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Finding points of intersection using vectors

  1. Feb 8, 2009 #1
    Find the points of intersection of the lines...

    L1: R = 2i + 3j + 3k + t(i - 2j + 5k)
    L2: (x + 3) / 2 = (y + 1) / 2 = -z

    (I assume the plural in points is wrong... since that would be impossible)
    R, i, j, k are vectors; x, y, z are not

    ---

    x = 2 + t
    y = 3 - 2t
    z = 3 + 5t
    for some value(s) of t. If this point also lies on L2, then (x + 3)/2 = -z, so that:
    (2 + t + 3)/2 = -(3 + 5t)
    5 + t = -6 - 10t
    11t = -11
    t = -1
    For this to really be on L2, you also need to have (y + 1)/2 = -z, or:
    (3 - 2t + 1)/2 = -(3 + 5t)
    4 - 2t = -6 - 10t
    8t = -10
    t = -5/4

    The two t's don't equal each other. What am I doing wrong?
     
  2. jcsd
  3. Feb 8, 2009 #2

    Mark44

    Staff: Mentor

    I parametrized the equations for L2 using a different parameter: w.

    x = 2w - 3
    y = 2w - 1
    z = -w

    For any point (x, y, z) to lie on L1 and L2, it must be that the x coordinates on both lines are equal as must be the y and z coordinates.

    So,
    2 + t = 2w + 3
    3 - 2t = 2w - 1
    3 + 5t = -w

    These are three equations in two unknowns, an overdetermined system.

    I ended up with no solution, which means (assuming my calculations are correct) that the two lines don't intersect.
     
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