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Equation of a line/planes chapter question

  1. Sep 19, 2009 #1
    1. The problem statement, all variables and given/known data

    Find the parametric equation and symmetric equation for the line :

    11) The line through (1,-1,1) and parallel to the line x + 2 = y/2 = z-3


    2. Relevant equations

    Parametric equation :

    x = x0 - at
    y = y0 - bt
    z = z0 - ct

    symmetric equations :

    (x - xo) / a = (y - yo)/b = (z - zo) /c


    3. The attempt at a solution

    I just need to find a parallel vector to the line x + 2 = y/2 = z-3.
    Then I could use that vector with the initial point to find the parametric equation and the
    symmetric equation.

    I am just not sure how to find the parallel vector. Looking at the book for answer,
    I see that the parallel vector is <2,3,2>, but I can't figure out how to obtain that answer.
     
  2. jcsd
  3. Sep 19, 2009 #2

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    One way to convert from "symmetric form" (x- x0)/a= (y- y0)/b= (z- z0)/c to "parametric form" is to use that "common value" as your parameter. That is, (x- x0)/a= (y- y0)/b= (z- z0)/c= t so you have (x- x0)/a= t, (y- y0)/b= t, and (z-z0)/c= t. Solve those for x, y, and z to get parametric equations for the same line.

    Of course, once you have parametric equations, x= At+ x0, y= Bt+ y0, z= Ct+ z0, a vector parallel to the line is <A, B, C>.

    (No, <2, 3, 2> is NOT parallel to x+ 2= y/2= z- 3. for example, on point that satisfies those is (-1, 2, 4) (I took x+2= y/2= z- 3= 1) and another is (0, 4, 5) (x+2= y/2= z- 3= 2). The vector from one of those to the other is NOT parallel to <2, 3, 2>.)
     
  4. Sep 19, 2009 #3
    I though that the equation given "x + 2 = y/2 = z-3" looked like the symmetric form,
    with a = b = c = 1. So <A,B,C> = <1,1,1> ?

    The from that I could find the displacement vector of <1,-1,1> , <1,1,1>

    Then use that vector, which must be parallel to find the answer?
     
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