When does $\vec{a} \langle \vec{b}, \vec{c} \rangle = \langle \vec{a}, \vec{b} \rangle \vec{c}$ hold? 
When does $\vec{a} \langle \vec{b}, \vec{c} \rangle = \langle \vec{a}, \vec{b} \rangle \vec{c}$, where $\langle \vec{a}, \vec{b} \rangle$ is the dot product of $\vec{a}$ and $\vec{b}$?

I can see that it holds when $\vec{a}$ and $\vec{c}$ are perpendicular to $\vec{b}$, because then $\langle \vec{a}, \vec{b} \rangle = \langle \vec{b}, \vec{c} \rangle = 0$, but how do I make sure if there are any other possibilities? 
 A: An interesting identity of vector arithmetic is the "bac cab rule"
$$ \vec{a} \times (\vec{b} \times \vec{c}) = 
\vec{b} \langle \vec{a}, \vec{c} \rangle 
- \vec{c} \langle \vec{a}, \vec{b} \rangle $$
Thus,
$$\vec{a} \langle \vec{b}, \vec{c} \rangle  = \langle \vec{a}, \vec{b} \rangle  \vec{c} 
\qquad \text{ if and only if }\qquad 
\vec{b} \times (\vec{a} \times \vec{c}) = 0
$$
Since, for nonzero $\vec{v}$ and $\vec{w}$, we have that $\vec{v} \times \vec{w} = 0$ if and only if they are parallel, and otherwise $\vec{v} \times \vec{w}$ is perpendicular to the plane they span, we conclude:
Theorem: We have $\vec{a} \langle \vec{b}, \vec{c} \rangle  = \langle \vec{a}, \vec{b} \rangle  \vec{c} $ if and only if one (or more) of the following hold:


*

*one or more of the vectors are zero,

*$\vec{b}$ is perpendicular to the plane spanned by $\vec{a}$ and $\vec{c}$,

*$\vec{a}$ and $\vec{c}$ are parallel.

A: We can consider case by case.


*

*If $(b,c)\neq 0$ and $(a,b)\neq 0$, then $a$ and $c$ are parallele.

*If, say, $(b,c) = 0$, then you have the zero vector on LHS. Thus $(a,b)=0$ or $c=0$.

A: If $\langle\vec{a},\,\vec{b}\rangle=0$, we require $\vec{a}=0\lor\langle\vec{b},\,\vec{c}\rangle=0$. We can deal with the case $\langle\vec{b},\,\vec{c}\rangle=0$ similarly, so hereafter assume neither is true.
We need $\vec{a}$ to be parallel to $\vec{c}$, where I call the zero vector parallel to everything, and indeed if either $\vec{a}$ or $\vec{c}$ vanishes the condition holds. Otherwise parallelism can be stated as $\vec{a}=k\vec{c}$ for $k\ne 0$, so the given condition reduces to $k\vec{c}\langle\vec{b},\,\vec{c}\rangle=\langle k\vec{c},\,\vec{b}\rangle\vec{c}$. For a real-valued inner product, this holds for any $\vec{b},\,\vec{c}$. For a complex-valued inner product linear in its first argument, we also need $\langle\vec{b},\,\vec{c}\rangle\in\mathbb{R}$. For a complex-valued inner product antilinear in its first argument, we instead have the final requirement $k\langle\vec{b},\,\vec{c}\rangle\in\mathbb{R}$.
