Find parametric question for the plane

  • Thread starter Thread starter bonbon
  • Start date Start date
  • Tags Tags
    Parametric Plane
Click For Summary

Homework Help Overview

The discussion revolves around finding parametric equations for the plane defined by the equation 2x-3y+z-6=0. Participants explore the concept of direction vectors in relation to the plane.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the definition of direction vectors in the context of a plane versus a line. There is an exploration of how to derive two independent vectors that lie within the plane, as well as the relationship between the normal vector and the direction vectors.

Discussion Status

Some participants have provided insights into the relationship between the normal vector and the direction vectors, suggesting methods to find independent vectors. There is an acknowledgment of different interpretations of the term "direction vector" and its application to planes versus lines.

Contextual Notes

There is a potential confusion regarding terminology, particularly the use of "direction vector" in the context of planes, which may differ from its use in line equations. The discussion also hints at the need for two parameters in the parametric equations for a plane.

bonbon
Messages
5
Reaction score
0

Homework Statement


Give parametric questions for the plane : 2x-3y+z-6=0


The Attempt at a Solution



i know that the normal is (2,-3,1)
how do i find the direction vector of the plane?
 
Physics news on Phys.org
The plane has two linearly independent 'direction vectors'. Find any two independent vectors perpendicular to your normal.
 
Dick is assuming that by 'direction vector', you mean two independent vectors in the plane. I've only seen the term used with lines.

I wonder if you are not confusing this with "find the parametric equations for a line". To do that you would find 'direction vector' for the line. Parametric equations for a plane will involve two parameters. Here, since you can write z as a function of x and y, z= 6- 2x+ 3y, you can use x and y themselves as parameters or, if you prefer distinct variables, x= u, y= w, z= 6- 2u+3w.
 
That's a much more direct way. I was fixated on the 'direction vector' (basis) picture.
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
2K
Replies
8
Views
3K
  • · Replies 7 ·
Replies
7
Views
3K
Replies
10
Views
3K
  • · Replies 8 ·
Replies
8
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K