The dot product of two perpendicular vectors is not zero

Question

The radius vector for a helix is $$\mathbf{r}(t)=a\cos(t)\ \mathbf{i} + a \sin(t)\ \mathbf{j} +c\ t\ \mathbf{k}$$ and it's tangent vector is $$\mathbf{r}'(t)=-a\sin(t)\ \mathbf{i} + a \cos(t)\ \mathbf{j} +c\ \mathbf{k}.$$ They should be perpendicular to each other hence their dot product should be zero. But their dot product is $c^2t$. So they are not perpendicular unless $t=0$. How can it be?

Answer 1

The velocity is everywhere perpendicular to the position if and only if $$ r'(t) \cdot r(t) = 0 $$ $$ 2r'(t) \cdot r(t) = 0 \implies r'(t)\cdot r(t) + r(t) \cdot r'(t) = 0$$ $$ \frac{d}{dt} (r(t) \cdot r(t)) = 0 \implies ||r(t)||^2 = c$$ so the length of the position vector must be constant i.e. the motion is on a circle (in $\mathbb{R}^2$) or a sphere (in $\mathbb{R}^3$). The helix doesn't lie on a sphere, so it's not true that the velocity is perpendicular to the position. So if you had circular motion in $\mathbb{R}^2$, then yes it would be perpendicular, but your motion is in $\mathbb{R}^3$.