A quote within the question reciprocal vectors reads:
$$ {\bf A} = \frac{{\bf b}\times {\bf c}}{{\bf a}\cdot{\bf b}\times
{\bf c}}, $$plus cyclic permutations, are said to be reciprocal vectors.
To the OP's comment
…but I'm still not sure what they mean by reciprocal vectors
found under this answer, and interesting reply by† @KimJongUn says:
It's just a name. Looking at ${\bf A} = \frac{{\bf b}\times {\bf c}}{{\bf a}\cdot{\bf b}\times {\bf c}},$, if you "cancel" $\bf b\times \bf c$, then you are left with $\bf A=\frac{1}{\bf a}$, the "reciprocal" of 𝐚. Of course, you can't legitimately carry out this cancellation but it does help to explain the name.
Question: How much like reciprocals are reciprocal vectors? Is there a matrix division that allows $\mathbf{A} = 1 / \mathbf{a}$ in three or two dimensions?
I think we can write it like that at least for 1D, but I'm not sure if there are issues with the sign.
†okay maybe it's not them
Best Answer
The reciprocal vectors are defined if the three given vectors are linearly independent, because otherwise $\mathbf{a}\cdot\mathbf{b}\times\mathbf{c}=0$ (and also for all permutations). If the three vectors are linearly independent, then $\mathbf{b}\times\mathbf{c}$ is orthogonal to both $\mathbf{b}$ and $\mathbf{c}$, so it cannot be orthogonal also to $\mathbf{a}$.
If we take the three coordinate vectors $\mathbf{i},\mathbf{j},\mathbf{k}$, the first of the reciprocal vectors is $$ \dfrac{\mathbf{j}\times\mathbf{k}}{\mathbf{i}\cdot\mathbf{j}\times\mathbf{k}}=\mathbf{i} $$ and similarly you get $\mathbf{j}$ and $\mathbf{k}$ for the other two.
More generally, if you take the scalar products of the first reciprocal vector with the vectors $\mathbf{a},\mathbf{b},\mathbf{c}$ you get $$ \frac{\mathbf{b}\times\mathbf{c}}{\mathbf{a}\cdot\mathbf{b}\times\mathbf{c}}\cdot\mathbf{a}=1, \quad \frac{\mathbf{b}\times\mathbf{c}}{\mathbf{a}\cdot\mathbf{b}\times\mathbf{c}}\cdot\mathbf{b}=0, \quad \frac{\mathbf{b}\times\mathbf{c}}{\mathbf{a}\cdot\mathbf{b}\times\mathbf{c}}\cdot\mathbf{c}=0 $$ and similarly for the cyclic permutations. So the set of reciprocal vectors is nothing else than the dual basis with respect to the identification of the dual space with $\mathbb{R}^3$ itself by means of the scalar product.