Given two skew lines $a, b$ forming an angle $\theta$, the length of their common perpendicular $A^{\prime} A$ is $d$. On lines $a, b$, take points $E, F$ respectively, with $A^{\prime} E=m, A F=n$. Find $E F$. ($A^{\prime}$ is on line $a \perp, A$ is on line $\left.b \perp\right)$
$\sqrt{d^{2}+m^{2}+n^{2} \pm 2 m n \cos \theta}$
II. [Analysis and Solution] Let the plane $\alpha$ pass through $b$ and be parallel to $a$, and let the plane $\beta$ pass through $a$ and $A A^{\prime}$. Let $\alpha \cap \beta=c$, then $c \parallel a$. Therefore, the angle between $b$ and $c$ is $\theta$, and $A A^{\prime} \perp c$. Since $A A^{\prime} \perp b$, it follows that $A A^{\prime} \perp \alpha$.
According to the theorem for determining perpendicular planes, $\beta \perp \alpha$.
In the plane $\beta$, construct $E G \perp c$, then $E G \perp \alpha$.
Connect $F G$, then $E G \perp F G$. In the right triangle $F E G$, $E F^{2}=E G^{2}+F G^{2}$.
Since $A G=m$, in $\triangle A F G$, $F G^{2}=m^{2}+n^{2}-2 m n \cos \theta$.
Since $E G^{2}=A A^{\prime 2}=d^{2}$, it follows that $E F^{2}=d^{2}+m^{2}+n^{2}-2 m n \cos \theta$.
If point $F$ (or $E$) is on the other side of point $A$ (or $A^{\prime}$),
then $E F^{2}=d^{2}+m^{2}+n^{2}+2 m n \cos \theta$. Therefore, $E F=\sqrt{d^{2}+m^{2}+n^{2} \pm 2 m n \cos \theta}$.