Properties about unit tangent vector and unit normal vector Why is the unit tangent vector $T$ always perpendicular to the unit normal vector $N$?
$T=\frac{r'(t)}{|r'(t)|}$ and $N=\frac{T'(t)}{|T'(t)|}$
 A: The unit normal vector $\bf N$ is by definition ${\bf T}'\over \Vert {\bf T}'\Vert$. 
Since $\bf T$ has constant norm one, it is orthogonal to $\bf T'$: 
$${\bf T}\cdot {\bf T}'=0$$ 
(if $\bf X$ has constant norm, then $0={d\over dt}({\bf X}\cdot{\bf X })=
{\bf X}'\cdot{\bf X }+{\bf X}\cdot{\bf X }'  =2{\bf X}' \cdot{\bf X} $)$^\dagger$. 
This implies ${\bf T}\cdot {{\bf T}'\over \Vert {\bf T}'\Vert}=0$; thus $\bf N$ is normal to $\bf  T$.



$^\dagger$ This is the real "reason" why $\bf N$ is orthogonal to $\bf T$. To repeat myself: if ${\bf X} $ has constant norm, then ${\bf X} (t)$ is orthogonal to ${\bf X}'(t)$. Though the product rule for differentiating a dot product is a perfectly fine way to see this, I prefer the following "proof":

Suppose ${\bf X}(t)$ has constant norm $a$ and consider it to be a position vector. Then as $t$ varies, ${\bf X}(t)$ describes a path on the surface of a sphere of radius $a$. We know that ${\bf X'}(t_0)$ is tangent to the path traced out by ${\bf X}(t)$ at the point ${\bf X}(t_0)$; so, ${\bf X'}(t_0)$ is tangent to the aforementioned sphere  and, hence, orthogonal to ${\bf X}(t_0)$.
