The set of all points for which two spheres intersect orthogonally

Click For Summary

Homework Help Overview

The discussion revolves around finding the set of points (a,b,c) for which two spheres intersect orthogonally. The first sphere is defined by the equation x² + y² + z² = 1, while the second sphere is defined as (x-a)² + (y-b)² + (z-c)² = 1. The problem involves understanding the geometric and mathematical relationships between the two spheres.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the relevance of the gradient vectors of the spheres and their orthogonality. There are questions about whether to examine the intersection of the functions or to compute the dot product of the gradients. Some suggest focusing on the geometry of the situation before delving into the gradients.

Discussion Status

Participants are exploring various approaches to the problem, with some suggesting that understanding the intersection points of the spheres is crucial before considering the gradients. There is a mix of ideas being shared, with no explicit consensus reached on a single method or solution.

Contextual Notes

Some participants note that any sphere with a radius of 1 intersecting with the first sphere must have its center within a larger sphere defined by x² + y² + z² = 9. There is also a discussion about the implications of the orthogonality of the tangent planes at the intersection points.

twotwelve
Messages
9
Reaction score
0
Apostol 281, 4.



Homework Statement


Find the set of points (a,b,c) for which the spheres below intersect orthogonally.
sphere 1: f(x,y,x):x^2+y^2+z^2=1
sphere 2: g(x,y,z):(x-a)^2+(y-b)^2+(z-c)^2=1




The Attempt at a Solution


II know that the gradient vector, \nabla f, is normal to the surface determined by f, I'm just unclear on creating the connection to the either the gradient or surface of g. Clarification would be great.

Thanks
 
Physics news on Phys.org
twotwelve said:
Apostol 281, 4.



Homework Statement


Find the set of points (a,b,c) for which the spheres below intersect orthogonally.
sphere 1: f(x,y,x):x^2+y^2+z^2=1
sphere 2: g(x,y,z):(x-a)^2+(y-b)^2+(z-c)^2=1




The Attempt at a Solution


II know that the gradient vector, \nabla f, is normal to the surface determined by f, I'm just unclear on creating the connection to the either the gradient or surface of g. Clarification would be great.

Thanks

What you want are all the intersection points at which the gradient for the sphere whose center is at the origin is orthogonal to the gradient of the sphere whose center is at (a, b, c).
 
Does this suggest that I should examine the intersection f-g?

Or should I find \nabla f \cdot \nabla g?

I am genuinely unclear...
 
twotwelve said:
Does this suggest that I should examine the intersection f-g?
No.
twotwelve said:
Or should I find \nabla f \cdot \nabla g?
Yes, but first things first. Find the points of intersection of the two spheres. Unless I am mistaken, there are three cases: the spheres intersect at every point on each; they intersect in a circle; they intersect at a single point (a degenerate circle of radius 0).

After you find the intersection points, then check that the gradients are othogonal. That would be how I would go about it.
twotwelve said:
I am genuinely unclear...
 
Hmm, after deliberating over this problem for a good bit of time, the most that I can figure is any sphere having radius 1 and intersecting with f must have a center within the outer sphere h:x^2+y^2+z^2=9.

I know that the gradient vectors of f are all orthogonal to f(x,y,z)=1. I also know that the gradient vectors of g are orthogonal to g(x,y,z)=1. I believe that at any point that these spheres are orthogonal to each other, than the gradient vectors at that point will also be orthogonal.

I understand that I need to first account for this intersection...yet I don't understand how.
 
Last edited:
twotwelve said:
Hmm, after deliberating over this problem for a good bit of time, the most that I can figure is any sphere having radius 1 and intersecting with f must have a center within the outer sphere h:x^2+y^2+z^2=9.
Well, that's nice, but I don't see how it's helpful. IMO, you should quit thinking about the functions f and g for awhile, and focus on the geometry of this situation, and find the intersection points. Then you can go to town with gradients and such.
twotwelve said:
I know that the gradient vectors of f are all orthogonal to f(x,y,z)=1. I also know that the gradient vectors of g are orthogonal to g(x,y,z)=1. I believe that at any point that these spheres are orthogonal to each other, than the gradient vectors at that point will also be orthogonal.

I understand that I need to first account for this intersection...yet I don't understand how.
 
Yes, I was heading in the wrong direction.

My solution:
Any points (x_0,y_0,z_0) meeting our conditions will also have orthogonal tangent planes at those points satisfying (x_0,y_0,z_0) \cdot (a-x_0,b-y_0,c-z_0)=0. However, they also satisfy both f and g. Solving for all three yields a^2+b^2+c^2=2, a sphere of radius 2 centered at the origin.

Thanks for the assistance.
 
x2 + y2 + z2 = 2 would be a sphere centered at the origin, of radius sqrt(2), but your equation in a, b, and c is just a relationship between three parameters.
 

Similar threads

  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 13 ·
Replies
13
Views
1K
  • · Replies 21 ·
Replies
21
Views
4K
Replies
3
Views
2K
Replies
5
Views
1K
Replies
7
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K