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Distance through a cube (Posted on 2022-10-20) Difficulty: 3 of 5
Let point A be on a diagonal of one side of a cube. Let point B be on a diagonal of an adjacent side of the cube such that the two diagonals do not meet at a vertex.

Minimize the distance AB.

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soln Comment 3 of 3 |
Solution the 3rd way, finding the minimum where the first derivative is 0.

On a unit cube, point N is an arbitrary distance n up the 1st diagonal, and M is a distance m up the second diagonal. Call the distance^2 between the points D2.  We have the function D2(n,m). Find m(n) that minimizes D2 for any n. Then find the minimum of this function, D2[n,m(n)], over n.    
 Using coordinates (x,y,z)

N = (n/sqrt2, n/sqrt2, 0), and
 
M = (1, m/sqrt2, m/sqrt2)

D2(n,m) = del_x^2 + del_y^2 + del_z^2 

             = 1 + n^2 - m n - sqrt2 n + m^2                 (eqn 1)

(d/dm) D2(n,m) = 2 m - n = 0 at the minimum 

m(n) = n/2

Plugging m=n/2 into eqn 1 gives the minimum separation for any n:  

D2(n) = 1 + n^2 - (n^2)/2 - sqrt2 n + (n^2)/4 
D2(n) = 1 + 3/4 n^2 - sqrt2 n                            (eqn 2)

(d/dn) D2(n) = (3/2) n - sqrt2 = 0 at the minimum

n = (2/3) sqrt2

Plugging this into eqn 2 give the minimum distance: 

D2 = (1/3), Dmin = sqrt(1/3)

And, yes, there is symmetry: one point is 1/3 the way up the diagonal, m = (1/3) sqrt2, and the other is 1/3 the way down, n = (2/3) sqrt2 


Edited on October 20, 2022, 11:57 pm
  Posted by Steven Lord on 2022-10-20 13:07:24

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