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 $\ge 2$? 



Is geodesic plane field a Killing field?
(I assume by a plane field you mean a [distribution](http://en.wikipedia.org/wiki/Distribution_%28differential_geometry%29).)


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. 


