Let $ \vec x(t)$ be a path of class $C^1$ that does not pass through the origin in $R^3$. If $\vec x(t_0)$ is the point on the image of $\vec x$ closest to the origin and $\vec x'(t_0) \ne 0$, show that the position vector $\vec x(t_0)$ is orthogonal to the velocity vector $\vec x'(t_0)$.
My goal is:$$ \vec x(t_0)\cdot \vec x'(t_0)=0$$
Since $\vec x(t_0)$ is the closest point to the origin we must have:$$2x(t_0)x'(t_0)+2y(t_0)y'(t_0)+2z(t_0)z'(t_0)=0$$ where $x,y,z$ are the vector components of $\vec x$.
The square from the distance from the origin is given by $$d(t)^2=\vec x(t)\cdot \vec x(t).$$ Since the path is $C^1$, the minimum will be attained when $$\frac{d}{dt}(d^2)(t_0)=0 \Rightarrow \frac{d}{dt}(\vec x\cdot \vec x)(t_0)=0 \Rightarrow2\vec x(t_0)\cdot \vec x'(t_0)=0.$$ So, the velocity is orthogonal to the position at $t_0$.