Double Dot Product: Solving 3D Vector Problem

[Naake]

Hi,
I have following problem of double dot product (\vec a \cdot \vec b)(\vec a^* \cdot \vec c), and I have expected rusult |a|^2(\vec b \cdot \vec c), 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 \vec a is complex and \vec b, \vec c are real 3D vectors. Maybe can help, that all vectors lie in the plane. So is it true that
(\vec a \cdot \vec b)(\vec a^* \cdot \vec c) =? |a|^2(\vec b \cdot \vec c)?
Thanks,
Michal

 

[matteo137]

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 $\beta(\vec a^* \cdot \vec c)=(\beta\vec a^*) \cdot \vec c$, ...

 

[HallsofIvy]

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 \vec a is complex and \vec b, \vec c are real 3D vectors."
There is no standard notation in which "\vec{a}" would be used to denote a complex number. Did you mean that \vec{a} 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".

 

[mathman]

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.