 Two-family inextensible-fibre pure stretch

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 $U=\mathrm{diag}({\lambda }^{-1/2}\alpha ,{\lambda }^{-1/2}{\alpha }^{-1},\lambda )$, 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 $\lambda >\sin 2\theta$. The smallest possible axial stretch is $\lambda =\sin 2\theta$, at which ${\alpha }^2=\tan \theta$ and $(U{l}_{+})\cdot (U{l}_{-})=0$. Since both image fibres have unit length, their mutual area cannot exceed one; their area is $\sin 2\theta /\lambda$, which gives the same lower bound. Coincident or antiparallel fibres at the angle endpoints do not give the two-family orthogonality conclusion.