3. In tetrahedron $ABCD$, $\triangle ADB$ is an isosceles right triangle, $AD=1, \angle ADB=90^{\circ}, \angle BDC=$ $\angle ADC=60^{\circ}$. Then the distance between the skew lines $AB$ and $CD$ is $\qquad$
3. $\frac{1}{2}$.

As shown in Figure 2, take the midpoint $E$ of edge $AB$, and draw $EF \perp CD$ at point $F$.

From the given conditions, we know that line segment $EF$ is the desired segment, and
$$
\begin{array}{l}
\cos \angle CDE \\
=\frac{\cos \angle CDB}{\cos \angle EDB} \\
=\frac{\cos 60^{\circ}}{\cos 45^{\circ}}=\frac{\sqrt{2}}{2} .
\end{array}
$$

Thus, $\angle CDE=45^{\circ}$.
Also, $DE=\frac{\sqrt{2}}{2}$, so $EF=\frac{\sqrt{2}}{2} \sin \angle CDE=\frac{1}{2}$.