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I Scalar product

  1. May 1, 2016 #1
    Why A.A = ||A||^2 , I know that from product rule we can prove this where theta =0 , I am asking this because I have seen many proves for A.B = ||A||||B||cos(theta) and to prove this they have used A.A = ||A||^2, how can they use this , this is the result of dot product formula. I havee seen every where even on Wikipedia they have used this method only
     
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  3. May 2, 2016 #2

    blue_leaf77

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    That's the definition of a norm.
     
  4. May 2, 2016 #3
    Okk , can u give me the proof of vector product,or tell me the about the difficulty or knowlege requires to prove this

     
  5. May 2, 2016 #4

    blue_leaf77

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    Do you mean "scalar product"?
     
  6. May 2, 2016 #5
    No ,vector product, A X B = ||A||||B||sin(theta)
     
  7. May 2, 2016 #6

    blue_leaf77

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    Do you want to prove that the magnitude of ##\vec{A}\times \vec{B}## is equal to ##|\vec{A}| |\vec{B}| \sin \theta##?
     
  8. May 2, 2016 #7
    Yess
     
  9. May 2, 2016 #8

    blue_leaf77

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    The proof is tedious and might look unattractive if one were to outline it here, for this purpose I will simply refer you to this link.
     
  10. May 2, 2016 #9

    Mark44

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    This formula is not correct. On the left side, A X B is a vector. On the right side ##|A||B|\sin(\theta)## is a scalar. As already noted, ##|A||B|\sin(\theta)## is equal to the magnitude of A X B; that is, |A X B|.
     
  11. May 2, 2016 #10

    mathman

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    http://clas.sa.ucsb.edu/staff/alex/DotProductDerivation.pdf

    Above may help.
     
  12. May 3, 2016 #11

    SammyS

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    If you want to prove either:
    AB = |A| |B| cos(θ)​
    or
    |A×B| = |A| |B| sin(θ)​
    , you need to give the definitions you are using for scalar product and vector product.
     
  13. May 3, 2016 #12
    I forget to write it as |A X B|
     
  14. May 4, 2016 #13

    Mark44

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    If you forget to write the absolute value symbols, you will mistake a vector for a scalar (a number).
     
  15. May 4, 2016 #14
    I know that

     
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