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?
Take any non-integrable distribution $Q\subseteq TM$ of codimension $1$, for example the kernel of a contact form 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.
Consider $S^3$ endowed with the Hopf fibration. The fibres of the fibration are periodic orbits of a flow. Take any metric on $S^3$ and consider the distribution orthogonal to the fibres of the Hopf fibration it is not integrable. If it was integrable, you would have a Reeb component (Novikov) and any flow transverse to the Reeb component has non periodic orbits since an orbit which meets the boundary torus cannot leave the Reeb component and is not closed. For more details see:
Sue Goodman
Vector fields with transverse foliations
Topolopy Vol. 24. No. 3. pp. 333-340. 1985.
