Q:

Acceleration & the non-uniqueness of the unit normal vector to a curve at a point

Knowing the equation of a curve in three-dimensional space in the following parametric form $$x=x(u), y=y(u), z=z(u),$$ we can determine the tangent vector $\hat{T}$ at a point $(x,y,z)$ using the formula $$\hat{T}(u)=\frac{\vec{r}^\prime(u)}{\sqrt{x^{\prime2}(u)+y^{\prime2}(u)+z^{\prime2}(u)}}.$$ Having obtained $\hat{T}(u)$, a fair task is to obtain a unit normal vector $\hat{N}(u)$ at that point. But the unit vector $\hat{N}(u)$ does not seem to be unique. The direction of $\hat{N}(u)$ can have any direction on the plane transverse to $\hat{T}(u)$. This troubles me! Let me say why.
From kinematics, the instantaneous acceleration of a particle moving along a trajectory $\vec{r}(t)$ are respectively given by $$\vec{a}(t)=\dot{v}(t)\hat{T}(t)+\kappa v^2(t)\hat{N}(t)$$ which tells that acceleration will not be unique unless $\hat{N}(t)$ is uniquely determined. 

But how is this possible that the acceleration to a given trajectory is not unique? What am I missing?

A:

$\hat N(t)$ refers to the principal normal; the normal in the direction $\dfrac{d\hat T}{dt}$. More precisely,
$$
\hat N(t)=\frac{\frac{d\hat T}{dt}}{\left\lVert\frac{d\hat T}{dt}\right\rVert}
$$

