Equivalence of two 1-form defining the same hyperplane field on a manifold. I have tried my best to explain my question.
Suppose M is n dimensional smooth manifold and $\xi$ is a smooth hyperplane field define on it. In addition, M has two 1-form define on it, $\alpha_1$ and $\alpha_2$ such that kernel($\alpha_1$)=kernel($\alpha_2$)= $\xi$.
Since $\xi$ can also be thought as smooth subbundle of the smooth tangent bundle, we can (pointwise) factor  the tangent bundle $TM = \xi \oplus \xi^{\bot}$. Therefore, these 1-form has action (effective) on the 1-dimensional bundle $\xi^{\bot}$. At a point $p \in M$ one can associate a real values depending on $p$ such that ${\alpha_1}_p = \lambda(p) {\alpha_2}_p$ and since p is arbitrary choosen. We define a non-zero real valued function $\lambda: M \to \mathbb{R}-\{0\}$.
Now I dont know, how to prove that this function $\lambda$ is smooth.

You can find a section $X$ of $\xi^{\perp}$ on an open subset which contains $U$. The function $f_i(x)=\alpha_i(X(x))$ is differentiable and not zero ${{f_1}\over{f_2}}$ is differentiable.
