If you have an equation $$r(t)=\langle 1,\cos(t),\sin(t)\rangle$$ then $$r'(t)=\langle 0,-\sin(t),\cos(t)\rangle $$ and $$\|r'(t)\|=\sqrt{\sin^2(t)+\cos^2(t)}=1$$ thus $T(t)=\langle 0,-\sin(t),\cos(t)\rangle$.
Intuitively, it makes sense that finding the rate of change of $T(t)$, aka $T'(t)$ should, would always be normal to the $T(t)$. This is simply explained that as the magnitude of $r'(t)$ never changes, the acceleration must always only be acting to change direction hence being perpendicular.
However, if you have an equation
$$r(t)=\langle t,2t^2,t^2/2\rangle$$ then $$r'(t)=\langle 1,4t,t\rangle$$ and $$\|r'(t)\|=\sqrt{1+17t^2}$$
In this case, the magnitude of $r'(t)$ is changing so why should the rate of change of $T(t)$ be perpendicular to the path of $T(t)$ and $r'(t)$? In physical scenarios, the only circumstances which acceleration is perpendicular to velocity are where the velocity magnitude is unchanging.