18. (9th "Hope Cup" Invitational Competition Question) Given a tetrahedron $ABCD$ where $AD=BC=1$, $E$ and $F$ are points on $AB$ and $CD$ respectively, and $BE:EA=CF:FD=1:2, EF=a(a>0)$, then the angle $\theta$ formed by $AD$ and $BC$ is $\qquad$.
18. When $\frac{1}{3}<a \leqslant \frac{\sqrt{5}}{3}$, $\theta=\arccos \frac{5-9 a^{2}}{4}$; when $\frac{\sqrt{5}}{3}<a<1$, $\theta=\pi-\arccos \frac{5-9 a^{2}}{4}$. Reason: Draw $E G / / B C$ intersecting $A C$ at $G$, and connect $G F$. It is easy to see that $G F / / A D$, and $E G=\frac{2}{3}, G F=\frac{1}{3}$. Thus, $\angle E G F$ is the angle formed by $A D$ and $B C$. We have $\cos \angle E G F=\frac{\left(\frac{2}{3}\right)^{2}+\left(\frac{1}{3}\right)^{2}-a^{2}}{2 \cdot \frac{2}{3} \cdot \frac{1}{3}}=\frac{5-9 a^{2}}{4}$. Clearly, it must be that $\left|\frac{5-9 a^{2}}{4}\right|<1$, yielding $\frac{1}{3}<a<1$.