Vector Equation in Plane Space

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Homework Help Overview

The discussion revolves around determining a vector equation for a plane that contains two parallel lines represented by their vector equations. The original poster expresses uncertainty about how to proceed after identifying a position vector.

Discussion Character

  • Exploratory, Conceptual clarification, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the need to find two vectors in the plane, with one participant suggesting the use of directional vectors from the lines. There is a focus on identifying an additional vector to define the plane due to the linear dependence of the directional vectors from the parallel lines.

Discussion Status

Participants are actively exploring different approaches to find the necessary vectors for the plane equation. Guidance has been provided regarding the need for independent vectors and the importance of selecting points from the lines to establish these vectors. The conversation indicates a productive direction, with participants clarifying concepts and correcting misunderstandings.

Contextual Notes

There is an emphasis on the need for two independent vectors to define the plane, and participants are navigating the challenge of identifying appropriate points on the given lines. The original poster's initial attempts reflect a common struggle with the problem setup.

Hollysmoke
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I know it's not a difficult question, but I don't really know how to solve the question, and the textbook doesn't help all that much:

Determine a vector equation for a plane containing two parallel lines-
r=(0,1,3) + t(-6,-3,6)
r=(-4,5,-4)+ s(4,2,-4)

I used (0,1,3) as the position vector but where do I go from there?
 
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One of the ways to find the equation of a plane is to first find two vectors that are in that plane.

Can you find those two vectors from your two lines?
 
You mean directional vectors, right?
 
Yes.

I guess in this case the two lines are actually parallel so the two directional vectors are linerly dependent so you'll need to find another vector that would lie in the plane.

Basically in the end it will be just like the equation of a line but instead of only one parameter the plane will have two parameters since it's 2-dimensional.

For a line:
[tex]\vec{r} = \vec{u} + s\vec{v}[/tex]
and similarly for a plane:
[tex]\vec{r} = \vec{u} + s\vec{v} + t\vec{w}[/tex]

Of course you'll also need a point on a plane along with the two vectors to define it.

Edit: http://en.wikipedia.org/wiki/Plane_(mathematics) If you scroll down to "Planes embedded in R3 you can read up on that to get some more details regarding other situations dealing with planes should they arise.
 
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I'm not getting the right answer and I think I'm starting it wrong...
 
Can you show what you are doing?
 
r=sa+tb

I got as far as r=(0,1,3) + s(2,1,-2) but I don't know how to figure out the directional vector for b
 
You just need two independent vectors that are on the plane. You already found one of them by using the movement vector for the lines. The second one can be found by finding two points [tex]\vec{a}[/tex] and [tex]\vec{b}[/tex] on the plane and finding the vector from one to the other, [tex]\vec{ab}[/tex]
 
So I should find the vector between (2,1,-2) and (-6,-3,6)?
 
  • #10
Pick a point on line 1 and another point on line 2 and find the vector from one to the other.

(2,1,-2) and (-6,-3,6) are not points on the line; they're the movement vectors of the lines.
 
  • #11
OH! OMG I GET IT!

So I can use the Position vectors, and I'll get (4,-4,7)!

0-(-4)=4
1-5=-4
3-(-4)=7
 
  • #12
Thank you! 8D
 

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