OK I got confused really bad yesterday. The set $ \{\xi \in N_{t}\gamma \mid A_{\xi}=0 \}$ is indeed one dimensional. Without loss of generality, assume $\gamma$ be unit-speed, and $\overline{D}_{t}\dot{\gamma}$ never zero. Then $\overline{D}_{t}\dot{\gamma}(t)$ is a non-zero vector in the normal space $N_{t}\gamma$ of $\gamma$ at $t$. Choose a smooth orthonormal frame $(\xi_{1}, \dotsc, \xi_{N-1})$ for the normal bundle $N\gamma$ such that $\xi_{1} = \overline{D}_{t}\dot{\gamma}$. Then $\overline{D}_{t}\xi_{k} \cdot \dot{\gamma} = 0$ for all $k = 2, \dotsc, N-1$, and the claim follows.

OK I got confused really bad yesterday. The set $ \{\xi \in N_{t}\gamma \mid A_{\xi}=0 \}$ is indeed $(N-1)$-dimensional. Without loss of generality, assume $\gamma$ be unit-speed, and $\overline{D}_{t}\dot{\gamma}$ never zero. Then $\overline{D}_{t}\dot{\gamma}(t)$ is a non-zero vector in the normal space $N_{t}\gamma$ of $\gamma$ at $t$. Choose a smooth orthonormal frame $(\xi_{1}, \dotsc, \xi_{N-1})$ for the normal bundle $N\gamma$ such that $\xi_{1} = \overline{D}_{t}\dot{\gamma}$. Then $\overline{D}_{t}\xi_{k} \cdot \dot{\gamma} = 0$ for all $k = 2, \dotsc, N-1$, and the claim follows.