I have seen the definition of Hodge dual, but I have never understood what it is all about.

If a define a linear form [itex]\omega:\mathbb{R}^3\to\mathbb{R}[/itex] with formula

[tex]

\omega = \omega_i e_i^T = \omega_1 e_1^T + \omega_2 e_2^T + \omega_3 e_3^T

[/tex]

then its Hodge dual with respect to the metric [itex]g=\textrm{id}_{3\times 3}[/itex] is

[tex]

*\omega = (*\omega)_{ij} e_i^T\otimes e_j^T

[/tex]

with coefficients

[tex]

((*\omega)_{ij})_{1\leq i,j,\leq 3} = \left(\begin{array}{ccc}

0 & \omega_3 & -\omega_2 \\

-\omega_3 & 0 & \omega_1 \\

\omega_2 & -\omega_1 & 0 \\

\end{array}\right)

[/tex]

So calculating the coefficients of the Hodge dual did not give me two vectors that would have spanned the orthogonal complement of given one dimensional space. But apparently it did give me a set of three vectors, which span the two dimensional orthogonal complement...