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Math Help - Proof of Midpoint and Vectors in 3 Dimensions

  1. #1
    Junior Member
    Joined
    May 2009
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    Proof of Midpoint and Vectors in 3 Dimensions

    Let Q be the midpoint of PR and and T be the midpoint of SU of vectors, not coplanar in R^3 (3 dimensions). If the dot product of PT and TR = the dot product of SQ and QU = 0, show that the magnitude of PR is equal to the magnitude of SU.

    I've been trying to rearrange the equations by using the distance formula and the dot product properties, like A dot B = magnitude of A * magnitude of B * cos(theta). Since the dot product of the new lines is 0, I know that these lines must make a 90 degree angle, but am still at a loss over how to show the lengths of PR and SU are the same.

    I don't need a solution, just a way to get started because I am out of ideas.
    Thanks in advance!
    Last edited by JoAdams5000; May 21st 2009 at 05:32 AM.
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  2. #2
    Senior Member
    Joined
    Apr 2009
    From
    Atlanta, GA
    Posts
    408

    Simple Geometry

    Don't worry about the equations - draw a picture!

    You have two right triangles, \Delta PTR and \Delta UQS, where T bisects US and Q bisects PR.

    Draw this and see if you can convince yourself that \Delta TQR is congruent to \Delta QTS
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