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Vector equation of a plane containing this line.

  1. Jun 27, 2012 #1
    1. The problem statement, all variables and given/known data

    What is the vector equation of a plane that contains the line y=(3,2,1)+x(1,0,2) and is parallel to the z-axis?

    2. Relevant equations

    Equation of a plane: Ax + By + Cz + D = 0

    3. The attempt at a solution

    Ok so i substitute the values into the plane,
    A(1)+B(0)+C(2)+D=0

    But then I am left with 3 variables. Can you please offer some help?

    Kind Regards,
     
  2. jcsd
  3. Jun 27, 2012 #2

    tiny-tim

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    welcome to pf!

    hi manmachine! welcome to pf! :smile:

    sorry, but i don't understand what you're doing at all :redface:

    the easiest way of finding the equation is probably first to find the normal to the plane …

    what will that be? :wink:

    (and i think the question probably means the parametric equation v = (3,2,1) + t(1,0,2))
     
  4. Jun 27, 2012 #3
    Re: welcome to pf!

    Yes but I don't know what the plane is. :redface: I know that the normal is the coefficients if the plane. Please help.
     
  5. Jun 27, 2012 #4

    HallsofIvy

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    Why do you substitute those values? Those would be the components of a vector in the plane, not the coordinates of any point in the plane. (3, 2, 1) is a point in the plane. Setting x= 3, y= 2, z= 1, we have 3A+2B+ C+ D= 0.

    Also, it is not a good idea to use "x" and "y" in that equation for the plane. In that vector equation "x" is a parameter, not the x- coordinate and "y" is a vector. Better to use, say, r= (3, 2, 1)+ t(1, 0, 2).

    In that form, you can always divide through by any non=zero coefficient so you really have only two variables left. To eliminate one, use the fact that the plane is parallel to the z-axis. That means that the normal to the plane is also normal to the z-axis. That is, <0, 0, 1>.<A, B, C>= C= 0.
     
  6. Jun 27, 2012 #5
    And yes, but how can I find the normal if don't have to vector equations IAXBI?
     
  7. Jun 27, 2012 #6

    tiny-tim

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    hi manmachine! :smile:
    yes, but you do know the directions of two lines that the normal is perpendicular to, don't you? :wink:
     
  8. Jun 27, 2012 #7
    There is only one line Tim! I can't use cross product :(
     
  9. Jun 27, 2012 #8

    tiny-tim

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    z-axis? :smile:
     
  10. Jun 27, 2012 #9
    oh fudge :P (bangs head on wall)
    So it is 0,1,0 by cross product right?
     
  11. Jun 27, 2012 #10

    tiny-tim

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    he he! :biggrin:
    uhh? :confused:
     
  12. Jun 27, 2012 #11
    Isn't that the normal to plane?
     
  13. Jun 27, 2012 #12

    tiny-tim

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    (0,1,0) is the y-axis, and is normal to the xz-plane :confused:
     
  14. Jun 27, 2012 #13
    If I take the cross product of the z-axis (being 0,0,1) and our vector (1,0,2) this is what i get.
     
  15. Jun 27, 2012 #14

    tiny-tim

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    oh i see!! :rolleyes:

    ok, now you know the normal to the plane, and you know at least one point on the plane,

    so the vector equation of the plane is … ? :smile:
     
  16. Jun 27, 2012 #15
    Am i right :)?

    Take coefficents
    0a+1b+0c+D=0
    (0)(3)+(1)(1)+D=0
    d=-1

    This equation of our is plane is this:
    b+-1=0 ?
     
  17. Jun 27, 2012 #16

    tiny-tim

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    shouldn't there be a 2 in there somewhere? :confused:
    what happened to x y and z ? :confused:

    don't make it so complicated

    you know it's vertical, and parallel to the xz-plane, so it must be y = constant

    ok, what is obviously that constant?? :smile:
     
  18. Jun 27, 2012 #17
    I have no idea :p
     
  19. Jun 27, 2012 #18
    What do I do next . I am sorry I just dont get it
     
  20. Jun 27, 2012 #19

    HallsofIvy

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    This is very peculiar. You say that you know that any plane can be written as Ax+ By+ Cz+ D= 0. It has already been pointed out that you an divide through by any number, say, D, to reduce to three unknown constants, A'x+ B'y+ C'z+ 1= 0 (with A'
    = A/D etc.). If the line given by the vector formula r= (3,2,1)+t(1,0,2) or parametric equations x= 3+ t, y= 2, z= 1+ 2t. In particular, taking t= 0, (3, 2, 1) is a point in the plane so we must have 3A'+ 2B'+ C'+ 1= 0.

    Also, taking t= 1, (3+ 1, 2+ 0, 1+ 2)= (4, 2, 3) is also a point in the plane. so we must have 4A'+ 2B'+ 3C'+ 1= 0.

    The last condition is that the plane is parallel to the z-axis. That means that its normal vector, <A, B, C> or <A', B', C'> is normal to <0, 0, 1>. Their dot product gives C= C'= 0 and so we have the two equations -3A'- 2B'- 1= 0 and 4A'+ 2B'+ 1= 0.

    By the way, I find it simpler to use the fact that if a plane has normal vector <A, B, C> and contains point [itex](x_0, y_0, z_0)[/itex], then the plane has equation [itex]A(x- x_0)+ B(y- y_0)+ C(z- z_0)= 0[/itex].
     
  21. Jun 27, 2012 #20

    tiny-tim

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    you know that the plane is y = constant

    you also know that (3,2,1) lies on the plane

    soooo, y must be … ? :smile:
     
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