Geodesics on a sphere lie on a plane I have a question concerning closed geodesics on a sphere. I know that non-constant closed geodesics on a sphere are great circles. Hence if choosing one the image lies on a plane. This is easily seen if we know that closed geodesics are great circle. But can one show this without knowing about this? 
The geodesic equation is given by $\ddot\gamma \bot T_\gamma S^2$, which yiels also that we have
$$
\ddot \gamma = |\dot \gamma|^2 \gamma.
$$
From the above equations we find that $\gamma \cdot \dot\gamma = 0$ and $\ddot \gamma \cdot \dot\gamma =0$. If we can show that $\dot\gamma  \times \ddot\gamma =0$ then we're done (right?), as the torsion would be $0$.
Could you help me out with this a litte? Maybe my thought are also completely wrong. If you have another idea to show that if one chooses a closed goedesic on $S^2$ then it lies on a plane with the constraint that one can't use that it is a great circle, I would be very happy. 
Thanks in advance!

It suffices to show that $\dot{\gamma}$ is normal to a constant vector $n$. Then it has to move in a plane orthogonal to $n$. The vector $n=\gamma\times \dot{\gamma}$ is normal to $\dot{\gamma}$ and  $\dot{n}=\gamma\times \ddot{\gamma}= 0$ by the geodesic equation.
