Position vector perpendicular to the derivative

Let $ a: I \rightarrow \mathbb{R}^3 $ , where $ I $ is an open interval of the real line endowed with the usual topology, be a smooth , parametrized curve that does not pass through the origin (that is the point $O(0, 0, 0) $ ) . If $ a(t_0) $ is the shortest to the origin point of the curve $ a$ and $ a'(t_0) \neq 0 $ , then prove that the position vector $ a(t_0) $ is perpendicural to $ a'(t_0) $.

Solution

According to the hypothesis we have that  

$$\left \| a(t) \right \|\geq \left \| a(t_0) \right \|>0 \implies \left \| a(t) \right \|^2\geq \left \| a(t_0) \right \|^2 >0 \;\; \forall t \in I$$

We observe that the real-valued function $||a||^2:I\rightarrow \mathbb{R}$ has a minimum value $t_0\in I$, where $I$ is an open interval. Also, this function is differentiable in $I$ with

$$\frac{\mathrm{d}}{\mathrm{d}t}\left \| a(t) \right \|^2=\frac{\mathrm{d}}{\mathrm{d}t}\langle a(t),a(t)\rangle=2\langle a(t),a'(t) \rangle, \; t \in I$$

So, according to Ferma's theorem, we have that

$$\left[\dfrac{\mathrm{d}}{\mathrm{d}t}\left \| a(t) \right \|^2\right]_{t=t_0}=0\implies \langle a(t_0),a'(t_0) \rangle=0$$

which means that the position vector of $a(t_0)$ is perpendicular to the position vector of $a'(t_0)$ .