How to prove that exist one scalar function $f$ such that $\omega=f (\eta\wedge \theta)$. The question is: let $\eta,\theta\in \Omega^1(A)$ where $A\subset \mathbb R^3$ open set, where $\eta\wedge \theta(x)\neq 0 $ for all $x\in A$. If $\omega$ is two form that satisfy $\omega\wedge \eta= \omega\wedge \theta=0$. Show that exist $f:A\to \mathbb R$ such that $\omega=f(\eta\wedge\theta)$.
We try the next: choose $B_i=\{x\in \mathbb R^3: \eta_i\wedge\theta_i(x)\neq 0\}$. Hence $\eta\wedge \theta(x)\neq 0 $ then $R^3= \cup_{i=1}^3 B_i$ so if I will defined $f_i= \frac{\omega_i}{\eta_i\wedge\theta_i}$ and maybe this work? but i dont use the hypothesis $\omega\wedge \eta= \omega\wedge \theta=0$.
the other side i thought too in to do this: like $\omega\wedge \eta= \omega\wedge \theta=0$ multiply with $wedge$ product in both side in the above equality then $\omega\wedge \eta\wedge \theta= \omega\wedge \theta\wedge \theta=0$ so when I will applicated $x\in A$ we have
$\omega \wedge (\eta\wedge\theta)(x)=0$ but $\eta\wedge\theta(x)\neq 0$ so using properties of the wedge product we have $\omega =c (\eta\wedge\theta)$. In this case maybe $f=c$. Is possible? I have two ideas but I not sure if both are right, or maybe just one, or maybe need other side, somebody can help me please? Thank you so much.
This is true more generally than in $\Bbb R^3$. An approach you should consider is to start with the linear algebra, thinking just at a point. Then you can make this argument work locally on a manifold (or on an open subset of $\Bbb R^n$) and check smoothness.
If $\eta(p)$ and $\theta(p)$ are linearly independent elements of $\Bbb R^{n*}$ (so, $1$-forms at a point), then you can extend to a basis. Then, knowing what a basis for $\Lambda^2(\Bbb R^{n*})$ is, you write $\omega(p)$ as a linear combination of basis elements and deduce from $\omega(p)\wedge\eta(p)\wedge\theta(p)=0$ that $\omega(p)$ must be a scalar multiple of $\eta(p)\wedge\theta(p)$.
(Notationally, if will probably be easier to write $\eta=\theta_1$ and $\theta=\theta_2$, extend to a basis $\theta_1,\dots,\theta_n$, and then recall that $\theta_i\wedge\theta_j$, with $i<j$, form a basis for $\Lambda^2(\Bbb R^{n*})$. Then you write $\omega = \sum\limits_{i<j}c_{ij}\theta_i\wedge\theta_j$. Then show that only $c_{12}\ne 0$; for example, to see $c_{13}=0$, consider $0=\omega\wedge\theta_2 = -c_{13}\theta_1\wedge\theta_2\wedge\theta_3 + \dots$, and use linear independence of $\theta_i\wedge\theta_j\wedge\theta_k$ for $i<j<k$.)