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Homework Help: Prove distance between lines

  1. May 25, 2010 #1
    1. The problem statement,
    all variables and given/known data

    There are 4 vectors a,b,u,v in space and {u,v} are linearly independen
    The groups A={a+s*u|s E R} B = {b+t * v|t E R}are two lines in space

    The Two A , B lines does not meet and the "part/line/segment (I dont know how it called ) " PQ is the smallest part that is ends are [tex]Q \in B ,P \in A [/tex] show that the length of PQ is"
    [tex]\frac{|(u \times v)\bullet(a-b)|}{||u \times v||}[/tex]



    2. Relevant equations

    all calc/..

    3. The attempt at a solution
    I mange to get to the point of
    [tex]u \times v =(s)[/tex]
    and then
    [tex]\frac{L(s)*L(a-b)*cos(w)}{L(s)}=L(a-b)*cos(w) [/tex]
    bUT I dont know whats next
     
  2. jcsd
  3. May 26, 2010 #2

    gabbagabbahey

    User Avatar
    Homework Helper
    Gold Member

    Huh?:confused: How are you managing to get to this point? What do [itex]w[/itex] and [itex]L[/itex] represent?

    What is the general formula for the distance between two points [itex]\textbf{r}_1[/itex] and [itex]\textbf{r}_2[/itex]? What if the points lie on the lines given in the problem (i.e. [itex]\textbf{r}_1=\textbf{a}+s\textbf{u}[/itex] and [itex]\textbf{r}_2=\textbf{b}+t\textbf{v}[/itex])? How would you go about minimizing that distance?
     
  4. May 26, 2010 #3
    Yes I've managed to do it...
    I've done:
    [tex]u \times v[/tex] and got the vector that is vertical to both u , v
    Then I've made a plane that use this vector and use point a, then I've used the forumala of plane distance from point in the first line.
    Thank you/
     
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