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Let $M$ be a closed orientable Riemannian manifold. Recall that a plane field on a Riemannian manifold is said to be geodesic if any geodesic tangent to the plane field at one point is tangent to it at every point. Is it true that if $E\subset TM$ a one-dimensional geodesic plane field, then there exists $X\in \Gamma(E)$ such that $X$ is a Killing vector field? where $\Gamma$ means the section of a bundle.

How about $E$ has dimension $\ge 2$?

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(I assume by a plane field you mean a distribution.)

No, take $\mathbb{T}^3$ parametrized as $(x,y,z) \in [0,2\pi)^3$. The field $v = \partial_x + \sin(z) \partial_y$ is geodesic. But for any $\phi(x,y,z)$ the deformation tensor $\mathcal{L}_{\phi v} g$ can be computed to be $$ \begin{pmatrix} 2 \phi_x & \phi_y + \phi_x \sin(z) & \phi_z \newline \phi_y + \phi_x \sin(z) & 2\phi_y \sin(z) & \phi_z \sin(z) + \phi \cos(z) \newline \phi_z & \phi_z \sin(z) + \phi \cos(z) & 0 \end{pmatrix} $$ which can only vanish if $\phi_x = \phi_y = \phi_z = 0$ (from the (1,1), (2,2) and (1,3) components) and $\phi \equiv 0$ (from the (2,3) component).

For distributions of higher dimension, you can observe that for any complete Riemannian manifold $TM$ as a top-dimensional distribution is geodesic. So it suffices to find a Riemannian manifold of the appropriate dimension that does not admit any Killing vector field to get a counterexample. Alternatively you can also modify the above one dimensional construction in the obvious way.

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