# Find the basis of a plane

## Homework Statement

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

## 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]

LCKurtz
Homework Helper
Gold Member
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.

## 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.

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
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$^\bot$ ("W perp") is therefore a one-dimensional subspace of R3. The normal to the plane is in the same direction as W$^\bot$.

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