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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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i am asking abut the unit tangent and unit normal of the curve and not just the normal. –  Mathematics Jan 13 '12 at 16:17
The length of the normal vector does not affect whether it is orthogonal to the tangent vector or not. –  JavaMan Jan 13 '12 at 16:18
i know,but it is not enough to conclude that they are perpendicular –  Mathematics Jan 13 '12 at 16:24
@Mathematics Orthogonal = perpendicular. They mean the same thing. –  Alex Becker Jan 15 '12 at 3:46
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1 Answer

up vote 2 down vote accepted

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 ${\bf X}(t_0)$.

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