# Find the basis of a plane

• memo_juentes
In summary, the question is asking for a basis for the orthogonal complement of the plane W, which is a one-dimensional subspace of ℝ3. This can be found by taking the normal vector to the plane [3,2,-1] as the basis vector for W perpendicular.

## 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]

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.

## 1. What is the definition of a basis of a plane?

A basis of a plane is a set of two linearly independent vectors that span the entire plane. This means that any point in the plane can be reached by a linear combination of these two vectors.

## 2. How do you find the basis of a plane?

To find the basis of a plane, you can use the cross product of two non-parallel vectors in the plane. The resulting vector will be perpendicular to both of the original vectors and will serve as the third basis vector. Then, you can use Gaussian elimination to reduce the three vectors to two, which will be the basis of the plane.

## 3. Can the basis of a plane be unique?

Yes, the basis of a plane can be unique. If the plane is defined by two non-parallel vectors, then the basis will also be unique. However, if the plane is defined by three or more vectors, there may be multiple ways to reduce them to a basis, resulting in non-unique solutions.

## 4. What is the significance of finding the basis of a plane?

Finding the basis of a plane is important because it allows us to understand the structure and properties of the plane. It also allows us to perform calculations and transformations on the plane, such as finding the area or volume of a shape in the plane.

## 5. Can the basis of a plane change?

Yes, the basis of a plane can change depending on the choice of vectors used to define the plane. For example, if a different set of non-parallel vectors is used, the resulting basis will also be different. However, the basis will always consist of two linearly independent vectors that span the plane.