Dot product equality of three different vectors Given three vectors $A$, $B$ and $C$ of n dimension. Also $a_i$, $b_i$ and $c_i$ are complex non zero values.
Can $A \cdot B$ = $C \cdot B$ and $A \neq C$ ?
On others words can $(A-C) \cdot B = 0$, being $ A \neq C $ ?
If this is answered somewhere or is too silly, please, point me out the reference since I wasn't able to find it.
Thanks in advance!
Edit: I made a typo on the first equation.
 A: An example:
$$\begin{pmatrix} 1\\0\\0 \end{pmatrix}.\begin{pmatrix} 0\\1\\0 \end{pmatrix} = 0$$
$$\begin{pmatrix} 1\\0\\0 \end{pmatrix}.\begin{pmatrix} 0\\0\\1 \end{pmatrix} = 0$$
Therefore the answer is yes.

What you probably want to know is if you can solve equations involving dot products like:
$$\mathbf{x}.\mathbf{a} = b$$
where $\mathbf{x}$ is an unknown vector, $\mathbf{a}$ is a known vector,  and $b$ is a constant.
Eg. $$\mathbf{x}.\begin{pmatrix} 1\\2\\3 \end{pmatrix} = 2$$
With an operation like multiplication, you can find a unique solution for equations in that form. For example, $3x = 5$ has a unique solution for $x$.
This isn't necessarily true for dot products. In fact, if $\mathbf{x}$ and $\mathbf{a}$ are 3 dimensional vectors, then there are a whole plane of solutions for $\mathbf{x}$ which is why that is one of the standard notations for the equation of a plane.
Vector equation of a plane in 3 dimensions:
$$\mathbf{r}.\mathbf{n} = D$$
where $\mathbf{r}$ is a variable vector, $\mathbf{n}$ is a constant vector and $D$ is a constant.
A: Yes. 
$$(A-C)\cdot B = 0$$
Just means that $A-C$ and $B$ are orthogonal ("perpendicular"), which can definitely happen. For instance, let $A = [2,2,1]$, $C = [1,1,3]$, and $B = [1,1,1]$. 
A: For any vector $\mathbf{D}$,
$$\mathbf{C} =
\mathbf{A\times D}+\frac{(\mathbf{A\cdot B})\mathbf{A}}{A^2}
\implies \mathbf{A\cdot C} = \mathbf{A\cdot B}$$
