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Intersection of Lines

  1. Oct 7, 2008 #1

    Determine wether the lines L1, L2 are parallel, skew or intersecting. If they intersect, find the point of intersection.

    L1 = [tex]\left(\begin{array}{ccc}x\\y\\z\end{array}\right) =[/tex] [tex]\left(\begin{array}{ccc}1\\2\\0\end{array}\right)[/tex] [tex]+ t \left(\begin{array}{ccc}1\\-1\\3\end{array}\right)[/tex]

    L2 = [tex]\left(\begin{array}{ccc}x\\y\\z\end{array}\right) =[/tex] [tex]\left(\begin{array}{ccc}2\\1\\4\end{array}\right)[/tex] [tex]+ s \left(\begin{array}{ccc}-1\\2\\1\end{array}\right)[/tex]

    For some [tex]t, s \in R[/tex]

    3. The attempt at a solution

    I believe L1 & L2 are not parallel because their direction vectors aren't multiples of each other.

    The dot product of their direction vectors are zero; this tells us that they are orthogonal/skew and not parallel.

    To find the intersection point we say: L1 = L2

    x1 = x2

    [tex]\left(\begin{array}{ccc}1\\2\\0\end{array}\right)[/tex] [tex]+ t \left(\begin{array}{ccc}1\\-1\\3\end{array}\right) =[/tex] [tex]\left(\begin{array}{ccc}2\\1\\4\end{array}\right)[/tex] [tex]+ s \left(\begin{array}{ccc}-1\\2\\1\end{array}\right)[/tex]

    yields the following system of linear equations;

    1+t = 2-s
    2-t = 1+2s
    3t = 4+s

    I think maybe it can be re-written as:

    t+s = 1
    -t-2s = -1
    3t-s = 4

    Is my working correct so far? If yes, can you guys show me how to solve this set of simultaneous equations to find the actual point of intersection please. Thanks.

  2. jcsd
  3. Oct 7, 2008 #2


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    Hi roam! :smile:
    Perfect! :biggrin:

    And the trick now is to solve two of the equations (I'd suggest the first two :wink:), and see whether the result fits the third one.

    Have a go! :smile:
  4. Oct 7, 2008 #3
    Thanks for checking my work. :smile: :smile:

    I don't know how to solve this simultaneous equation but I'll have a go;

    (1) t+s = 1
    (2) -t-2s = -1
    (3) 3t-s = 4

    If (2) is true then t = 1-2s

    And if (1) is true then t+s = 1 => t = 1-s

    If (1) and (2) are both true then;

    1-2s = 1-s
    s = 0

    Now from equation (1)
    t+s = 1
    t = 1

    You see? My solutions never fits the third one, what did I do wrong? :rolleyes:
  5. Oct 7, 2008 #4


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    oooh, this is so long-winded.

    First, you don't have to say "if (2) is true …": just say "from (2) …"

    Second, how about just adding (1) and (2)? :smile:
    Nothing! :rolleyes:

    Read the question
    so … ? :smile:
  6. Oct 7, 2008 #5
    So, what can we conclude from this? Does this tell us the system has no solution ie. the system is inconsistent?

    Therefore there is no intersection. The lines are skew (they are orthogonal but do not meet).

  7. Oct 7, 2008 #6


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    :biggrin: Woohoo! :biggrin:
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