5. In a regular tetrahedron $S-ABC$, the edge length is $a$, and $E$, $F$ are the midpoints of $SA$ and $BC$ respectively. Find the angle formed by the skew lines $BE$ and $SF$.

In a regular tetrahedron $S-ABC$, the edge length is $a$, and $E$, $F$ are the midpoints of $SA$ and $BC$ respectively. Find the angle formed by the skew lines $BE$ and $SF$.
5. $\overrightarrow{B E} \cdot \overrightarrow{S F}=(\overrightarrow{B S}+\overrightarrow{S E}) \cdot(\overrightarrow{S B}+\overrightarrow{B F})=\overrightarrow{B S} \cdot \overrightarrow{S B}+\overrightarrow{B S} \cdot \overrightarrow{B F}+\overrightarrow{S E} \cdot \overrightarrow{S B}+\overrightarrow{S E} \cdot \overrightarrow{B F}$,

Since $\overrightarrow{B S} \cdot \overrightarrow{S B}=-a^{2}, \overrightarrow{B S} \cdot \overrightarrow{B F}=\frac{1}{2} a^{2} \cos 60^{\circ}=\frac{1}{4} a^{2}, \overrightarrow{S E} \cdot \overrightarrow{S B}=\frac{1}{2} a^{2} \cos 60^{\circ}=\frac{1}{4} a^{2}$,
$\overrightarrow{S E} \cdot \overrightarrow{B F}=\frac{1}{2} \overrightarrow{S A} \cdot \frac{1}{2} \overrightarrow{B C}=\frac{1}{4} \overrightarrow{S A} \cdot(\overrightarrow{A C}-\overrightarrow{A B})=\frac{1}{4}(\overrightarrow{S A} \cdot \overrightarrow{A C}-\overrightarrow{S A} \cdot \overrightarrow{A B})=0$
Therefore, $\overrightarrow{B E} \cdot \overrightarrow{S F}=-a^{2}+\frac{1}{4} a^{2}+\frac{1}{4} a^{2}=-\frac{1}{2} a^{2}$,
Thus, $\cos \langle\overrightarrow{B E}, \overrightarrow{S F}\rangle=\frac{\overrightarrow{B E} \cdot \overrightarrow{S F}}{|\overrightarrow{B E}||\overrightarrow{S F}|}=\frac{-\frac{1}{2} a^{2}}{\left(\frac{\sqrt{3}}{2} a\right)^{2}}=-\frac{2}{3}$,
Therefore, the angle between the skew lines $B E$ and $S F$ is $\arccos \frac{2}{3}$.