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Double dot product

  1. Mar 1, 2015 #1
    Hi,
    I have following problem of double dot product [tex](\vec a \cdot \vec b)(\vec a^* \cdot \vec c),[/tex] and I have expected rusult [tex]|a|^2(\vec b \cdot \vec c),[/tex] but I don't know if it is the exactly result (I am unable to find any appropriate identity or proove it), or it is just an approximation... where [tex]\vec a[/tex] is complex and [tex]\vec b, \vec c[/tex] are real 3D vectors. Maybe can help, that all vectors lie in the plane. So is it true that
    [tex](\vec a \cdot \vec b)(\vec a^* \cdot \vec c) =? |a|^2(\vec b \cdot \vec c)?[/tex]
    Thanks,
    Michal
     
  2. jcsd
  3. Mar 1, 2015 #2
    Approach 1 : write down ##(\vec a \cdot \vec b)(\vec a^* \cdot \vec c) = (a_1 b_1 + ...)(a_1^\ast c_1 + ...)## and go through the calculation.

    Approach 2 : Each dot product is a scalar e.g. ##(\vec a \cdot \vec b) = \beta##. You know that ##\beta(\vec a^* \cdot \vec c)=(\beta\vec a^*) \cdot \vec c##, ...
     
  4. Mar 1, 2015 #3

    HallsofIvy

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    There may be a language problem here. I do not understand your "but I don't know if it is the exactly result (I am unable to find any appropriate identity or prove it), or it is just an approximation... where [itex]\vec a[/itex] is complex and [itex]\vec b[/itex], [itex]\vec c[/itex] are real 3D vectors."
    There is no standard notation in which "[itex]\vec{a}[/itex]" would be used to denote a complex number. Did you mean that [itex]\vec{a}[/itex] might be a vector over the complex field? In either case, the "dot product" is only defined for vectors in the same vector space. If any of the vectors were in a vector space over the complex numbers then then they would all have to be- possibly with the complex part of their components equal to 0.

    I can't imagine why you would think this was "just an approximation".
     
  5. Mar 1, 2015 #4

    mathman

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    In general your result is wrong. Example b=c, b perpendicular to a. a.b=0, a.c=0, but |a|2b.c not 0.
     
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