,y1,z1), point 2 (x2,y2,z2), and point 3 (x3,y3,z3) are known. I tried to solved them simultaneously using Cramer's rule but without any success. The equations get more complex.
Date: 06/05/2003 at 09:36:30 From: Doctor George Subject: Re: Intersection of three spheres Hi Rob, Thanks for writing to Doctor Math. I would recommend a vector approach to finding the intersection points of three spheres. The basic strategy has two steps. 1. Find circle C that is the intersection of spheres 1 and 2. 2. Find the intersection of circle C with sphere 3. The last section in the following answer from the Archives shows a vector approach to finding the intersection of two circles. Intersecting Circles http://mathforum.org/library/drmath/view/51710.html The method described here can be used to find the center point and radius of circle C. In step 2, look at the intersection of the plane of circle C and the third sphere. You should see that step 2 can be reduced to the intersection of two circles. Does that make sense? Write again if you need more help. - Doctor George, The Math Forum http://mathforum.org/dr.math/
Date: 06/05/2003 at 09:58:57 From: Doctor Jerry Subject: Re: Intersection of three spheres Hi Rob, The intersection of two spheres is a circle, and if you throw in a third sphere you'll often find two points as the intersection of all three. Cramer's Rule won't work because it applies to linear systems, not quadratic. You can subtract equation 1 from eq 2, finding a linear relation in x,y, and z. Do the same for eq 1 and eq 3. Solve the linear system of two equations for x and y in terms of z. Substitute these results in any of the equations of the spheres. This will give an equation in terms of z alone. It will be quadratic. Solve it for z. Generate the corresponding x and y by going back to the linear system. For x1 = y1 = z1 = 0; r1 = 3; x2 = 1; y2 = 1; z2 = 1; r2 = 2; x3 = -1; y3 = 1; z3 = 2; r3 = 2 I found the linear system to be -8 + 2 x + 2 y + 2 z = 0 -11 - 2 x + 2 y + 4 z = 0 and solving this for x and y, x = (1/4)*(-3 + 2*z), y = (1/4)*(19 - 6*z) Substituting into the first sphere equation, I found z = (1/14)*(30 - Sqrt), z = (1/14)*(30 + Sqrt) - Doctor Jerry, The Math Forum http://mathforum.org/dr.math/
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