# The Darboux vector is defined by $D = \tau T + \kappa B$. Show that $T' = D \times T$

The Darboux Vector is defined as $D = \tau T + \kappa B$. Show that for a unit speed curve

$$T' = D \times T \hspace{1cm} ...$$

Here, the $...$ represents the fact that there are a few more equations I need to show, but I think with the hints for this first one, I can do the rest. For a space curve $\gamma$, $T$ is the unit tangent, i.e $T = \frac{\gamma'}{|\gamma'|}$. If $P$ is the principal normal, i.e $P = \frac{T'}{|T'|}$, then $B$ is the Binormal, i.e $B = T \times P$. $\tau$ is the torsion and $\kappa$ is the curvature.

I know that as its unit speed, $|\gamma ' | = 1$. The method I want to show is that, I know from the Serret-Frenet equations that $T' = \kappa P$, where $P$ is the principal normal, i.e $P = \frac{T'}{|T'|}$. So I want to show that $D \times T = \kappa P$, but the bit I am stuck on is how to do the cross product of $(\tau T + \kappa B) \times \gamma'$.

I could use the fact that $T = P \times B$, but even then, I wouldn't know how to do $D \times (P \times B)$.

Can someone give me some tips please. Thank you

$$(\tau T + \kappa B) \times T= \kappa B \times T = \kappa P$$
• How can I tell the directions of these vectors like this? Or is it just the general rule that $T$ and $\gamma'$ are in the same direction due to the fact that they are in a SF basis? – Kaish Feb 13 '13 at 23:05
• You can tell by the way you define $T$. It is just $\gamma^{\prime}$ divided by its lenght. So $T$ is by definition a unit vector in the direction of $\gamma^{\prime}$. – PAD Feb 13 '13 at 23:12