6. In the quadrilateral pyramid $P-ABCD$, it is known that $AB // CD$, $AB \perp AD$, $AB=4$, $AD=2\sqrt{2}$, $CD=2$, $PA \perp$ plane $ABCD$, $PA=4$. Let $Q$ be a point on the line segment $PB$, and the sine of the angle formed by line $QC$ and plane $PAC$ is $\frac{\sqrt{3}}{3}$. Then
$\frac{PQ}{PB}$ is $\qquad$
6. $\frac{7}{12}$.

From $\frac{C D}{A D}=\frac{A D}{A B} \Rightarrow \triangle A C D \backsim \triangle D B A$
$\Rightarrow \angle D A C+\angle A D B=90^{\circ} \Rightarrow A C \perp B D$.
Since $P A \perp$ plane $A B C D \Rightarrow P A \perp B D$.
Therefore, $B D \perp$ plane $P A C$.
Let $B D$ intersect $A C$ at point $O$. Connect $P O$, and draw $Q H / / B O$, intersecting $P O$ at point $H$. Then $Q H \perp$ plane $P A C$.

Connect $H C$. Then $\angle Q C H$ is the angle between line $Q C$ and plane $P A C$.
$$
\text { Let } \frac{P Q}{P B}=\lambda(0 \leqslant \lambda \leqslant 1) \text {. }
$$

Then $Q H=\lambda B O=\frac{4 \sqrt{6}}{3} \lambda$.
In $\triangle P B C$, by the cosine rule,
$$
\cos \angle P B C=\frac{\sqrt{6}}{6}, Q B=(1-\lambda) 4 \sqrt{2} \text {. }
$$

In $\triangle Q B C$, by the cosine rule,
$$
Q C=\sqrt{32(1-\lambda)^{2}+16 \lambda-4} .
$$

In Rt $\triangle Q C H$,
$$
\frac{Q H}{C Q}=\frac{\sqrt{3}}{3} \text {. }
$$

Combining equations (1), (2), and (3), we get $\lambda=\frac{7}{12}$.