3. As shown in Figure 2, in the sector $A O B$, $\angle A O B=90^{\circ}$, $O A=2$, point $P$ is on the chord $A B$, and $A P=1$, point $Q$ is on $\overparen{A B}$, and $P Q / / O B$. Then $P Q=$ $\qquad$ .
3. $\frac{\sqrt{8 \sqrt{2}-2}-\sqrt{2}}{2}$.

As shown in Figure 4, extend $Q P$ to intersect $A O$ at point $D$, and connect $O Q$. Then
$$
\begin{array}{l}
A D=D P=\frac{1}{\sqrt{2}}, \\
D O=O A-A D \\
=2-\frac{\sqrt{2}}{2} .
\end{array}
$$

By the Pythagorean theorem, we get
$$
\begin{array}{l}
D Q=\sqrt{Q O^{2}-D O^{2}}=\sqrt{2 \sqrt{2}-\frac{1}{2}} \text {. } \\
\text { Therefore, } P Q=D Q-D P=\frac{\sqrt{8 \sqrt{2}-2}-\sqrt{2}}{2} \text {. }
\end{array}
$$