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Cartesian equation of plane that i perpendicular to plane and contains line

  1. Oct 29, 2008 #1
    1. The problem statement, all variables and given/known data
    Question states "The plane that contains the line r=<-2,4,3>+t<3,2-1> and is perpendicular to the plane r=<5,0,0>+s<2,1,0>+t<-1,0,1> is:"

    Answer is y+2z=10


    2. Relevant equations

    Cross product and dot product of vectors


    3. The attempt at a solution

    I found a vector normal to the plane r=<5,0,0>=s<2,1,0>+t<-1,0,1>

    by doing the cross product of the two direction vectors is <2,1,0> x <-1,0,1>
    getting <1,-2,1>

    than apply rule of n.v=0 ie <1,-2,1> . <x-(-2), y-4, z-3>
    to get x+2-2y+8+z-3=0
    and so x-2y+z = -7

    Not sure what I have done wrong here, could someone explain please?
     
  2. jcsd
  3. Oct 29, 2008 #2

    HallsofIvy

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    You found a plane that has normal vector <1, -2, 1>? Then you found a plane that is paralllel to r, not perpendicular to it!
     
  4. Oct 29, 2008 #3
    By doing the cross product of the direction vectors of r=<5,0,0>+s<2,1,0>+t<-1,0,1> am I not find a vector perpendicular to the plane?
     
  5. Oct 29, 2008 #4

    HallsofIvy

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    Yes, that is correct. But in your form x-2y+z = -7, the vector of coefficients, <1, -2, 1> is perpendicular that this new plane. Since it was also perpendicular to the original plane, the two planes are parallel.
     
  6. Oct 29, 2008 #5
    Should I than from finding vector <1, -2, 1> do the cross product of it and the direction vector of the line <3,2-1> and than try and find the Cartesian equation of plane ?
     
  7. Oct 30, 2008 #6

    gabbagabbahey

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    That should work.
     
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