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Distance between two lines in 3D

  1. Dec 1, 2008 #1
    1. The problem statement, all variables and given/known data
    L2=(x,y,z)=(0,0,4)+ s(-2,-3,-1)

    Find the shortest distance between the two lines

    3. The attempt at a solution
    First, I did the cross product of the two directional vectors from both lines:
    (-2,3,-2) x (-2,-3,-1) = (-9,-2, 12)
    And its unit vector is:

    Then I found the distance between two points on the lines:
    R = (0,0,4)-(-2,-4,-3) = (2,4,7)

    And then I just tried to project the R onto the unit vector and got 58/sqrt(229), but that's not right. Help?
  2. jcsd
  3. Dec 1, 2008 #2


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    that should be correct.

    EDIT: yeah Dick is right...Sorry about that
    Last edited: Dec 1, 2008
  4. Dec 1, 2008 #3


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    I get that the cross product is (-9,+2,+12).
  5. Dec 1, 2008 #4
    Ooohhh!!! I see, I'm just retarded and forgot about the negative space in the second row. OOOOH. This all makes sense now!

    Thank you!
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