# planar curve if and only if torsion

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 "$k$"?

-
Any plane with normal vector $\vec{v}_0$ satisfies an equation of the form $\vec{x}\cdot\vec{v}_0 = k$. The quantity $k$ basically measures how far the plane is from the origin - it doesn't have to be $0$. –  anon Aug 7 '11 at 21:47
Daniel, I attempted to edit your post so it was readable, but I had difficulty understanding your writing in some parts. You might want to check it - and make sure you communicate more clearly in the future. I'll leave figuring out a meaningful title up to you or someone else. –  anon Aug 7 '11 at 21:55

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 $\alpha$.)