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Describe each plane, 8x-5y=-40z

  1. Jun 14, 2008 #1
    Apologies for this being a kind of stupid question...

    1. The problem statement, all variables and given/known data

    Describe each plane

    (a) [tex]8x-5y=-40z[/tex]

    2. Relevant equations

    Probably have to find the x, y, and z intercepts.

    3. The attempt at a solution

    I would assume that in order to find the x intercept you set y and z equal to zero. And the same idea for the other two.

    THe answer that I got was (0, 0, 0). Am I doing this wrong?
     
  2. jcsd
  3. Jun 14, 2008 #2

    tiny-tim

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    Hi paralian! :smile:

    No, you're completely correct …

    the plane intersects all three axes at the origin, (0, 0, 0). :smile:

    erm … wasn't that obvious? :rolleyes:
     
  4. Jun 15, 2008 #3

    HallsofIvy

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    Since there are an infinite number of planes that contain (0,0,0), you might want to give some other points also to "describe" the plane. Three points determine a plane.
     
  5. Jun 15, 2008 #4

    Defennder

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    What is required to answer such a question?
     
  6. Jun 15, 2008 #5

    tiny-tim

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    Hi Defennder! :smile:

    Hint: Forget about axes … what defines a particular plane through a given point? :smile:
     
  7. Jun 15, 2008 #6

    Defennder

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    The dot product of the normal vector to the plane and a vector lying on the plane through that point?
     
  8. Jun 16, 2008 #7

    tiny-tim

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    Well, that will always be zero …

    but what defines the plane? :smile:

    (what makes one plane different from another plane?)
     
  9. Jun 16, 2008 #8

    HallsofIvy

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    More specifically, what do you or your teacher mean by "describe the plane". You said you were looking for the three axis-intercepts but found that the plane passes through (0,0,0). My point before, based on what you said you were doing, was that three points "determine" a plane. Did you intend to "describe" the plane by giving the 3 intercepts? If so that is the same as giving three points. But they don't have to be the intercepts. You know one point, (0,0,0). You can find two more points in the plane by picking any two of x, y, or z to be some simple numbers (simplicity is, after all, is the whole point of the intercepts) and then solve for the third component.

    It is also true that a single point (which you already have) and a normal vector "determine" a plane. Would you accept that as a "description" of the plane?
     
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