Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Closest possible points on skew lines

  1. Nov 24, 2008 #1
    1. The problem statement, all variables and given/known data

    Find points P,Q which are closest possible with P lying on line:
    x=7-5t, y=-5+11t, z=-3-1t
    and Q lying on line:
    x=-354-8t, y=-194+12t, z=-73+7t

    *the line joining P + Q is perpendicular to the two given lines.

    2. Relevant equations

    Projection formula, cross product....

    3. The attempt at a solution

    So, this is the first time I've seen a problem with skew lines, so am a bit confused how to go about this one.

    I wrote the equations to be :
    X_p = [7,-5,-3] + [-5,11,-1]s
    X_q = [-354,-194,-73] + [-8,12,7]t

    Therefore, the direction vectors are:
    d_p = [-5,11,-1] and
    d_q = [-8,12,7]

    I took the cross product: [-5,11,-1] x [-8,12,7] to get the normal that is perpendicular to both lines, to be:
    [89,43,28]

    I figured I would take a point on Line P, and a point on Line Q, to get an arbitrary vector connecting the two lines; I took the points given in the equation:
    [7,-5,-3] - [-354,-194,-73] to get v = [361,-189,70]

    I then projected this onto the normal, so took:
    proj_n_v


    Am I correct in doing this? Would this proj_n_v be the distance from P to Q? If it is, what do I do now to get the two points P and Q.

    Am I going about this correctly; I've never encountered this type of problem before, but see it in the practice problems on projections....


    Thanks so much.
     
  2. jcsd
  3. Nov 24, 2008 #2

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Yes, the projection will give you the distance. To get the actual two points, write W=[89,43,28] (your cross product). Then the difference of the two points must be parallel to W. So write X_p(s)-X_q(t)=u*W. That gives you three equations in three unknowns, s,t and u. Solve them. You could also do this more directly by minimizing (X_p(s)-X_q(t))^2. (Take the two partial derivatives wrt s and t and set them to zero.).
     
  4. Nov 24, 2008 #3
    Thanks for the reply Dick!

    Ahh k, got it now!! )
     
    Last edited: Nov 24, 2008
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook