Example 1 See Figure $15-1$, after joining a set of triangular boards, let the triangular board $BCD$ stand up along $BC$, making the planes of the two triangular boards perpendicular to each other. Find the angle formed by the skew lines $AD$ and $BC$.
Given $\angle B A C=\angle B C D=90^{\circ}, \angle C B A=45^{\circ}$ and $\angle B D C=60^{\circ}$
Taking the midpoint $O$ of $B C$ as the origin, establish a spatial rectangular coordinate system, and let $O C=1$
Then point $C(0,1,0)$, point $A(1,0,0)$, point $D\left(0,1, \frac{2}{\sqrt{3}}\right)$
So $\overrightarrow{A D}=\left(0,1, \frac{2}{\sqrt{3}}\right)-(1,0,0)=\left(-1,1, \frac{2}{\sqrt{3}}\right)$
$\begin{aligned} \overrightarrow{A D} \cdot \overrightarrow{O C} & =1 \\ \text { So } \cos \langle \overrightarrow{A D}, \overrightarrow{O C}\rangle & =\frac{\overrightarrow{A D} \cdot \overrightarrow{O C}}{|\overrightarrow{A D}| \cdot|\overrightarrow{O C}|}=\frac{1}{\sqrt{(-1)^{2}+1^{2}+\left(\frac{2}{\sqrt{3}}\right)^{2}}}=\sqrt{\frac{3}{10}}\end{aligned}$ Therefore, the angle formed by the skew lines $A D$ and $B C$ is $\arccos \sqrt{\frac{3}{10}}$