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Scalar triple product invariance under circular shift proof

  1. Apr 26, 2015 #1
    1. The problem statement, all variables and given/known data
    Prove that for any three vectors ##\hat a, \hat b ## and ## \hat c##, ##\hat a \cdot (\hat b \times \hat c)## = ##(\hat a \times \hat b) \cdot \hat c ##

    2. Relevant equations

    ## \hat i \cdot \hat i = \hat j \cdot \hat j = \hat k \cdot \hat k = (1)(1)\cos(0) = 1 ##
    ## \hat i \cdot \hat j = \hat i \cdot \hat k = \hat j \cdot \hat k = (1)(1)\cos(90) = 0 ##

    3. The attempt at a solution

    I tried this procedure:

    1. calculate the inner cross product ##(\hat b \times \hat c)## and ##(\hat a \times \hat b) ##
    2. dotting the resulting vector with the third remaining vector (## \hat a ## and ## \hat c ## respectively)
    3. I expected that the resulting vectors were identical in terms of their components, but they aren't.


    These are the calculations for the LHS of my thesis:

    1. ##(\hat b \times \hat c) ##= ##(b_y c_z - b_z c_y)\hat i## + ##(b_z c_x - b_x c_z)\hat j## + ##(b_x c_y - b_y c_x)\hat k##

    2. dotting with ## \hat a ##:

    ## (a_x \hat i + a_y \hat j + a_z \hat k)## ## \cdot [(b_y c_z - b_z c_y)\hat i## + ##(b_z c_x - b_x c_z)\hat j## + ##(b_x c_y - b_y c_x)\hat k]## =
    ##(a_x b_y c_z - a_x b_z c_y)\hat i## + ##(a_y b_z c_x - a_y b_x c_z)\hat j## + ##(a_z b_x c_y - a_z b_y c_x)\hat k##

    Now the calculations for the RHS of my thesis
    :

    1. ##(\hat a \times \hat b) ##= ##(a_y b_z - a_z b_y)\hat i## + ##(a_z b_x - a_x b_z)\hat j## + ##(a_x b_y - a_y b_x)\hat k##

    2. dotting with ## \hat c ##:

    ## (c_x \hat i + c_y \hat j + c_z \hat k)## ## \cdot [(a_y b_z - a_z b_y)\hat i## + ##(a_z b_x - a_x b_z)\hat j## + ##(a_x b_y - a_y b_x)\hat k]## =
    ##(a_y b_z c_x - a_z b_y c_x)\hat i## + ##(a_z b_x c_y - a_x b_z c_y)\hat j## + ##(a_x b_y c_z - a_y b_x c_z)\hat k##


    Clearly,

    ##(a_x b_y c_z - a_x b_z c_y)\hat i## + ##(a_y b_z c_x - a_y b_x c_z)\hat j## + ##(a_z b_x c_y - a_z b_y c_x)\hat k##

    is not equal to:

    ##(a_y b_z c_x - a_z b_y c_x)\hat i## + ##(a_z b_x c_y - a_x b_z c_y)\hat j## + ##(a_x b_y c_z - a_y b_x c_z)\hat k##

    What am I missing here?

    Thank you in advance
     
  2. jcsd
  3. Apr 26, 2015 #2

    Orodruin

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    You can never get a vector as a result from a scalar product. Your result should be a number.
     
  4. Apr 27, 2015 #3
    Thank you, I got it! If I don't consider the results as vectors but as normal algebraic expressions, they are perfectly equivalent.
     
  5. Apr 27, 2015 #4

    Orodruin

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    This rings a warning bell for me. It is not a matter of considering the result as a vector or not, the result of the triple product is a scalar and not a vector.
     
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