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Vector Equation in Plane Space

  1. Apr 2, 2006 #1
    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?
     
  2. jcsd
  3. Apr 2, 2006 #2

    dav2008

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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?
     
  4. Apr 2, 2006 #3
    You mean directional vectors, right?
     
  5. Apr 2, 2006 #4

    dav2008

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    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.
     
    Last edited: Apr 2, 2006
  6. Apr 2, 2006 #5
    I'm not getting the right answer and I think I'm starting it wrong...
     
  7. Apr 2, 2006 #6

    dav2008

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    Can you show what you are doing?
     
  8. Apr 2, 2006 #7
    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
     
  9. Apr 2, 2006 #8

    dav2008

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    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]
     
  10. Apr 2, 2006 #9
    So I should find the vector between (2,1,-2) and (-6,-3,6)?
     
  11. Apr 2, 2006 #10

    dav2008

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    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.
     
  12. Apr 2, 2006 #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
     
  13. Apr 2, 2006 #12
    Thank you! 8D
     
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