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Equation of a line from Parametric equations

  1. Sep 16, 2004 #1
    How does one find the equation of a line from parametric equations?

    In spefiic i'm looking at this: x(t) = 1+2t , y(t) = -1 + 3t , z(t) = 4+t.... I think i gotta use something liek x-1/a = y-1/b=z-1/c or something like that. If what i just said is true, then i'm lost on what to do next.


    Thanks
     
  2. jcsd
  3. Sep 16, 2004 #2

    Tide

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    t = (x-1)/2 = (y+1)/3 = z-4
     
  4. Sep 16, 2004 #3
    or wait.... do i uses this:

    <x_0,y_0,z_0>+ t<a,b,c> ...


    now i'm really confused.
     
  5. Sep 16, 2004 #4
    Yeah tide, i did that, but then isn't there something else i gotta do? I dont know i'm kinda confused about this stuff. Also allow me to add the rest of the question. IT's find the equation of a plane containing that line, and the point (1,-1,5). Sooo i need the N =<a,b,c> vector and a point ot get my equation of a plane.. So could i use use the second half of that equation i posted aboce ( t<a,b,c>) to get what i need for my N vector?
     
    Last edited: Sep 16, 2004
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