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Equation of 3-dimensional plane

  1. Sep 13, 2008 #1
    1. The problem statement, all variables and given/known data

    Find the equation of the plane through vectors u=(1,1,2) parallel to the plane containing v=(2,2,-1), w=(1,-1,0), and the origin.

    2. Relevant equations



    3. The attempt at a solution

    I did the cross product of v x w = (-1,-1,-4).
    Thus, my equation became -x-y-4z=-9. Is this correct? Thanks.
     
  2. jcsd
  3. Sep 13, 2008 #2
    bump...can anyone please see if I am right or wrong??
     
  4. Sep 13, 2008 #3

    gabbagabbahey

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    I'm not sure I understand the question here. Are you trying to find the equation of one plane (containing u) that is parallel to another plane (containing v,w. and the origin (0,0,0))? Or is there just one plane involved? Could you maybe repost the question in its original wording?
     
  5. Sep 13, 2008 #4

    Dick

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    Yes, vxw=(-1,-1,-4) is the normal to the plane. So the equation of the plane is -x-y-4z=C for some constant C. If I put (1,1,2) into that, I don't get -9.
     
  6. Sep 14, 2008 #5
    Ok thanks! Yeah, it was a typo, I meant -10.
     
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