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Non collinear vectors

  1. Jan 4, 2009 #1
    I kinda remember some identity which goes as follows:

    If [tex]a,b,c[/tex] are coplanar, non collinear vectors then

    [tex]\alpha a + \beta b + \gamma c = 0[/tex]
    => [tex]\alpha + \beta + \gamma = 0[/tex]

    or something like this. Can someone help me remember.
     
  2. jcsd
  3. Jan 4, 2009 #2

    radou

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    Re: Vectors

    Well, this isn't quite correct.

    If a, b, and c are three coplanar vectors, they are for sure linearly dependent in the plane, since, if a, b are two non colinear non zero vectors in a plane, they form a basis, i.e. [tex]\alpha a + \beta b = 0[/tex] => [tex]\alpha = \beta = 0[/tex]. Every third vector can be representet uniquely as a linear combination of the basis vectors a and b.
     
  4. Jan 4, 2009 #3

    mathman

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    Re: Vectors

    You need to replace "a, b, and c are coplanar" by "a, b, and c are NOT coplanar".
     
  5. Jan 4, 2009 #4

    rock.freak667

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    Re: Vectors

    Also for coplanar vectors
    [tex]a\cdot(b\times c)= 0[/tex]
     
  6. Jan 5, 2009 #5
    Re: Vectors

    I am sorry but I really mean [tex]\alpha + \beta + \gamma = 0[/tex] and not [tex]\alpha = \beta =\gamma = 0[/tex]
     
  7. Jan 5, 2009 #6

    radou

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    Re: Vectors

    Well, if [tex]\alpha = \beta =\gamma = 0[/tex], then most certainly [tex]\alpha + \beta + \gamma = 0[/tex]. :wink:
     
  8. Jan 5, 2009 #7

    HallsofIvy

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    Re: Vectors

    No, if they were not coplanar, that statement would not be true.
     
  9. Jan 5, 2009 #8

    mathman

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    Re: Vectors

    In his original statement he had all coef = 0, not the sum. Obviously changing the question would usually lead to a change in the response.
     
  10. Jan 5, 2009 #9

    HallsofIvy

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    Re: Vectors

    Oh, thanks. I hate it when people edit their post after there have been responses!
     
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