What is the basis for a plane perpendicular to 3x + 2y − z = 0?

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Homework Statement


Let W be the plane
3x + 2y − z = 0 in ℝ3.
Find a basis for W perpendicular

Homework Equations


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?
 
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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.
 
memo_juentes said:

Homework Statement


Let W be the plane
3x + 2y − z = 0 in ℝ3.
Find a basis for W perpendicular

I think the question is asking for a basis of the orthogonal complement of W, usually denoted W^{\perp} 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.
 
spamiam said:
I think the question is asking for a basis of the orthogonal complement of W, usually denoted W^{\perp} 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.

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?
 
No, I don't think that's what they're saying at all. W is the plane, a two-dimensional subspace of R3. W^\bot ("W perp") is therefore a one-dimensional subspace of R3. The normal to the plane is in the same direction as W^\bot.
 
Mark44 said:
No, I don't think that's what they're saying at all. W is the plane, a two-dimensional subspace of R3. W^\bot ("W perp") is therefore a one-dimensional subspace of R3. The normal to the plane is in the same direction as W^\bot.

ohh ok so then the answer is going to be just the vector of the plane itself?

[3,2,-1]
 
memo_juentes said:
ohh ok so then the answer is going to be just the vector of the plane itself?

[3,2,-1]

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