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Parametars B and D of line

  1. May 16, 2008 #1
    1. The problem statement, all variables and given/known data

    For which B and D the line [tex]\left\{\begin{matrix}
    x-2y+z-9=0 & \\
    3x+By+z+D=0 &
    \end{matrix}\right.[/tex] is in xy-plane ?


    2. Relevant equations



    3. The attempt at a solution

    I found the line:

    [tex]\frac{x-\frac{-D+9}{2}}{2-B}=\frac{y}{2}=\frac{z-\frac{D+27}{2}}{B+6}[/tex]

    What should I do next?
     
  2. jcsd
  3. May 16, 2008 #2

    tiny-tim

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    Hi Physicsissuef! :smile:

    Hint: you want z to be constant, but you don't want y to be constant.

    So … ? :smile:
     
  4. May 16, 2008 #3
    If it is in the xy-plane probably B+6=0.
     
  5. May 16, 2008 #4

    tiny-tim

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    But how do you stop y being infinite, then? :smile:
     
  6. May 16, 2008 #5
    A point in xy plane have coordinates (x,y,0). A line in xy plane probably have [itex]a(a_1,a_2,0)[/itex]. a is parallel vector to the line, in this case [itex]a(2-B , 2 , B+6)[/itex]. And I don't understand you what are you talking to me... :smile:
     
  7. May 16, 2008 #6

    HallsofIvy

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    I think you are putting far too much emphasis on plugging into formulas rather that thinking about what the problem says. In the xy plane, z is not just constant- it is specifically 0. If z= 0 then your equations become 3x- 2y- 9= 0 and 3x+ By+ D= 0. In the xy plane, those are each the equation of a line. For what B and D are those the same line?
     
  8. May 16, 2008 #7
    B=-6 and D=-27. But why they need to be same?
     
  9. May 16, 2008 #8

    HallsofIvy

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    Reread your original question! The equations represent a single line.
     
  10. May 16, 2008 #9
    Ohh.... I see, thanks. And is it correct that "a" should have coordinates [itex]a_1,a_2,0[/itex]?
     
  11. May 16, 2008 #10

    tiny-tim

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    I was referring to the fact that your equation y/2 = (z - (D+27)/2)/(B + 6) seems to make y fixed and infinite if B + 6 = 0, while you want y to be able to take any value. :smile:
    Well, yes, the line is parallel to a line through the origin with (a_1,a_2,0).

    But you're concentrating too much on formulas, as HallsofIvy said.

    btw, his method is a lot easier than the one you chose … if you do that in the exam, the examiner wil get the impression you're using the formula without really understanding it.
     
  12. May 16, 2008 #11
    Hehe... I used the formula because I was desperate, I forgot that z=0. Thanks for the help.
     
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