$i^*(\omega)$ vanish on S $\Leftrightarrow$ A is tangent to $S$ If $\omega\in \Omega^2(\mathbb{R}^3)$ is a 2-form: $\omega=f\;dx\wedge dy+g\;dx\wedge dz+h\;dy \wedge dz$, and $S$ is a surface in $\mathbb{R}^3$, prove that $i^*(\omega)$ vanish on $S$ $\iff$ $A=h\frac{\partial}{\partial x}-g\frac{\partial}{\partial y}+f\frac{\partial}{\partial z}$ is tangent to $S$.
The only clue I have been able to find is that there exists this isomorphism:
$$ \phi:\mathfrak{X}(\mathbb{R}^3)\longrightarrow \Omega^2(\mathbb{R}^3)$$
$$ A \longmapsto \phi(A),$$
where $\phi(A)(B,C)=(B\times C)\cdot A$. In this way $\phi(A)=h\;dx\wedge dy-g\;dx\wedge dz+f\;dy \wedge dz$, which is similar to what we are working about. Any idea?

Remark that if $\Omega$ is the determinant form on $\mathbb{R}^3$, $\Omega(A,..)=\omega$.
Suppose that the restriction of $\omega$ to $S$ is  zero and $A$ is not tangent to $S$, there exists $x\in S, u,v\in T_xS$ such that $A(x),u,v)$ is a basis of $\mathbb{R}^3$, this impllies that $\Omega(A(x),u,v)\neq 0$ contradiction.
On the other hand, suppose that $A(x)\in T_xS, x\in S$ we deduce that for every $u,v$ in $T_xS$, the dimension of $Vec(A(x),u,v)$ is at most 2 this implies that $\Omega(A(x),u,v)=\omega_x(u,v)=0$.
