In vector analysis, a vector-valued function $\mathbf{a}(t)$ doesn't change its direction iff $\mathbf{a}'(t) \times \mathbf{a}(t) = \mathbf{0}$ holds.
When I try to prove this, I use an extra assumption that $\mathbf{a}(t)$ vanishes nowhere. I wonder if it is necessary for a proof.
Here is my approach.
Suppose $\mathbf{a}'(t)\times\mathbf{a}(t)=\mathbf{0}$, then $\mathbf{e}(t) = \dfrac{{\mathbf{a}(t)}}{{|\mathbf{a}(t)|}}=\dfrac{{\mathbf{a}(t)}}{{f(t)}}$ is unit vector, where $f(t)=|\mathbf{a}(t)|$ vanishes nowhere. Only need to show $\mathbf{e}(t)$ is constant. Since $\mathbf{e}'(t)\cdot \mathbf{e}(t)=0$, $\mathbf{e}'(t)\cdot \mathbf{a}(t)=0$. It follows \begin{eqnarray*} \mathbf{a}(t)&=&f(t)\mathbf{e}(t),\\ \mathbf{a}'(t)&=&f'(t)\mathbf{e}(t)+f(t)\mathbf{e}'(t),\\ \mathbf{a}'(t)\times\mathbf{a}(t)&=&f(t)\mathbf{e}'(t)\times\mathbf{a}(t)=\mathbf{0}, \end{eqnarray*} Therefore $\mathbf{e}'(t)\times\mathbf{a}(t)=\mathbf{0}$. Set $\mathbf{e}'(t)=\lambda(t)\mathbf{a}(t)$, $$\mathbf{e}'(t)\cdot \mathbf{a}(t)=\lambda(t)\mathbf{a}(t)\cdot \mathbf{a}(t)=\lambda(t)f^2(t)=0,$$ It follows that $\lambda(t)=0$, $\mathbf{e}'(t)=0$, $\mathbf{e}(t)$ is constant.
Any help will be appreciated.