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Parallelogram in 3D space

  1. Sep 19, 2005 #1
    I wasn't sure what category to put this in, but here's my problem.

    There is a parallelogram with 3 given points for its corners (each with 3 different coordinates). The idea is to find the forth point. I've tried subbing (x,y,z) for the unknown point, and creating 4 vectors. I tried equating the cross product of two opposite vectors to zero as well as absolute value of opposite vectors to each other. This left me with alot of algebra after I combined both methods and then cheated to get a decimal answer which I'm not even sure is correct. Is there an easier way to do this problem that I'm missing? Thanks.
     
  2. jcsd
  3. Sep 19, 2005 #2
    What was the geometric motivation for adding vectors in the manner that we do ?
     
  4. Sep 19, 2005 #3

    mathman

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    Given three points, there are three possible solutions to you problem, depending on which point is the "middle" point. Let the three points be A,B,C and let A be the "middle". The the fourth point is given by A+(B-A)+(C-A)=B+C-A.
     
  5. Sep 20, 2005 #4
    hmm.. that sounds much easier. Well say there are three points A,B,C and point D is diagonally opposite A, does that mean that D is the point that does not connect to A? Thanks for the help by the way.

    That also gave me another similar idea. Shouldn't A + B + C + D = 0, and therefore the components of each should sum to zero?

    EDIT: just realized that that's what you did :rolleyes:
     
    Last edited: Sep 20, 2005
  6. Sep 20, 2005 #5

    mathman

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    A+D=B+C is the result I have where A is opposite D.

    One easy way to see it is by considering the point H at the center of the parallelogram.

    H=(A+D)/2
    H=(B+C)/2
     
    Last edited: Sep 20, 2005
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