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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)|}

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    @Mathematics Orthogonal = perpendicular. They mean the same thing.2012-01-15

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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 ${\bf X}(t_0)$.