= Two-family inextensible-fibre pure stretch
{title2=$\alpha^2=\frac{\lambda\pm\sqrt{\lambda^2-\sin^22\theta}}{2\cos^2\theta}$}
For two distinct reference fibre directions $l_\pm=(\cos\theta,\pm\sin\theta,0)$, with $0<\theta<\pi/2$, consider the volume-preserving deformation with <right stretch tensor> $U=\operatorname{diag}(\lambda^{-1/2}\alpha,\lambda^{-1/2}\alpha^{-1},\lambda)$, with positive $\lambda,\alpha$. The <inextensible fibre constraint> gives $\alpha^2\cos^2\theta+\alpha^{-2}\sin^2\theta=\lambda$. Setting $b=\alpha^2$ gives a quadratic with the displayed roots. There are two distinct positive branches for $\lambda>\sin2\theta$. The smallest possible axial stretch is $\lambda=\sin2\theta$, at which $\alpha^2=\tan\theta$ and $(Ul_+)\cdot(Ul_-)=0$. Since both image fibres have unit length, their mutual area cannot exceed one; their area is $\sin2\theta/\lambda$, which gives the same lower bound. Coincident or antiparallel fibres at the angle endpoints do not give the two-family orthogonality conclusion.
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