# Math Help - Minimal Distance Between Two Spheres

1. ## Minimal Distance Between Two Spheres

The minimal distance between any point on the sphere1 $(x-2)^{2}+(y-1)^{2}+(z-3)^{2}=1$ and $(x+3)^{2}+(y-2)^{2}+(z-4)^{2}=4$ (sphere2) is...?

Looking at the range of values
$1 \leq x \leq 3,
0 \leq y \leq 2,
2 \leq z \leq 4$
for sphere1 and
$-5 \leq x \leq -1,
0 \leq y \leq 4,
2 \leq z \leq 6$
for the sphere2, I decided that the minimal distance would occur at x=1 for sphere1 and x=-1 for sphere2 because these are the x-coordinates that are closest while still being on their respective spheres.

If I plug those in, it gives me
sphere1: $(y-1)^{2}+(z-3)^{2}=0$ and
sphere2: $(y-2)^{2}+(z-4)^{2}=0$. I think I should be doing something with these equations but I'm not sure what that would be.

I also decided that the minimal distance must lie along the line connecting the centers of the circles, so I used (2,1,3) - (-3,2,4) = <5,-1,-1> as a position vector and the coordinates of the center of sphere1 to give me an equation of a line: (2,1,3) + t<5,-1,-1>.

This is where I'm stuck. Using x=1 and then x=-1, I solved x=2+5t for t, then plugged these values back into the equations y =1-t and z=3-t to get the y and z coordinates, but the distance between the two points is incorrect (I have the correct answer; it's $3(\sqrt{3}-1)$. I calculated $(1, \frac{6}{5},\frac{16}{5})$ and $(-1, \frac{8}{5},\frac{18}{5})$ for the points, and the distance between them is $\frac{6}{5}\sqrt{3}$.

2. Originally Posted by StarKid
The minimal distance between any point on the sphere1 $(x-2)^{2}+(y-1)^{2}+(z-3)^{2}=1$ and $(x+3)^{2}+(y-2)^{2}+(z-4)^{2}=4$ (sphere2) is...?
The distance between the centers is $\sqrt{27}$.
The sum of the radii is $r_1+r_2=1+2=3$.
So what is the minimum distance between the spheres?

3. Originally Posted by Plato
The distance between the centers is $\sqrt{27}$.
The sum of the radii is $r_1+r_2=1+2=3$.
So what is the minimum distance between the spheres?
I see...the minimum distance is the distance between the centers minus the sum of the radii.