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Math Help - Intersection of Vector Lines

  1. #1
    Del
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    Intersection of Vector Lines



    Find the point of intersection, , of the lines and . Please help!
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    Lord of certain Rings
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    Quote Originally Posted by Del View Post


    Find the point of intersection, , of the lines and . Please help!
    <br />
r_1 = r_2
    (3,5,-9) + t(0,2,0) = (-1,4,-8)+t(-4,2,1)
    (3,5,-9) - (-1,4,-8) = t(-4,2,1) - t(0,2,0)
    (4,1,-1) = t(-4,0,1)

    No real t can satisfy this. Hence the lines do not intersect.
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    Quote Originally Posted by Del View Post


    Find the point of intersection, , of the lines and . Please help!
    set each of the coordinates equal to find values for t

    x: 3+0t=-1-4t \iff t=-1

    y: 5+2t=4+2t \iff 1=0 \mbox{ never true}

    we don't need to check z becuase the lines do not ever intersect because the y coordinates are never equal.
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  4. #4
    Del
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    The program that I'm inputting my answers to accepted as the correct answer.. I'm confused but thanks!
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    Lord of certain Rings
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    Quote Originally Posted by Del View Post
    The program that I'm inputting my answers to accepted as the correct answer.. I'm confused but thanks!
    Yes you are right.
    I made a dumb mistake of assuming t must be same.Sorry

     (3,5,-9) + t_2(0,2,0) = (-1,4,-8)+t_1(-4,2,1)

     (3,5,-9) - (-1,4,-8) = t_1(-4,2,1) - t_2(0,2,0)

     (4,1,-1) = (-4t_1,2t_1 - 2t_2,t_1)

    So -4t_1 = 4 \Rightarrow t_1 = -1

    And  2t_1 - 2t_2 = 1 \Rightarrow t_2 = -\frac32

    Thus (-1,4,-8)+t_1(-4,2,1) = (-1,4,-8) - (-4,2,1) = (3,2,-9) is the right answer.
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  6. #6
    Behold, the power of SARDINES!
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    I did the same thing sorry
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