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Finding the parametric equations of the line L

  1. Apr 18, 2012 #1
    1. The problem statement, all variables and given/known data
    ∏1: 2x + y - z = 4
    ∏2: 3x - 2y +z = 6

    Find the parametric equation of the line L...

    3. The attempt at a solution

    So I make them simultaneous equations...

    2x + y = 4 - t (1)
    3x - 2y = 6 - t (2)

    multiple equation 1 by 2.

    4x + 2y = 8 - 2t (1)
    3x - 2y = 6 - t (2)

    Solve for 'x'

    7x = 14 - 3t

    x = 2 - 7/3t

    ...

    It just doesn't look right to me..
     
  2. jcsd
  3. Apr 18, 2012 #2

    tiny-tim

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    hi dimens1 :smile:
    a parametric equation is of the form (x,y,z) = (f(t),g(t),h(t)) :wink:

    (or three equations, one for each of x y and z)
     
  4. Apr 18, 2012 #3
    I thought it was...
    x=n + t
    y=n+t
    z=n+t
     
  5. Apr 18, 2012 #4

    tiny-tim

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    (i assume you mean different RHSs ?)

    same thing

    so where's your y and z equations?​
     
  6. Apr 18, 2012 #5
    I was gonna substitute y into x then z=t.
     
  7. Apr 18, 2012 #6

    tiny-tim

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    should work :confused:

    oh, wait a mo …
    … should be 4 plus t ! :cry:
     
  8. Apr 18, 2012 #7
    ∏1: 2x + y - z = 4
    ∏2: 3x - 2y +z = 6


    2x + y = 4 + t
    3x - 2y = 6 - t

    (2x + y = 4 + t)*2

    4x + 2y = 8 + 2t
    3x - 2y = 6 - t

    7x = 14 + t


    x = 2 + t/7

    subt x = 2 + t/7

    ... 2(2+t/7) - 2y = 6 - t
    4 + 2t/7 - 2y = 6 - t
    -2y = 2 - 5t/7
    y = - 1 + 10t/7

    z=t

    so...
    x = 2 + t/7
    y = -1 + 10t/7
    z = t

    .. answer says its

    x = 2 + t
    y = 5t
    z = 7t

    :s
     
  9. Apr 18, 2012 #8

    HallsofIvy

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    Good so far!

    The equation was 3x- 2y= 6- t

     
  10. Apr 18, 2012 #9
    Still that answer wont match..
     
  11. Apr 18, 2012 #10

    HallsofIvy

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    Since you don't say what equations you got it is impossible to say whether your answer is correct or not. The fact that you get different equations than given as the answer is irrelevant. A single line can have an infinite number of different representations as parametric equations.

    Let t= 0 and calculate x, y, z. Do they satisfy both equations of the original given planes? If so, that is a point on the line of intersection. Let t= 1 and calculate x, y, z. Do they satisfy both equations of the original given planes? If so that is also a point on the line of intersection? If so your equations give two points on the line and so are parametric equations for the line.
     
  12. Apr 18, 2012 #11
    Yeah I picked up on that but still it doesn't match... x is meant to be..

    x = 2 + t not x = 2 + t/7

    weird.
     
  13. Apr 20, 2012 #12
    bump?
     
  14. Apr 20, 2012 #13
    Anyone? :(
     
  15. Apr 22, 2012 #14
    triple bump... lol
     
  16. Apr 22, 2012 #15

    HallsofIvy

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    That's a good way to get yourself banned!

    By the way, your original statement of the problem was
    "∏1: 2x + y - z = 4
    ∏2: 3x - 2y +z = 6

    Find the parametric equation of the line L..."
    which makes no sense because you haven't said what L is! We "guessed" that you mean that L is the line of intersection of the two planes but you should have said that!

    You eventually arrived at x= 2+ t/7 and then 3(2+t/7) - 2y = 6 - t, which gives 6+ 3t/7- 2y= 6- t. Then 2y= 3t/7+ t= 10t/7 so y= 5t/7.

    I said that there are an infinite number of possible parametric equations for any line or curve. Here, if you choose s= t/7, you get parametric equations x= 2+ s, y= 5s.

    Of course, it doesn't matter what letter you use for the parameter so those are exactly the same as x= 2+ t, y= 5t.
     
  17. Apr 22, 2012 #16

    tiny-tim

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    i always thought that one bump every 24 hours was considered acceptable? :confused:
    ok, so put T = t/7 :smile:

    then

    x = 2 + T
    y = -1 + 10T
    z = 7T

    which is the given answer (apart from the y, which i assume is fixable :rolleyes:)

    btw, can you se that if you'd eliminated x at the start (instead of y), you'd immediately have got 7y=5z ? :wink:
     
  18. Apr 22, 2012 #17
    Cheers. Just one more question... Do I always solve for Y instead of X? Or vise versa
     
  19. Apr 23, 2012 #18

    tiny-tim

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    no, it just happens to be a lot quicker in this case

    (and in the exam, saving a few minutes may get you extra marks on other questions :wink:)

    (oh, and it's "vice versa" :smile:or is it "verse vica"? :rolleyes:)
     
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