Again I have a question, it's about a proof, if the torsion of a curve is zero, we have that
$$ B(s) = v_0,$$
a constant vector (where $B$ is the binormal), the proof ends concluding that the curve $$
\alpha \left( t \right)
$$
is such that $$
\alpha(t)\cdot v_0 = k
$$
and then the book says, "then the curve is contained in a plane orthogonal to $v_0$." It's a not so important detail but …. that angle might not be $0$, could be not perpendicular to it, anyway, geometrically I see it that $ V_0 $ "cuts" that plane with some angle.
My stupid question is why this constant $k$ must be $0$. Or just I can choose some $v_0 $ to get that "$k$"?
Best Answer
The constant $k$ need not be $0$; that would be the case where $\alpha$ lies in a plane through the origin. You have $k=\alpha(0)\cdot v_0$, so for all $t$, $(\alpha(t)-\alpha(0))\cdot v_0=0$. This means that $\alpha(t)-\alpha(0)$ lies in the plane through the origin perpendicular to $v_0$, so $\alpha(t)$ lies in the plane through $\alpha(0)$ perpendicular to $v_0$. (If $0$ is not in the domain, then $0$ could be replaced with any point in the domain of $\alpha$.)