When two spheres intervine - very hard 3d geometry and vectors problem

[Nikitin]

Homework Statement

Sphere with P as centre has equation (x-5)^2 + (y-9)^2 + z^2 -100 = 0
Sphere with Q as centre has equation (x-1)^2 + (y+3)^2 + (z-3)^2 -49 =0

These spheres intervine with each other. Find the surface area of the object limited by the two spheres.

The Attempt at a Solution

Allright, the spheres have centres in P and Q. |PQ|=|[-4,-12,3]|=13. Radius of sphere P is 10, of sphere Q 7.

Here's a picture of the situation http://cameroid.com/i/JMJ80-A1 . Just read the letters from right to left and u should be good. The yellowish thing is supposed to represent the intervined part of the two spheres.

I need to find the distance PS but I don't know how. I tried proving that the triangle contains 3 proportional triangles but I don't think that's the case..

 

[LCKurtz]

Homework Statement

Sphere with P as centre has equation (x-5)^2 + (y-9)^2 + z^2 -100 = 0
Sphere with Q as centre has equation (x-1)^2 + (y+3)^2 + (z-3)^2 -49 =0

These spheres intervine with each other. Find the surface area of the object limited by the two spheres.

The Attempt at a Solution

Allright, the spheres have centres in P and Q. |PQ|=|[-4,-12,3]|=13. Radius of sphere P is 10, of sphere Q 7.

Here's a picture of the situation http://cameroid.com/i/JMJ80-A1 . Just read the letters from right to left and u should be good. The yellowish thing is supposed to represent the intervined part of the two spheres.

I need to find the distance PS but I don't know how. I tried proving that the triangle contains 3 proportional triangles but I don't think that's the case..

Your picture isn't coming through. If you look at a cross section, you have two intersecting circles of radius 10 and 7. If you draw a line from the top to bottom of the common lens shaped object, it is perpendicular to the line of centers. I'm guessing that intersection point is what you are calling S. You have two right triangles with hypotenuses 10 and 7 and a common side. Call PS = x and SQ = 13 - x.

Just use the Pythagorean theorem to set the common sides equal to each other and solve for x which is the distance PS.

 

[Nikitin]

Thanks!