Let $M$ be a manifold and let $p \mapsto F\_p$ be a foliation, i.e. each $p$ is a point of $M$ and each $F\_p \subset T\_p M$ is a linear subspace of $T\_p M$.


Fix a Riemannian metric on $M$. Is the family of perpendicular subspaces $p \mapsto F\_p^\perp$ a foliation of $M$ as well?



Perpendicular foliation
Take any non-integrable distribution $Q\subseteq TM$ of codimension $1$, for example the kernel of a [contact form](https://en.wikipedia.org/wiki/Contact_geometry) on a $3$-manifold. The distribution $F:=Q^{\perp}$ has dimension $1$. Now in every $1$-dimensional distribution the local sections are closed under Lie bracket (compute $[fX,gX]$ where $F=\langle X \rangle$ locally), so by Frobenius theorem $F$ is integrable, i.e. a foliation. 


$F^{\perp}=Q$ is then a non-integrable distribution.


