Is it necessary for a normal vector to be perpendicular to the tangent vector?

Question

Is it necessary for a normal vector to be perpendicular to the tangent vector? I am confused whether there is a problem, especially when I deal with parametric equation for parabola. $$ \begin{align} r(t)=\langle 2t,t^2+1 \rangle \end{align} $$ $$ \begin{align} T(t)&=\frac{r'(t)}{|r'(t)|}=\frac{1}{|r'(t)|}\langle 2,2t \rangle\\ N(t)&=\frac{T'(t)}{|T'(t)|}=\frac{1}{|T'(t)|}\langle 0,2 \rangle\\ T(t)\cdot N(t)&=\frac{1}{|r'(t)|}\langle 2,2t \rangle\cdot \frac{1}{|T'(t)|}\langle 0,2 \rangle\\&=\frac{1}{|r'(t)||T'(t)|}(0+4t)\\&=\frac{4t}{|r'(t)||T'(t)|}\neq 0 \end{align} $$

Answer 1

Your problem is the magnitudes do matter here--they are not constant in terms of $t$. For example: $$r'(t) = \langle 2, 2t \rangle \implies |r'(t)| = \sqrt{4 + 4t^2}$$

So $$T(t) = \left\langle \frac{1}{\sqrt{1 + t^2}}, \frac{t}{\sqrt{1 + t^2}} \right\rangle \implies T'(t) = \left\langle -\frac{t}{\left(1 + t^2\right)^\frac{3}{2}}, \frac{1}{(1 + t^2)^{\frac{3}{2}}} \right\rangle$$

The magnitude of $T'(t)$ doesn't matter since we are not taking another derivative. We can see $T(t) \cdot T'(t) = 0$.