exists a 1-form given exterior 2-form, 1-form on $3$-dimensional space?

Question

Let $\alpha$ be an exterior $2$-form, and $\beta$ is a $1$-form on a $3$-dimensional space. Suppose that $\alpha \wedge \beta = 0$. How do I go about showing there exists a $1$-form $\gamma$ such that $\alpha = \beta \wedge \gamma$?

Answer 1

$($Assume $\beta \neq 0$.$)$ Let $\alpha = a_1x_2 \wedge x_3 + a_2x_3 \wedge x_1 + a_3x_1 \wedge x_2$ and $\beta = b_1x_1 + b_2x_2 + b_3x_3$. Then $\alpha \wedge \beta = (a_1b_1 + a_2b_2 + a_3b_3)\, x_1 \wedge x_2 \wedge x_3$ so$$a \wedge \beta = 0 \text{ } \Longleftrightarrow \text{ } \langle a, b\rangle = 0.$$This is equivalent to $a = b \times c$ for some $c \in \mathbb{R}^3$, i.e.$$a_1 = b_2c_3 - b_3c_2,\text{ }a_2 = b_3c_1 - b_1c_3, \text{ }a_3 = b_1c_2 - b_2c_1$$which says exactly that$$\alpha = (b_1x_1 + b_2x_2 + b_3x_3)\wedge(c_1x_1 + c_2x_2 + c_3x_3).$$

Answer 2

Let us choose/denote the basis 1-forms to be $x_1=\beta \neq 0$, $x_2$, and $x_3$; and write $\alpha = a_1x_2 \wedge x_3 + a_2x_3 \wedge x_1 + a_3x_1 \wedge x_2$. As $\alpha\wedge x_1=0$, we must have $a_1=0$ allowing us to rewrite $$ \alpha = a_2x_3 \wedge x_1 + a_3x_1 \wedge x_2 = x_1 \wedge (a_3x_2-a_2x_3). $$