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Find the basis of a plane

  1. Nov 18, 2011 #1
    1. The problem statement, all variables and given/known data
    Let W be the plane
    3x + 2y − z = 0 in ℝ3.
    Find a basis for W perpendicular


    2. Relevant equations



    3. The attempt at a solution
    I thought a basis for this plane could be generated just by letting x=0 and y=1, finding z and then doing the same thing but this time letting x=1 and y=0 and finding z. If you do that you get:

    [0].........[1]
    [1].........[0]
    [2].........[3]

    Apparently this is wrong, can anybody tell me what's wrong about this?
     
  2. jcsd
  3. Nov 18, 2011 #2

    LCKurtz

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    You have just given two points on the plane or, interpreting them as vectors, the position vectors to those points. You want the normal vector [3,2,-1]. At least that's what I think the question is asking for.
     
  4. Nov 18, 2011 #3
    I think the question is asking for a basis of the orthogonal complement of W, usually denoted [itex] W^{\perp} [/itex] and read "W perp." So you wouldn't be looking for a basis for W, but for the set of all vectors perpendicular to W.

    EDIT: LCKurtz beat me to it.
     
  5. Nov 18, 2011 #4
    so are you saying that after getting the two vectors that i got solving for z, i should find any 2 other vectors that make the dot product with the first ones zero?
     
  6. Nov 18, 2011 #5

    Mark44

    Staff: Mentor

    No, I don't think that's what they're saying at all. W is the plane, a two-dimensional subspace of R3. W[itex]^\bot[/itex] ("W perp") is therefore a one-dimensional subspace of R3. The normal to the plane is in the same direction as W[itex]^\bot[/itex].
     
  7. Nov 18, 2011 #6
    ohh ok so then the answer is going to be just the vector of the plane itself?

    [3,2,-1]
     
  8. Nov 18, 2011 #7

    LCKurtz

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    That isn't the "vector of the plane". It is a vector perpendicular to the plane.
     
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