3. As shown in Figure 2, in tetrahedron $ABCD$, $\triangle ADB$ is an isosceles right triangle, $\angle ADB=$ $90^{\circ}, AD=1$, and $\angle BDC=$ $\angle ADC=60^{\circ}$. Then the distance between the skew lines $AB$ and $CD$ is
3. $\frac{1}{2}$.
$$
\begin{array}{l} 
A B \cdot C D=(A D+D B) \cdot C D=A D \cdot C D+D B \cdot C D \\
=|A D| \cdot|C D| \cos \langle A D, C D\rangle+ \\
|D B| \cdot|C D| \cos \langle D B, C D\rangle \\
= C D \cos 60^{\circ}+C D \cos 120^{\circ}=0 .
\end{array}
$$

Thus, $\boldsymbol{A B} \perp \boldsymbol{C D}$. Therefore, $\sin \langle\boldsymbol{A B}, \boldsymbol{C D}\rangle=1$.
Let the distance between the skew lines $A B$ and $C D$ be $d$.
Since $A D=D B=1, A B=\sqrt{2}$, then
$$
V_{D-A B C}=\frac{1}{6} A B \cdot D C \cdot d \sin \langle A B, D C\rangle=\frac{\sqrt{2}}{6} D C \cdot d \text {. }
$$

And $V_{D-A B C}=\frac{1}{6} D A \cdot D B \cdot D C \sin 60^{\circ} \cdot \sin 60^{\circ} \cdot \sin \alpha$, where $\alpha$ is the plane angle of the dihedral angle $A-C D-B$,
$$
\cos \alpha=\frac{\cos \angle A D B-\cos \angle A D C \cdot \cos \angle B D C}{\sin \angle A D C \cdot \sin \angle B D C}=-\frac{1}{3} \text {. }
$$

Thus, $\sin \alpha=\frac{2 \sqrt{2}}{3}$.
Then $\frac{\sqrt{2}}{6} D C \cdot d=\frac{1}{6} \times 1 \times 1 \times D C \times \frac{\sqrt{3}}{2} \times \frac{\sqrt{3}}{2} \times \frac{2 \sqrt{2}}{3}$. Solving for $d$ gives $d=\frac{1}{2}$.