Does there exist any linear function $f$ such that $\forall a, b, c \in \mathbb{R}^n: (a-b)\cdot c=a\cdot f(a, b, c)$ ?
And if so, how can one such function be found?
"$\cdot$" denotes the scalar product. The closest I got is $f(a,b,c)=c-\frac{a}{a^2}(b\cdot c)$ which is obviously not linear in $a$ but I am looking for a function linear in $a$, $b$ and $c$.
Edit: Since it's not possible if $a$ can be the zero-vector, does adding the condition that $a\neq 0^n$ make the problem solvable? ($0^n$ being the n-dimensional zero-vector)
Element-wise, we have
$$\sum_{i=1}^N (a_i-b_i)c_i = \sum_{i=1}^N a_i f_i$$
First, assume that each $f_i$ only depend on $a_i,b_i$ and $c_i$. Then, we must have that:
$$f_i(a_i,b_i,c_i) = c_i - \frac{b_i c_i}{a_i}$$
which is what you had found before and not linear in $a_i$. This means that if there exists an $f$, linear in $a$, then each entry $f_i$ must be a function of some other entries of $a,b,c$.
However, since the RHS is a weighted sum of the $f_i$ with $a_i$ as weights, I don't see how there can be a general solution $f$ that depends linearly on $a$, unless you know something more about the vectors $a,b,c$.