I have the vector $r(t) = (\cos(t), \sin(t)) \ \forall \ t \in [0,\pi]$. The tangent vector to this vector is $r'(t) = (-\sin(t), \cos(t))$. What is the difference between the normal vector to $r(t)$, $(\sin(t), -\cos(t))$, and the normal vector to the tangent vector $r'(t)$, $(\cos(t), \sin(t))$?
It seems to me that there should be no difference since the tangent vector is parallel to the vector at any point, which therefore means that the normal vector is pointing in the same direction?
I would greatly appreciate it if people could please take the time to clarify any differences.
There is no such thing as a tangent vector to a vector. The tangent vector you give is the tangent vector to the curve $t\mapsto r(t)$. That is, the tangent vector at $t$ does not just depend on the value at $t$, but at the behaviour in a neighbourhood of $t$.
And it is not true that the tangent vector is parallel to the vector at any point. Quite the opposite: For this curve, both are always orthogonal to each other, as can be seen by calculating the dot product: $$r(t)\cdot r'(t)=(\cos t, \sin t)\cdot (-\sin t,\cos t) = -\cos t \sin t + \sin t \cos t = 0$$