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Midpoint in 3 Dimensions Question (EASY)

  1. Oct 30, 2009 #1
    1. The problem statement, all variables and given/known data

    http://carlodm.com/calc/123.png [Broken]

    My question is:

    Why can't we use a scalar multiple like (3/8) PQ instead of using the midpoint formula twice (to get 3/8) ?
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Oct 30, 2009 #2

    Mark44

    Staff: Mentor

    PQ = (8, -8, -8), so (3/8)PQ = (3, -3, -3), which doesn't get you a point on that line. What you can do, though, is calculate OP + (3/8)PQ = (-4, 2, 6) + (3, -3, -3) = (-1, -1, 3), which is the same as the answer shown in your problem.

    What I've done above is to look at things from a vector perspective. OP is a vector from the origin to point P. PQ is a vector from P to Q, but it is equivalent to the vector from the origin to the point (8, -8, -8). To get to a point 3/8 of the way from P to Q, we have go from the origin to point P, and then 3/8 of the distance from P to Q.

    Whoever solved the problem used the midpoint formula three times, not twice: the first time to get a point M halfway between P and Q; the second time to get N, a point halfway between P and M (and a quarter of the way from P to Q); and the third time to get a point R halfway between N and M (3/8 of the distance from P to Q).

    Maybe this "graphic" will help you understand better.
    P----------------------------------------------------Q
    P-----------------------M
    P---------N------------M
    P---------------R

    d(P,M) = (1/2)d(P,Q)
    d(P,N) = (1/4)d(P,Q) = d(N,M)
    d(P,R) = (1/4)d(P,Q) + (1/2)d(N,M) = (1/4)d(P,Q) + (1/8)d(P,Q) = (3/8)d(P,Q)
     
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