Example 21 (2003 National College Entrance Examination Question) As shown in Figure 8-11, in the right triangular prism $A B C-A_{1} B_{1} C_{1}$, the base is an isosceles right triangle, $\angle A C B=90^{\circ}$, the lateral edge $A A_{1}=2$, $D$ and $E$ are the midpoints of $C C_{1}$ and $A_{1} B$ respectively, and the projection of point $E$ on plane $A B D$ is $G$, which is the centroid of $\triangle A B D$. Find the angle between $A_{1} B$ and plane $A B D$.
Solve: As shown in Figure 8-11, establish a coordinate system. Since $E G \perp$ plane $A B D$, then $\angle E B G$ is the required angle, and $\overrightarrow{G E}$ is the normal vector of plane $A B D$.

Let $A C=a$, then $A(a, 0,0)$, $B(0, a, 0)$, $D(0,0,1)$, $E\left(\frac{a}{2}, \frac{a}{2}, 1\right)$, $G\left(\frac{a}{3}, \frac{a}{3}, \frac{1}{3}\right)$, hence $\overrightarrow{G E}=\left(\frac{a}{6}, \frac{a}{6}, \frac{2}{3}\right)$, $\overrightarrow{A D}=(-a, 0,1)$. From $\overrightarrow{G E} \cdot \overrightarrow{A D}=0$ we get $a=2$, then the normal vector of plane $A B D$ is $\overrightarrow{G E}=\left(\frac{1}{3}, \frac{1}{3}, \frac{2}{3}\right)$. And $\overrightarrow{B E}=(1, -1,1)$, so $\sin \angle E B G=\frac{|\overrightarrow{G E} \cdot \overrightarrow{B E}|}{|\overrightarrow{G E}||\overrightarrow{B E}|}=\frac{\sqrt{2}}{3}$, thus the required angle is $\arcsin \frac{\sqrt{2}}{3}$.