Why is the unit normal of plane curves defined to be rotated? Let $\gamma: \mathbb R \to \mathbb R^2$ be a regular smooth curve given as $\gamma (t) = (x(t), y(t))$ moving at unit speed. 
Then the unit normal $N$ is defined to be 
$$ N(t) = (-y'(t), x'(t))$$
that is, the "usual" unit normal,
$$ N(t) = \gamma^{''}(t)/\|\gamma(t)^{''}\|$$
rotated by ${\pi \over 2}$.
I don't understand why the case $n=2$ needs this different definition. 

Please could someone explain to me why we have to rotate the normal
  vector if $n=2$?

 A: The direction of the normal to a curve is given by the component of the second derivative of the position which is orthogonal to the tangent.
The planar case is simpler in that the curve torsion is always null, so that the normal is just the perpendicular to the tangent and can be computed from the first derivative instead.

In the 3D formula for the normal,
$$N=\frac{\dot r\times(\ddot r\times\dot r)}{||\dot r||\ ||\ddot r\times\dot r||},$$
the product $\ddot r\times\dot r$ is parallel to the $z$ axis and perpendicular to $\dot r$, so that a nice simplification is possible
$$N=\frac{\dot r\times u_z}{||\dot r||}.$$
This is indeed the rotated tangent vector.
A: Since $x'^2+y'^2=1$, we have that $x'x''+y'y''=0$. Thus, $(x'',y'')\perp(x'y')$. Since $n=2$, we can rotate by $\frac\pi2$ to get
$$
(-y',x')\parallel(x'',y'')
$$
and
$$
\|(-y',x')\|=1
$$
Therefore,
$$
(-y',x')=\pm\frac{(x'',y'')}{\|(x'',y'')\|}
$$
In $\mathbb{R}^2$ there is only one direction perpendicular to the tangent. In higher dimensions, the situation is not so simple.
