Let $\alpha, \beta$ $1$-forms in the open $U \subset \mathbb{R}^{3}$ such that $\alpha \wedge \beta \neq 0$ in every $x \in U$. If a form $\omega$, of degree $2$ in $U$ is such that $\omega \wedge \alpha = \omega \wedge \beta = 0$, prove that there is a function $f: U \to \mathbb{R}$ such that $\omega = f\cdot \alpha \wedge \beta$.
I'm trying to prove it without needing a lot of calculations (explicitly calculating the exterior products), but I'm not getting good ideas.
My attempt was to write $\omega = \phi \wedge \theta$ where $\phi, \theta$ are $1$-forms. I know that $\omega, \alpha$ and $\omega, \beta$ are LD, but $\alpha, \beta$ are LI so, it's intuitive to think that $\phi = g \alpha$ and $\theta = h \beta$. But, I cannot formalize this. Maybe it's just a little detail, but I'd like some help.
Fix a point $x$ and notice $\alpha(x)$ and $\beta(x)$ are linearly independent vectors of the cotangent space $T_x^\ast U$ because $\alpha(x)\wedge\beta(x) \neq 0$. Therefore, since $\dim T_x^\ast U =3$, there is $\gamma \in T_x^\ast U$ such that $\alpha(x),\beta(x), \gamma$ is basis of $T_x^\ast U$. With that in mind we can write $$\omega(x) = a\alpha(x) \wedge \beta(x)+b \beta(x) \wedge \gamma+ c\gamma \wedge \alpha(x)$$ with $a,b,c \in \mathbb R$.
Now, since $\omega(x)\wedge \alpha(x)=0$, multiplying the above identity by $\alpha(x)$ give us $$0=\omega(x)\wedge\alpha(x) =b \beta(x) \wedge \gamma \wedge \alpha(x)$$ and therefore $b=0$. Analogously, using $\omega(x)\wedge \beta(x) =0$ we obtain $c=0$. So, we have $$\omega(x) = a\alpha(x)\wedge \beta(x).$$
Therefore, for each $x$ there exists $f(x)\in \mathbb R$ such that $$\omega(x) = f(x)\alpha(x)\wedge \beta(x).$$
Notice that $f$ is also smooth. Fix $x\in U$. Since $\alpha\wedge \beta\neq 0$, there are vector fields $v,w$ such that $\alpha \wedge \beta (u(x),v(x)) \neq 0$. By continuity there is a neighborhood $W$ of $x$ such that $\alpha \wedge \beta (u(y),v(y)) \neq 0$ for all $y\in W$. Since $\omega$ and $\alpha\wedge \beta$ are differential forms and $$f(y) = \frac{\omega(u(y),v(y))}{\alpha \wedge \beta(u(y),v(y))}\quad \forall y \in V$$ we conclude $f$ is smooth.