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I have homework that I have to solve, and the my question is:
A cube has side length of 2 cm. One ant starts at corner A and travels to corner B.
Another ant starts at corner C and travels to corner D. The two ants leave at the same time and travel at the same speed. Find the shortest distance between the two ants.
Question figure
and my solution:
Solution Pic
I don't know if my answer is correct, is it?

O-BL
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1 Answers1

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The second ant is at $(2,2,-2t)$.

The squared distance is

$$(2t-2)^2+(-2)^2+(2t)^2=8(t^2-t+1).$$

By completing the square, the minimum is achieved at $t=\dfrac12$, when the ants are at the midpoints, and the distance is $\sqrt6$.

  • By the way, it is easier/safer to write the coordinates of the ants by simple inspection than by error-prone computation of the line coefficients. –  Oct 13 '17 at 08:32
  • Could you explain a little further? I believe that the second ant starts at C, so at t = 0 the z-coords should be = 2, and you say it is just zero! – O-BL Oct 13 '17 at 08:38
  • @O-BL: lookup your own answer. What's the $z$ of $C$ ? –  Oct 13 '17 at 08:41
  • let's see; hmm, so the second ant z equation is not correct since C is (2,2,0), I see; and it should be as you said in the answer, thank you a lot, that helped. – O-BL Oct 13 '17 at 08:56