The set of all points for which two spheres intersect orthogonally

[twotwelve]

$$Apostol 281, 4.$$

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

 

[Mark44]

$$Apostol 281, 4.$$

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).

 

[twotwelve]

Does this suggest that I should examine the intersection $$f-g$$?

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

I am genuinely unclear...

 

[Mark44]

Does this suggest that I should examine the intersection $$f-g$$?

No.

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.

I am genuinely unclear...

 

[twotwelve]

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.

 

[Mark44]

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.

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.

 

[twotwelve]

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.

 

[Mark44]

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.