I'm having trouble understanding this question.
If $r(s)$ is a unit speed curve with $\kappa > 0$ and it lies in its rectifying plane for all $s$, then $r \cdot T = s + c$ for some constant $c$ and $ r\cdot B$ is a non-zero constant and the torsion $\tau$ is non-zero.
So since it lies in the rectifying plane then $r$ is a linear combination of vectors in $B$ and $T$. So when you dot $r$ and $T$ does that mean it gives you the curve but at a different point? Someone, please help me understand!
Sure, you start by writing down $$r(s) = \alpha(s) T(s) + \beta(s)B(s)$$ for some scalar functions $\alpha, \beta$. Now differentiate both sides and apply the Frenet-Serret formulas: $$T(s) = \alpha'(s)T(s) + \beta'(s)B(s) + \alpha(s)\kappa(s)N(s) - \beta(s)\tau(s)N(s).$$ From this equation we extract $1 = \alpha'(s)$, $\beta'(s) = 0$, and $\alpha(s)\kappa(s) = \beta(s)\tau(s)$. Therefore $\alpha(s) = s + c$ for some constant $c$, $\beta(s)$ is a nonzero constant (since otherwise $\kappa=0$ in the third equation), and $\tau(s) = \kappa(s) (s+c)/\beta$ is nonzero (except at $s=-c$).
