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Normal plane

  1. Apr 5, 2008 #1
    1. The problem statement, all variables and given/known data

    Find the normal vector equation (the equation with red in the picture below) of the plane which is normal on the vector n=(2,-2,1), and it is on distance 5 from O (i.e p=5)

    2. Relevant equations

    On this image are all the relevant equations.
    http://img232.imageshack.us/img232/6030/28762075sm2.jpg

    3. The attempt at a solution

    I should find [tex]n_o[/tex], and then substitute in the equation.
    [tex]n_o=\frac{n}{|n|}[/tex]
    So n=2i-2j+k, |n|=3, p=5
    [tex]n_o=\frac{2i-2j+k}{3}[/tex]

    So the equation:

    [tex]r * (\frac{2i-2j+k}{3}) - 5 =0 [/tex]

    The problem is that in my book they don't divide it by 3? Why ? Is my way correct?
     
  2. jcsd
  3. Apr 5, 2008 #2

    Hootenanny

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    Does it ask you for the unit normal, or just the normal?
     
  4. Apr 5, 2008 #3
    The angle between the plane and the vector is 90 degrees. Nothing more.
     
  5. Apr 5, 2008 #4

    Hootenanny

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    No, my point is the question asks you to find the normal vector, not the unit normal. There is no need to divide by it's magnitude.
     
  6. Apr 5, 2008 #5
    Actually, we should find the unit normal. I think we should divide it by its magnitude, because the definition says [tex]|n_o|=1[/tex]
     
  7. Apr 5, 2008 #6

    Hootenanny

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    In the question you stated,
    there is no mention of a unit normal. Hence the question wants n not n0.
     
  8. Apr 5, 2008 #7
  9. Apr 5, 2008 #8

    Hootenanny

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    Yes but where is that image taken from, a course text, lecture notes?
     
  10. Apr 5, 2008 #9
    from my textbook. It also says that [itex]|n_o|=1[/itex]
     
  11. Apr 5, 2008 #10

    Hootenanny

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    Does it occur to you that you are not meant to use that equation in this question?
     
  12. Apr 5, 2008 #11
    Then, when I should use this equation?
     
  13. Apr 5, 2008 #12

    Hootenanny

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    Perhaps,

    [tex]r\cdot \boldmath{n} - p = 0[/tex]

    Since you are not asked for the unit normal, simply the normal vector.
     
  14. Apr 5, 2008 #13
    If we write
    [tex]r\cdot \boldmath{n} - p = 0[/tex]
    then it wouldn't be plane, because equation of plane is only when [itex]|n_o|=1[/itex].
     
  15. Apr 5, 2008 #14

    Hootenanny

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    Yes it is still an equation of a plane, the very same plane in fact! The equation of a plane can be derived from any normal vector, not just a normalised one!
     
  16. Apr 5, 2008 #15
    Now, I even doubt why p=5, it should be p=3, since OP=p[itex]n_o[/itex]. Well, I think I am tottaly confused about the whole thing...
     
  17. Apr 5, 2008 #16

    Hootenanny

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    Would it be possible for you to scan the actual diagram from your textbook and post it here?
     
  18. Apr 5, 2008 #17
  19. Apr 5, 2008 #18

    Hootenanny

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    Thanks for the pictures, could you also type out the question in full, with all the information given?
     
  20. Apr 5, 2008 #19
    Find the normal vector equation of the plane which is normal on the vector n=(2,-2,1), at distance 5 from (0,0,0) (the coordinate start, sorry if I mistranslated). You can see the task on the third picture 88 page/1 task.
     
  21. Apr 5, 2008 #20

    Hootenanny

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    Thanks, perhaps something is getting lost in translation here. Does the vector n=(2,-2,1) lie on the plane, i.e. is it part of the plane, or is it perpendicular to it?
     
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