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3 dimension distance

  1. Oct 1, 2005 #1
    Can someone help me with this:

    What is the shortest distance between the two lines:

    L1: r= (1,0,0) +t( 2,3,4)
    L2 u= (2,1,0) +s(1,2,0)

    Thank you very much for your help

  2. jcsd
  3. Oct 1, 2005 #2


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    There is probably a shorter way in terms of vectors and stuff, but here's a way:

    Build the function "distance squared btw the lines", defined as the norm squared of the distance vector function

    [tex]\vec{d}(s,t) = \vec{r}(t) - \vec{u}(s)[/tex]

    The reason we work with the function distance squared instead of the function distance itself is that the distance is minimum when the distance squared is minimum, and as you'll see, it is much easier to find when d² is minimum then when d is minimum.

    This function is a function of the two variables t and s. Use the usual method to find the extrema of the function. (no need to use the second derivative test to determine the nature of each extrema; yuo know the function distance quared as no maxima and only one minimum :wink:.
  4. Oct 2, 2005 #3
    OK thank you very much
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