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Finding a plane

  1. Nov 11, 2009 #1
    1. The problem statement, all variables and given/known data

    Find the equation of the plane which contains the line:

    (x,y,z) = (3,3,-3) + t(0,-3,-3)

    and is parallel to the line:

    (x,y,z) = (4,2,0) + t (3,-2,1)

    Write your answer in the form 2 x + B y + C z = D, and give the values of B, C and D as your answer


    2. Relevant equations
    Cross Product
    (x-x0, y-y0, z-z0) . (a, b, c) = 0


    3. The attempt at a solution

    So I need a point and a normal to determine the equation of the plane. Point: (3,3,-3). The normal I found was the cross product of (0,-3,-3) and (3,-2,1) = (-9,9,-9).
    I then used the equation (x-x0, y-y0, z-z0) . (a, b, c) = 0 and got -9x+9y-9z=27.
    Then i multiplied it by -2/9 since the answer needed to start with 2x, I get 2x-2y+2z=-6
    but its wrong..

    So any help would be appreciated!
    Thank You!
     
  2. jcsd
  3. Nov 11, 2009 #2
    i don't konw whether you did a mistake or me, when finding the normal vector on your plane because i got (-9,-9,9)

    and for the final answer i'm getting x+y-z=9. what's the answer on yor book/notes?
    look you might be getting a parallel plane, depending on what point you chose.
     
    Last edited: Nov 11, 2009
  4. Nov 11, 2009 #3
    I think your right. So was your finally answer 2x+2y-2z=18?
     
  5. Nov 11, 2009 #4
    yes, i got x+y-z=9, which is a parallel plane with yours.
     
  6. Nov 11, 2009 #5
    Nice! Thanks sutupidmath!
     
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