SOLUTION: Two vertices of a cube have coordinates A(5, 12, 12) and B(20, 28, 24). What is the smallest possible volume of the cube?

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Question 611580: Two vertices of a cube have coordinates A(5, 12, 12) and B(20, 28, 24). What is the smallest possible volume of the cube?
Found 2 solutions by Edwin McCravy, Alan3354:
Answer by Edwin McCravy(20056) About Me  (Show Source):
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A(5, 12, 12) and B(20, 28, 24). 
The cube will be smallest if AB is an edge rather than a diagonal.

The distance from A to B is found by the distance formula in 3-dimensions:

d =  

d = sqrt%28%2820-5%29%5E2%2B%2828-12%29%5E2+%2B+%2824-12%29%5E2%29

d = sqrt%28%2815%29%5E2%2B%2816%29%5E2+%2B+%2812%29%5E2%29

d = sqrt%28225%2B256+%2B+144%29

d = sqrt%28625%29

d = 25

So each edge of the cube will be 25, and the volume is
given by the formula:

V = E³ where E is the length of an edge, or

V = (25)³ = 15625 cubic units.

Edwin

Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
Two vertices of a cube have coordinates A(5,12,12) and B(20,28,24). What is the smallest possible volume of the cube?
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A and B can be on the same edge, diagonally on the same face, or diagonally thru the center of the cube.
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The distance AB is

AB = 25
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If AB is an edge, volume = 25^3 = 15625
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If AB is a diagonal on a face, the edges are 5sqrt(2)/2 --> Vol =~ 44.19
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If AB is a diagonal thru the center, the edges are 5sqrt(3)/3 --> Vol =~ 24.056