# Double dot product

1. Mar 1, 2015

### 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

2. Mar 1, 2015

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

3. Mar 1, 2015

### HallsofIvy

Staff Emeritus
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".

4. Mar 1, 2015

### 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.