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Thread: line in three dimensions

  1. #1
    Junior Member
    Apr 2008

    line in three dimensions

    I have an exam in a few hours and I would like someone to solve 2 questions and tell me each step they did along the way, 1 question i ahve done and the other is similar but i can seem to get the correct answer:

    Q1: L=(0,-1,0) M=6,5,3) P=(3,0,1)

    P is not on the line LM, find a point on line LM, called N, which is perpendicular to P.
    This one i can do and I get N=(2,1,1)

    Q2: x=2t y=-t z=4t.

    Find a point Q on this line closest to a point P, not on this line.

    I get the wrong answer for this 1, I think its because i am finding line PQ instead of QP. Is this where i am going wrong?, also if it is can someone tell me why it matters that QP is used and not PQ?
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  2. #2
    May 2008
    For Q2, I think you're missing some info - mainly what point P is.

    To find the shortest distance means find the point perpendicular on line Q to point P.
    In this case, you need to solve:
    PQ.u = 0
    u = (2, -1, 4) % taken from x=2t, y=-t & z=4t. i.e make t=1, point u therefore lies on line Q.
    PQ = P - Q
    Q = (2t, -t, 4t)
    P= ?? %need to have point P here. We can't use (x,y,z) as we already have unknown variable t.
    Unless you mean use point P from Q1. In that case
    P = (3, 0, 1)
    PQ = (3-2t, t, 1-4t)
    PQ.u = (6 - 4t) - t + (4 - 16t) = 0
    => -21t +10 = 0
    => t = 10/21
    Put this back into Q and you get:
    (20/21, -10/21, 40/21) %this is the closest point on line Q to point P.

    good luck with the exam!
    Were there any questions you found particularly difficult?
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