Example 6 In the rectangular prism $A B C D-A_{1} B_{1} C_{1} D_{1}$, $A B=2, A A_{1}=A D=1$. Find the angle formed by $A B$ and the plane $A B_{1} C$.
Solution: Establish a spatial rectangular coordinate system as shown in Figure 7. Then
$$
\begin{array}{l}
D(0,0,0) \text {, } \\
A(1,0,0) \text {, } \\
C(0,2,0) \text {, } \\
B_{1}(1,2,1) .
\end{array}
$$

Thus, $A C=(-1,2,0), A B_{1}=(0,2,1)$.
Let the normal vector of plane $A B_{1} C$ be $\boldsymbol{n}=(x, y, z)$.
Since $\boldsymbol{n} \perp \boldsymbol{A C}, \boldsymbol{n} \perp \boldsymbol{A B _ { 1 }}$, we get
$-x+2 y=0,2 y+z=0$.
Let $x=2$, then $y=1, z=-2$.
Therefore, $\boldsymbol{n}=(2,1,-2)$.
Also, $A B=(0,2,0), A B$ forms an angle with plane $A B_{1} C$ as
$$
\begin{aligned}
\theta & =\arcsin \frac{|A B \cdot n|}{|n||A B|} \\
& =\arcsin \frac{|(0,2,0) \cdot(2,1,-2)|}{2 \times 3}=\arcsin \frac{1}{3} .
\end{aligned}
$$

Hence, the angle formed by $A B$ and plane $A_{1} B_{1} C$ is $\arcsin \frac{1}{3}$.