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Homework Help: Vector parametric equation of a line

  1. Oct 5, 2008 #1
    Find a vector parametric equation of the line in [tex]R^{2}[/tex] with equation 2x-3y = 4

    Attempt at a solution

    I haven't seen this type of question before so I don't know where to start. I suppose that the equation 2x-3y = 4 is a vector equation of that line and is in the form x = x0 + tv. I just don't know how to split this vector equation up and try to write the parametric vector equation.

    Can someone please explain to me what to do. Many Thanks.

  2. jcsd
  3. Oct 5, 2008 #2


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    2x - 3y = 4 is the cartesian equation, not the vector equation.

    If a line has the vector equation r = x0 + tv, then it passes through the point x0 and is parallel to the vector v.

    Can you find a point that the line 2x - 3y = 4 passes through? Can you find a vector to which it is parallel?
  4. Oct 5, 2008 #3


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    Hi roam! :smile:

    Yes, that is the right form. :smile:

    x0 can be any point on the line.

    So just choose some point on the line (at random), and chug away. :wink:
  5. Oct 5, 2008 #4


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    One of your difficulties is that there are many different parametric equations for the same line.

    Yes, you want x= at+ b, y= ct+ d. But you can choose "t" anyway you want. You might, for example, choose t to be x itself: that is x= t. Now y= ?
  6. Oct 5, 2008 #5
    Hi Tim!

    I think I should even write the equation 2x-3y=4 in the form:

    [tex]y = \frac{2x -4}{3}[/tex] for some t

    And I choose some random point on the line, (2,0)

    Now for x = x0 + bt, y = y0+ bt

    If x = t, like Hall said, y = [tex]\frac{2t -4}{3}[/tex]

    is this the vector parametric equation of the line? Do I need to specify a "t" as well?
  7. Oct 5, 2008 #6
    Is it (t, [tex]\frac{2t -4}{3}[/tex]) ?
  8. Oct 5, 2008 #7


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    vector parametric equation

    Hi roam! :smile:
    So far, so good. :smile:

    (though you could have got that straight from 2x-3y=4, couldn't you? :wink:)

    At this point I'm going to disagree with HallsofIvy, and say that I think the question is asking for a pure vector equation, rather than a coordinate-based one.

    Coordinate-based would be of the form x = f(t), y = g(t).

    Pure vector would be of the form r(t) = a + bt, where a and b are both constant vectors.

    So the answer would be r(t) = (2,0) + (?,?)t. :smile:
  9. Oct 7, 2008 #8
    Re: vector parametric equation

    Hi Tim,

    How do we determine the direction vector? Thanks!
  10. Oct 7, 2008 #9


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    oh come on! :rolleyes:

    just look at the original equation: 2x-3y = 4 …

    if x goes up by 1, then y goes up by … ? :smile:
  11. Oct 7, 2008 #10


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    Very little difference between "parametric equations", x= f(t), y= g(t), as I gave, and "vector parametric equation", [itex]\vec{r}= f(t)\vec{i}+ g(t)\vec{j}[/itex].
  12. Oct 7, 2008 #11
    Would it be 3/2?
  13. Oct 7, 2008 #12
    Hmm, try again.

    You yourself said that y=[tex]\frac{2x-4}{3}[/tex]. What is the slope of that line? How would you express that slope in the form of a direction vector?
  14. Oct 7, 2008 #13

    m = 2/3 ?
  15. Oct 7, 2008 #14

    [tex]y = \frac{2x-4}{3}[/tex]
    if x = 1 then y = -10/3

    Is that it? can we plug it in the equation (2,0) + t(*,*)? :confused:
  16. Oct 7, 2008 #15


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    Yes! :smile:
    uh? wots happening? :confused:

    (x,y) = (2,0) + t(1,2/3), or perhaps more neatly (x,y) = (2,0) + t(3,2). :smile:
  17. Oct 7, 2008 #16

    You're golden.
  18. Oct 7, 2008 #17
    If x=1, then y=-2/3. Just for clarification. Where did you get the -10/3 from?
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