Example 18 (1991 National College Entrance Examination) Given that the side length of square $ABCD$ is $4, OG \perp$ plane $ABCD, OG=2, E, F$ are the midpoints of $AB, AD$ respectively, find the distance from point $B$ to plane $GEF$.
Solve: As shown in Figure 8-8, establish a spatial rectangular coordinate system. Then, $B(0,4,0)$, $E(2,4,0)$, $F(4,2,0)$, $G(0,0,2)$. The vectors are $\overrightarrow{E F}=(2,-2,0)$, $\overrightarrow{G E}=(2,4,-2)$, $\overrightarrow{B E}=(2,0,0)$. A normal vector to the plane $E F G$ is $\vec{n}=\left(y_{1} z_{2}-y_{2} z_{1}, z_{1} x_{2}-z_{2} x_{1}, x_{1} y_{2}-x_{2} y_{1}\right)=(4,4,12)$. For simpler calculation, take $\vec{n}=(1,1,3)$. Then, $\overrightarrow{B E} \cos \theta=\frac{\vec{n} \cdot \overrightarrow{B E}}{|\vec{n}|}=\frac{2 \sqrt{11}}{11}$. Therefore, the distance from point $B$ to the plane $E F G$ is $\frac{2 \sqrt{11}}{11}$.