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Vector identity

  1. Dec 22, 2009 #1
    1. The problem statement, all variables and given/known data
    This is a problem from a textbook, Riley Hobson and Bence 'Mathematical Methods for Physics and Engineering'. It asks to check the validity of a vector identity. If a, b and c are general vectors satisfying a x c = b x c, does this imply c . a - c . b = c|a-b|

    2. The attempt at a solution
    According to the book the identity is satisfied, but I think that there should be a plus/minus sign before the RHS, since we don't know whether the vectors a-b and c are in the same or opposite directions.
     
    Last edited: Dec 23, 2009
  2. jcsd
  3. Dec 22, 2009 #2

    LCKurtz

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    Nothing implies that because the left side is a scalar and the right side is a vector.
     
  4. Dec 22, 2009 #3

    cepheid

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    How is the right hand side a vector? There are absolute value signs around the only vectors in the expression.
     
  5. Dec 22, 2009 #4

    LCKurtz

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    The OP stated:
    "If a, b and c are general vectors satisfying a x c = b x c, does this imply c . a - c . b = c|a-b|"

    That makes the right side the vector c times the scalar |a-b|. That's how.
     
  6. Dec 23, 2009 #5
    LCKurtz, thanks for your interest in the topic, I would just like to mention that c means 'the vector c' while c means 'the magnitude of vector c'. The bold notation for vectors is widely used.
    I would be grateful if someone has an opinion on the problem. Thanks.
     
  7. Dec 23, 2009 #6

    LCKurtz

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    Perhaps, if you are using such a convention, you should then have bolded them when you posted "If a, b and c are general vectors".
     
  8. Dec 23, 2009 #7
    Yes, you're right, they should be bolded, thanks for that.
     
  9. Dec 23, 2009 #8

    cepheid

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    :rolleyes: <-- That is all I have to say about that. It was quite clear what the OP meant from the fact that he started using boldface for vectors. Your interpretation of his expression is silly.

    Okay! Are we done splitting hairs now? Ready to actually start helping this guy with his problem?

    Grand, one thing I can think of as a starting point is the fact that the dot product is distributive over vector addition, therefore it must be true that:

    c·a - c·b = c·(a - b) ​

    You also know that the definition of the dot product of two vectors is the product of the magnitudes of those two vectors multiplied by the cosine of the angle between them. Does that help?
     
  10. Dec 23, 2009 #9
    That's how I did it, and I would also explain why do I think that there should be a +/- sign in the RHS.

    Starting with the given relation:

    a x c = b x c
    a x c - b x c = 0
    (a - b) x c = 0
    (a - b) x c = 0 implies |a - b||c|sinD=0 where D is the angle between the vectors. Since the vectors are general and therefore non-zero, it follows that sinD=0 and this leaves us with 2 possible values for D - 0 and 180 degrees This means that these vectors have either the same or opposite directions.

    Now we continue with:
    c . a - c . b = c . (a - b) = |c||a - b| cosD = c|a - b| cosD

    However the two values for D give two different values for cosD : -1 and 1, and here's where the plus/minus sign comes from:
    c . a - c . b = +/-c|a - b|
     
  11. Dec 23, 2009 #10

    tiny-tim

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    Welcome to PF!

    Hi Grand! Welcome to PF! :smile:

    Yes, you're right …

    as a simple counter-example, if a = 0 and b = c, then you need the minus.
     
  12. Dec 23, 2009 #11

    LCKurtz

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    You can show there must be a - sign allowed by just taking an example:

    a=<1,0,0>
    b=<1,1,0>
    c=<0,1,0>

    a x c = b x c = <0,0,1>

    a . c - b . c = -1 = -|c||a-b|
     
  13. Dec 23, 2009 #12
    Thanks a lot for the immediate help, you're invaluable. I'm really grateful.
     
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