How far apart are these two circumspheres?
The problem question: How far apart are these two circumspheres?
Here are 4 3-dimensional points on each sphere:
- Sphere 1: ⎝ ⎜ ⎜ ⎛ 2 6 0 1 1 0 2 6 0 4 5 2 − 1 5 8 − 7 8 − 1 5 8 − 7 8 2 3 1 2 3 1 3 5 1 3 5 1 ⎠ ⎟ ⎟ ⎞
- Sphere 2: ⎝ ⎜ ⎜ ⎛ 1 5 2 3 0 2 1 5 2 3 4 4 6 4 2 7 2 2 6 4 2 2 8 2 3 5 1 3 5 1 4 7 1 3 5 1 ⎠ ⎟ ⎟ ⎞
You will need to find the spheres' centers and radii. Since spheres look the same in all directions, the remainder of the problem can be solved without calculus to compute the minimum distance! But, the surface to surface minimum distance is desired as the answer.
The answer is 0.000.
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I will work this problem twice, once using the full power of Wolfram Mathematica 11.3, which I used to generate and verify this problem. I am retired and my only relationship with Wolfram is as a customer.
s1 = Circumsphere [ p1 ] ⟹ Sphere [ { 2 8 1 , 6 2 , 2 9 1 } , 2 2 9 ]
s2 = Circumsphere [ p2 ] ⟹ Sphere [ { 3 2 3 , 5 0 2 , 4 1 1 } , 2 2 9 ]
EuclideanDistance [ { 2 8 1 , 6 2 , 2 9 1 } , { 3 2 3 , 5 0 2 , 4 1 1 } ] ⟹ 4 5 8
4 5 8 − ( 2 2 9 + 2 2 9 ) ⟹ 0
The second method uses multiple equations in multiple unknowns, which I will do using determinants. The x , y and z columns are the respective point's coordinates, the x 2 + y 2 + z 2 column is that computation done on the points' coordinates and the last column of 1 s is just to make the matrix square.
This method is by W. H. Beyer .
D e t [ ⎝ ⎜ ⎜ ⎜ ⎜ ⎛ x 2 6 0 1 1 0 2 6 0 4 5 2 y − 1 5 8 − 7 8 − 1 5 8 − 7 8 z 2 3 1 2 3 1 3 5 1 3 5 1 x 2 + y 2 + z 2 1 4 5 9 2 5 7 1 5 4 5 2 1 5 7 6 5 3 3 3 5 8 9 1 1 1 1 1 ⎠ ⎟ ⎟ ⎟ ⎟ ⎞ ] = = 0 ⟹ 3 2 8 3 2 0 0 x 2 − 1 8 4 5 1 5 8 4 0 0 x + 3 2 8 3 2 0 0 y 2 − 4 0 7 1 1 6 8 0 0 y + 3 2 8 3 2 0 0 z 2 − 1 9 1 0 8 2 2 4 0 0 z + 3 7 7 7 1 5 7 4 4 0 0 0 = 0
If you examine this closely, then you might recognize that this is the formula for the circumsphere of those points.
By completing the square, this becomes more obvious.
( x − 2 8 1 ) 2 + ( y − 6 2 ) 2 + ( z − 2 9 1 ) 2 = 2 2 9 2
By apply the same procedure to the other set of points.
D e t [ ⎝ ⎜ ⎜ ⎜ ⎜ ⎛ x 1 5 2 3 0 2 1 5 2 3 4 4 y 6 4 2 7 2 2 6 4 2 2 8 2 z 3 5 1 3 5 1 4 7 1 3 5 1 x 2 + y 2 + z 2 5 5 8 4 6 9 7 3 5 6 8 9 6 5 7 1 0 9 3 2 1 0 6 1 1 1 1 1 1 ⎠ ⎟ ⎟ ⎟ ⎟ ⎞ ] = = 0 L o n g r i g h t a r r o w 8 3 2 3 2 0 0 x 2 − 5 3 7 6 7 8 7 2 0 0 x + 8 3 2 3 2 0 0 y 2 − 8 3 5 6 4 9 2 8 0 0 y + 8 3 2 3 2 0 0 z 2 − 6 8 4 1 6 7 0 4 0 0 z + 3 9 3 5 3 1 7 1 6 1 6 0 0 = 0
( x − 3 2 3 ) 2 + ( y − 5 0 2 ) 2 + ( z − 4 1 1 ) 2 = 2 2 9 2
And, the last step is the same as in the first method.
Now, the meta-problem: how did I set this problem so that every number was an integer: What is the diagonal of rectangular parallelepiped with side of 3, 4 and 12? Hint: there are two Pythagorean triangles involved: 3,4 and 5 and 5, 12 and 13.