= Director kinematics of a rod
{title2=$\mathbf d_i'=\mathbf u\times\mathbf d_i$}
For a differentiable <material frame of a rod>, there is a unique vector $\mathbf u$ such that $\mathbf d_i'=\mathbf u\times\mathbf d_i$. To see this, form the <orthogonal matrix> $R=(\mathbf d_1,\mathbf d_2,\mathbf d_3)$. Differentiation of $RR^T=I$ shows that $R'R^T$ 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 $\mathbf d_3'=u\mathbf d_1$ and $\mathbf d_1'=-u\mathbf d_3$.
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