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Distance between two skew lines given by parametric equations (and more!)

  1. Jan 27, 2009 #1
    1. The problem statement, all variables and given/known data
    L1 L2
    x= 1+t x= 1+2s
    y= 1+6t y= 5+15s
    z= 2t z= -2+6s

    My professor said to find a normal vector and project the lines, but I'm new to this calculus and all these words are just a fuzzy cloud over my head.

    2. Relevant equations

    Projection of c onto L1 -> (C dot L1 / magnitude C) (L1 / magnitude L1)
    Or (C dot L1)C/(magnitude C^2)

    3. The attempt at a solution

    L1xL2= a vector of <6,2,3> = C
    The direction vector for L1 = <1,6,2>
    L2 = <2,15,6>

    So I think I need to project one onto the other....
    But when I do that I get another vector, that's <144/7,48/7,72/7>. Do I find the magnitude of this, and that should be the distance? I'm lost. The magnitude of that vector is 24, which is a clean answer so it could be right.....
     
    Last edited: Jan 27, 2009
  2. jcsd
  3. Jan 27, 2009 #2
    You normal vector is wrong. If c is normal (perpendicular) to L1 or L2 then
    c.L1 = 0
    c.L2 = 0
     
  4. Jan 27, 2009 #3
    But, I think here's what you are doing
    [​IMG]
     
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