Geometrical interpretation of dual basis in Euclidean 2-dim and 3-dim space In this Wikipedia entry there is an example and a formula to calculate the dual basis for a basis in $3-$dim Euclidean space. I copy here for convenience:


For example, the standard basis vectors of $R^2$ (the Cartesian plane) are
$\displaystyle \{\mathbf {e} _{1},\mathbf {e} _{2}\}=\left\{{\begin{pmatrix}1\\0\end{pmatrix}},{\begin{pmatrix}0\\1\end{pmatrix}}\right\}$
and the standard basis vectors of its dual space $R^{2*}$ are
$\displaystyle \{\mathbf {e} ^{1},\mathbf {e} ^{2}\}=\left\{{\begin{pmatrix}1&0\end{pmatrix}},{\begin{pmatrix}0&1\end{pmatrix}}\right\}{\text{.}}$
In 3-dimensional Euclidean space, for a given basis $\{e_1, e_2, e_3\}$, you can find the biorthogonal (dual) basis $\{e^1, e^2, e^3\}$ by formulas below:
$\displaystyle \mathbf {e}^{1}=\left({\frac {\mathbf {e}_{2}\times \mathbf {e} _{3}}{V}}\right)^{\text{T}},\ \mathbf {e}^{2}=\left({\frac {\mathbf {e} _{3}\times \mathbf {e}_{1}}{V}}\right)^{\text{T}},\ \mathbf {e} ^{3}=\left({\frac {\mathbf {e}_{1}\times \mathbf {e} _{2}}{V}}\right)^{\text{T}}.\tag 1$
where T denotes the transpose and
$\displaystyle V\,=\,\left(\mathbf {e} _{1};\mathbf {e} _{2};\mathbf {e} _{3}\right)\,=\,\mathbf {e} _{1}\cdot (\mathbf {e} _{2}\times \mathbf {e} _{3})\,=\,\mathbf {e} _{2}\cdot (\mathbf {e} _{3}\times \mathbf {e} _{1})\,=\,\mathbf {e} _{3}\cdot (\mathbf {e} _{1}\times \mathbf {e} _{2})$
is the volume of the parallelepiped formed by the basis vectors $\displaystyle \mathbf {e} _{1},\,\mathbf {e} _{2}$  and $\displaystyle \mathbf{e}_{3}.$


To avoid misunderstandings, Eq.$(1)$ says that the cross product of every combination of the basis vectors (for example $\mathbf e_2 \times \mathbf e_3$, which would yield a vector with magnitude equal to the surface of the parallelogram defined by the vectors $\mathbf e_2$ and $\mathbf e_3$), scaled down by the volume of the parallelepiped $V$ will result in a different basis vectors element for the dual space (after transposing).
If all this is true, then the question is,
What is the geometrical interpretation in the case of the standard basis both for 2D and 3D, as well as with any basis vectors in 3D? I imagine this could be a representation of the covectors of a basis in 3D...


This picture is in reference to a comment, and portrays the basis covectors from two views:

 A: Given the standard vector basis
$\displaystyle \{\mathbf {e} _{1},\mathbf {e} _{2}\}=\left\{{\begin{pmatrix}1\\0\end{pmatrix}},{\begin{pmatrix}0\\1\end{pmatrix}}\right\}$ we can interpret the dual basis 
$\displaystyle \{\mathbf {e} ^{1},\mathbf {e} ^{2}\}=\left\{{\begin{pmatrix}1&0\end{pmatrix}},{\begin{pmatrix}0&1\end{pmatrix}}\right\}{\text{.}}$
as two families of straight lines orthogonal to the vectors of the basis, as in this figure (from here). 

In this interpretation  the linear functional $e^1(\vec v)$ is a number that count the number of lines crossed by the vector.
In a similar fashion we ca interpret a dual basis and a linear functional in three dimensions, as you can see in the figure at this page of Wikipedia (note that a one-form is essentially the same as a linear functional).
I think that, looking at these pictures, you can interpret the definition of the dual basis that you have cited.

I use the term ''interpret'' because the elements of the dual space $V^*$ of a vector space $V$ are not the vectors of $V$, but the linear functionals on this space.  So  your picture is correctonly if we interpret the elements $e^i$ not as vectors, but as direction in which  the projections of vectors of $V$ are measured. This is the reason because I prefer to think at the linear functionals as a family of lines (or planes, in a $3D$ space), that are othogonal to the vectors $e^i$ in your figure.  Clearly the vector used in your figure identify exactly the plane of the linear form, so they identify  a basis if they are linearly independent. In other word, the the operation of transposition, from a vector $ v$ to $v^T$ is interpreted as the change from a vector that define a direction and a unit of mesure in such direction, to a set of planes orthogonal to such direction, whose distance is measured in the same unity (note that also the figure in the comment of @amd use a set of plane for the dual basis).
Last: this interpretation of the dual space is useful becausecanbe extended  from the 1-forms ( that are the same as the linear functionals) to n-forms as is suggested here (and you can see better reading the answers here)
