Determine the largest volume between two spheres

Question

I came across this problem which seemed interesting. Let me know what your approach was.

Given two spheres, $B_1,B_2$

$(x-3)^2 + (y-2)^2 + z^2= 100$, $(x-7)^2 + (y-4)^2 + (z+1)^2= 81$

Question: Find the largest sphere contained in their intersection.

My approach:

1. Find vector between center of spheres, $\vec{v_{B_1 , B_2}}$
2. Normalize this vector. $\hat{v}$
3. Scale the normal vector by the radius of $B_1$, call it $\vec{w_1} = 10 \hat{v}$
4. The vector $\langle{3,2,0} \rangle + \vec{w_1}$ is the vector that points to the center of the other sphere and stops at the boundary of the first sphere.
5. We can use the negative of our normal vector and do steps 3-4 to find the vector that points in the other direction. $\langle{7,4,-1} \rangle + \vec{w_2} = \langle{7,4,-1} \rangle -9\hat{v}$
6. Use the midpoint formula to find the center of the new sphere.

$ \frac{ (\langle{3,2,0} \rangle + \vec{w_1} ) - (\langle{7,4,-1} \rangle + \vec{w_2}))}{2}$

7. The distance between the center of the new sphere and either $\langle{3,2,0} \rangle + \vec{w_1} \textrm{ or } \langle{7,4,-1} \rangle + \vec{w_2}$ will be the radius of the new sphere.

Is this approach right? What else could I have done to solve this problem?

Work for each step:

1. $\vec{v_{B_1,B_2}} = \langle{7-3, 4-2, -1 - 0}\rangle =\langle{4, 2, -1}\rangle$
2. $\hat{v} = \langle{\frac{4}{\sqrt{21})}, \frac{2}{\sqrt{21}}, \frac{-1}{\sqrt{21}} }\rangle$
3. $\hat{w_1} = \langle{\frac{40}{\sqrt{21})}, \frac{20}{\sqrt{21}}, \frac{-10}{\sqrt{21}} }\rangle$
4. This is the vector $\langle{3,2,0} \rangle + \hat{w_1} = \langle{3+ \frac{40}{\sqrt{21})}, 2+ \frac{20}{\sqrt{21}}, \frac{-10}{\sqrt{21}} }\rangle$
5. This is the vector $\langle{7,4,-1} \rangle + \hat{w_2} = \langle{7+ \frac{40}{\sqrt{21})}, 4+ \frac{20}{\sqrt{21}}, -1 +\frac{-10}{\sqrt{21}} }\rangle$
6. The midpoint is situated at $ \langle{5+ \frac{40}{\sqrt{21}}, 3+ \frac{20}{\sqrt{21}}, \frac{-1}{2} +\frac{-10}{\sqrt{21}} }\rangle$
7. The distance between our center and the vector from step 4 is $\sqrt{ ((5+ \frac{40}{\sqrt{21}}) - (3+ \frac{40}{\sqrt{21}}) )^2 + ( (3+ \frac{20}{\sqrt{21}}) - (2+ \frac{20}{\sqrt{21}})) ^2 + ( (\frac{-1}{2} +\frac{-10}{\sqrt{21}}) - \frac{-10}{\sqrt{21}})^2 } = \sqrt{ (2)^2 + (1)^2 + (\frac{1}{2})^2 } = \frac{\sqrt{21}}{2} $

Putting all this together: We get that the largest sphere contained in the intersection of our two spheres is $(x-(5+\frac{40}{\sqrt{21}}))^2 + (y - (3+ \frac{20}{\sqrt{21}}))^2 + (z - ( \frac{-1}{2} +\frac{-10}{\sqrt{21}}))^2 = \frac{25}{16}$

Answer 1

I think your approach is correct. However, If all you need is the volume, notice that the diameter of the largest sphere you can inscribe in the common region is $R_1+R_2-d$ where $R_1$ and $R_2$ are the radii and $d$ is the distance between the centers. In your case $d=\sqrt{21}$, and the radii are $9$ and $10$. So the diameter is $19-\sqrt{21}$. This value does not match the one in your answer, so I wonder if there is some computational error.