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

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Homework Help Overview

The discussion revolves around finding a basis for a plane defined by the equation 3x + 2y − z = 0 in ℝ3, with a focus on understanding the relationship between the plane and its perpendicular vector space.

Discussion Character

  • Conceptual clarification, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants explore the idea of generating a basis for the plane by substituting values for x and y to find corresponding z values. There is a question about the correctness of this approach.
  • Some participants clarify that the normal vector [3, 2, -1] is relevant to the problem, suggesting that the original poster may be misunderstanding the question's intent.
  • There is a discussion about whether the task is to find a basis for the plane itself or for its orthogonal complement, W^{\perp}.
  • One participant questions if the approach should involve finding additional vectors that are orthogonal to the initial vectors derived from the plane's equation.

Discussion Status

The conversation is active, with participants providing insights into the nature of the problem and clarifying the distinction between the plane and its normal vector. There is no explicit consensus yet, but several productive lines of inquiry are being explored.

Contextual Notes

Participants are grappling with the definitions of the plane and its orthogonal complement, as well as the implications of dimensionality in ℝ3. The original poster's understanding of the problem setup is being questioned, particularly regarding the interpretation of the basis being sought.

memo_juentes
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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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