Director kinematics of a rod (${\mathbf{d}}_i^{\prime }=\mathbf{u}\times {\mathbf{d}}_i$)

For a differentiable material frame of a rod, there is a unique vector $\mathbf{u}$ such that ${\mathbf{d}}_i^{\prime }=\mathbf{u}\times {\mathbf{d}}_i$. To see this, form the orthogonal matrix $R=({\mathbf{d}}_1,{\mathbf{d}}_2,{\mathbf{d}}_3)$. Differentiation of $R{R}^T=I$ shows that $R^{\prime }{R}^T$ is a skew-symmetric matrix; every three-dimensional skew-symmetric matrix has the unique form $\mathbf{v}\mapsto \mathbf{u}\times \mathbf{v}$. Primes may denote differentiation with respect to reference or current arc length, but that choice must be stated. In a planar bend about ${\mathbf{d}}_2$, reference differentiation gives ${\mathbf{d}}_3^{\prime }=u{\mathbf{d}}_1$ and ${\mathbf{d}}_1^{\prime }=-u{\mathbf{d}}_3$.