3. Given in the tetrahedron $D-ABC$, $\angle ACB=\angle ABD=90^{\circ}$, $CA=CB$, $\angle BAD=30^{\circ}$. If the projection of point $C$ on the plane $ABD$ is exactly on $AD$, then the sine value of the dihedral angle $C-AB-D$ is $\qquad$.
Answer
As shown in the figure, $C E \perp$ plane $A B D, C A=C B$
$\Rightarrow E A=E B \Rightarrow E$ is the midpoint of $A D$,
Draw $E F \perp A B \Rightarrow F$ is the midpoint of $A B$
$\Rightarrow \angle C F E$ is the plane angle of the dihedral angle $C-A B-D$.
Let $E A=E B=1$, then $E F$
$=\frac{1}{2}, A B=\sqrt{3}$, thus $C F=\frac{\sqrt{3}}{2}, C E=\frac{\sqrt{2}}{2} \Rightarrow \sin \angle C F E=\frac{\sqrt{6}}{3}$.
Therefore, the sine value of the plane angle of the dihedral angle $C-A B-D$ is $\frac{\sqrt{6}}{3}$.