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

[memo_juentes]

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?

 

[LCKurtz]

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.

 

[spamiam]

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.

 

[memo_juentes]

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?

 

[Mark44]

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.

 

[memo_juentes]

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]

 

[LCKurtz]

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.