What are reciprocal vectors ? I did not undertand the definition in this question 
Two groups of vectors a, b, c, and a', b', c'
are called of reciprocal if a.a' = b.b' = c.c'
= 1 and whathever other mixed scalar product like a.b' = 0. Show that:

*

*c' = (a x b)/Q

*a' = (b x c)/Q

*b' = (c x a)/Q

Q = a . (b x c)

I understood that:

*

*a $\perp$ b $\perp$ c
and:

*

*a // a'

*b // b'

*c // c'

In this way, in three dimensions, we have that  b x c = a, so Q = 1. And it makes no sense, because this way would be a = a', b = b' and c = c'.
How to solve that ?

One way to do this is to show that

*

*The given formulas for $\mathbf{a}'$, $\mathbf{b}'$, $\mathbf{c}'$ satisfy the definition, and

*Show these reciprocal vectors are unique, i.e. if $\mathbf{a}''$, $\mathbf{b}''$, $\mathbf{c}''$ are an alternate set of reciprocal vectors to $\mathbf{a}$, $\mathbf{b}$, $\mathbf{c}$, then $\mathbf{a}' = \mathbf{a}''$, $\mathbf{b}' = \mathbf{b}''$, $\mathbf{c}' = \mathbf{c}''$.

The second step is important from a logical standpoint, but I have a feeling that whoever asked you the question was not really expecting you to complete it.

To complete the first step, recall the scalar triple product:
$$[\mathbf{p}, \mathbf{q}, \mathbf{r}] = (\mathbf{p} \times \mathbf{q}) \cdot \mathbf{r}.$$
We will need the following properties:

*

*$\displaystyle{[\mathbf{p}, \mathbf{q}, \mathbf{r}] = [\mathbf{q}, \mathbf{r}, \mathbf{p}] = [\mathbf{r}, \mathbf{p}, \mathbf{q}]}$

*If any two of $\mathbf{p}, \mathbf{q}, \mathbf{r}$ are the same, then $\displaystyle{[\mathbf{p}, \mathbf{q}, \mathbf{r}] = 0}$.

Then,
$$Q = [\mathbf{a}, \mathbf{b}, \mathbf{c}] = [\mathbf{b}, \mathbf{c}, \mathbf{a}] = [\mathbf{c}, \mathbf{a}, \mathbf{b}].$$
This means, for example,
$$\mathbf{c}' \cdot \mathbf{c} = \frac{\mathbf{a} \times \mathbf{b}}{Q} \cdot \mathbf{c} = \frac{[\mathbf{a}, \mathbf{b}, \mathbf{c}]}{Q} = 1,$$
and
$$\mathbf{a}' \cdot \mathbf{b} = \frac{\mathbf{b} \times \mathbf{c}}{Q} \cdot \mathbf{c} = \frac{[\mathbf{b}, \mathbf{c}, \mathbf{c}]}{Q} = 0.$$
The other $4$ calculations follow similarly.

To show uniqueness, we first show that $\mathbf{a}$, $\mathbf{b}$, $\mathbf{c}$ form a basis for $\Bbb{R}^3$. Since there are three such vectors, it suffices to show linear independence.
Suppose $d\mathbf{a} + e\mathbf{b} + f\mathbf{c} = \mathbf{0}$. Taking the inner product of both sides with, say, $\mathbf{a}'$, we get
\begin{align*}
&d(\mathbf{a} \cdot \mathbf{a}') + e(\mathbf{b} \cdot \mathbf{a}') + f(\mathbf{c} \cdot \mathbf{a}') = \mathbf{0} \cdot \mathbf{a}' \\
\implies &d \times 1 + e \times 0 + f \times 0 = 0 \\
&\implies d = 0.
\end{align*}
Taking dot products with $\mathbf{b}$ and $\mathbf{c}$ imply $e = 0$ and $f = 0$ respectively. Thus, $\mathbf{a}$, $\mathbf{b}$, $\mathbf{c}$ are linearly independent, and hence form a basis (in particular, they are spanning). One nice property is that, if a vector $\mathbf{v}$ is perpendicular to each of $\mathbf{a}$, $\mathbf{b}$, $\mathbf{c}$, then it must be $\mathbf{0}$.
Suppose $\mathbf{a}''$, $\mathbf{b}''$, $\mathbf{c}''$ and $\mathbf{a}'$, $\mathbf{b}'$, $\mathbf{c}'$ are both reciprocal to $\mathbf{a}$, $\mathbf{b}$, $\mathbf{c}$. Consider, for example, $\mathbf{v} = \mathbf{a}' - \mathbf{a}''$. Note that it is perpendicular to $\mathbf{a}$, as
$$\mathbf{a} \cdot \mathbf{v} = \mathbf{a} \cdot \mathbf{a}' - \mathbf{a} \cdot \mathbf{a}'' = 1 - 1 = 0,$$
and it is perpendicular to $\mathbf{b}$ (or similarly $\mathbf{c}$) as
$$\mathbf{b} \cdot \mathbf{v} = \mathbf{b} \cdot \mathbf{a}' - \mathbf{b} \cdot \mathbf{a}'' = 0 - 0 = 0.$$
Therefore, $\mathbf{a}' - \mathbf{a}'' = 0$, i.e. $\mathbf{a}' = \mathbf{a}''$. Similar arguments show $\mathbf{b}' = \mathbf{b}''$ and $\mathbf{c}' = \mathbf{c}''$ as well.
