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Math Help - finding the parametric equation of a line

  1. #1
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    finding the parametric equation of a line

    question: Find the parametric equation of the line through the points (0,.5,1) and (2,1,-3)

    work:r=r_0+tv the vector parallel to the line is <2-0,1-.5,-3-1> I thought I would take (0,.5,1) to be the r_0 but the text book didn't. How do I know which initial point to use?

    Am I doing this right? I mean is it necessary to find the vector equation and then from that take the parametric equation or is there a faster way?
    Last edited by superdude; November 8th 2009 at 09:44 PM. Reason: removed math tag so vectors appear in bold face
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  2. #2
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    Quote Originally Posted by superdude View Post
    question: Find the parametric equation of the line through the points (0,.5,1) and (2,1,-3)

    work:r=r_0+tv the vector parallel to the line is <2-0,1-.5,-3-1> I thought I would take (0,.5,1) to be the r_0 but the text book didn't. How do I know which initial point to use?

    Am I doing this right? I mean is it necessary to find the vector equation and then from that take the parametric equation or is there a faster way?

    the equation of a line passing through two points  a , b is

     r = ta + (1-t)b

     r = ( 2t , 5 - 4t , 1 - 4t )
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  3. #3
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    so your saying the problem can't be solved using the formula I was using?
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  4. #4
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    Hi

    Actually , we are talking the same method but my is the simplier expression :

    We can take either (0,5,1) or  (2,1,-3)
    to be the  r_0

    After we calculate the vector parallel to the required line  a - b = (0,5,1) - (2,1,-3) = ( -2,4,4)    , we put the parameter [tex] t ATH] next to the vector and add up with  a or  b , the sum is what we are looking for

    Don't believe ?

    if we take  (0,5,1 ) to be the initial point

    the required equation is  (0,5,1) + t (-2,4,4) = -2t , 5 + 4t , 1 + 4t )

    now if we take  (2,1,-3) to be the initial point ,
    the equation becomes    ( 2 - 2t , 1 + 4t , -3 + 4t )


    Well , obviously , they are totally different but if we change the parameter t by (t+1) , the equation (2) becomes

     ( 2(-t) , 1 + 4t +4 , -3 + 4 + 4t ) = ( - 2t , 5 + 4t , 1 + 4t )
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  5. #5
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    I don't think you found the direction vector right, it's point b-a not a-b
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