Question

Find the distance between spheres x^2+y^2+z^2=4 and (x-2)^2+(y-2)^2+(z-2)^2=23

Answer 1

How do you have a sphere with an imaginary radius?

Answer 2

assume it is 23 lol

Answer 3

That's what my book says

Answer 4

(The second equation is not a sphere in R3 space), but let us assume it is +23. Then you just have to think of this as two spheres with radius sqrt(4) and sqrt(23) with centers at (0,0,0) and (2,2,2). These spheres would necessarily overlap.

Answer 5

do you want it from center to center of sphere or from tip of one to tip of other. It a trick question where they kinda messed up i think.

If they want it from center to center it just be the distance formula of the origins of the 2 spheres i j k

Answer 6

Sorry about that, it should have a radius 1. I was looking at a different problem.

Answer 7

Oh, much better. So the spheres have radii sqrt(4) and sqrt(1). Do you see how I know that?

Answer 8

Yes. The points would be (0,0,0) and (2,2,2) with radius 2 and 1

Answer 9

So the distance between their centers is easy to find. The problem is probably asking for the closest distance between their surfaces which is just the distance between their centers minus the sum of their respective radii.

Answer 10

A bit lost there. Would I find the distance between the points and then subtract the sum of the radii (3)?

Answer 11

Yes, exactly. Just use Pythagoean theorum twice to get the distance.

Answer 12

So wouldn't it be sqrt(12)-3? The distance between the points is sqrt(2^2+2^2+2^2) and subtracting 3 would be sqrt(12)-3

Answer 13

Exactly.

Answer 14

Thanks!