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Math Help - Vectors: Finding an equation for plane that contains the two lines

  1. #1
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    Vectors: Finding an equation for plane that contains the two lines

    Find an equation for the plane that contains the two lines
    l1: x = t + 1, y = 5t 4, z = 4t + 3
    and
    l2: x = -2 t, y = 4t 19, z = -9 2t.

    I rewrote this in vector form
    L1 = (1,-4,3) + t(1,5,4)
    L2 = (-2,-19,-9) + t(-1,4,-2)

    Then i went to find the point of intersection
    x = 1+s = -2-t
    y = -4+5s = -19+4t
    z = 3+4s = -9-2t

    used x to find 's'

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

    plugged in 's' to z

    3+4(-3-t) = -9-2t
    3-12-4t=-9-2t
    9-4t=-9-2t
    -6t=0
    t = 0

    plugged 't' back into x

    1+s = -2-(0)
    s = -3

    I plugged in t into x,y and z and get the following
    x = -2-(0) = -2
    y = -19+4(0) = -19
    z = -9-2(0) = -9

    P = (-2,-19,-9)

    Then i went to find the cross product
    n = (1,5,4)x(-1,4,-2)
    n = (-26,2,9)

    Then finally went ahead to find the equation.
    -26(x-(-2)) + 2(y-(-19)) + 9(z-(-9)) = 0
    -26x-52 + 2y+38 + 9z+81 = 0
    -26x + 2y + 9z = -67


    but this is WRONG!!! what am i doing wrong? Thanks!!!
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  2. #2
    Junior Member
    Joined
    Oct 2009
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    silly mistake. a friend of mine helped me out. seems i accidentally did my cross product wrong. it was supposed to be (-26,-2,9) instead.
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