# A space curve with non-vanishing curvature is planar iff its torsion is 0

Intuitively this is simple and to prove the backwards direction: $\tau = 0 \Rightarrow \mathbf{b'}=0 \Rightarrow \mathbf{b}$ is constant.

Then letting $\gamma$ be the parametrisation of the curve and $s$ its arc-length, $\frac{d}{ds} (\gamma \cdot \mathbf{b}) = \dot{\gamma} \cdot \mathbf{b} + \gamma \cdot \mathbf{b'} = \dot{\gamma} \cdot \mathbf{b} = \mathbf{t} \cdot \mathbf{b} = 0$

Hence $\gamma \cdot \mathbf{b}$ is constant and so the space curve is contained within the plane.

But how could I go about showing the opposite direction?

• The opposite direction is the "easy" direction. :) At worst, run your argument backward. – Andrew D. Hwang May 1 '16 at 16:06
• Interesting, thank you! Out of curiosity, what would be at best? I still think I'm missing something obvious – user336138 May 1 '16 at 20:18
• One approach is to let $\nu$ denote a unit normal to the plane of $\gamma$, so that $(\gamma(t) - \gamma(t_{0})) \cdot \nu = 0$ for all $t$. Differentiating twice tells you the binormal of $\gamma$ is $\pm \nu$, hence constant. – Andrew D. Hwang May 1 '16 at 20:47