(5) In a regular tetrahedron $ABCD$, $M$ and $N$ are the midpoints of $BC$ and $DA$ respectively. The cosine of the angle formed by the lines $AM$ and $BN$ is ( ).
(A) $\frac{1}{3}$
(B) $\frac{1}{2}$
(C) $\frac{2}{3}$
(D) $\frac{3}{4}$
(5) C Hint: Suppose the edge length of the regular tetrahedron is 1, then
$$
\begin{array}{ll}
\overrightarrow{A B} \cdot \overrightarrow{A B}=1, & \overrightarrow{A B} \cdot \overrightarrow{D B}=\frac{1}{2}, \\
\overrightarrow{A C} \cdot \overrightarrow{A B}=\frac{1}{2}, & \overrightarrow{A C} \cdot \overrightarrow{D B}=0 .
\end{array}
$$

Now $\overrightarrow{A M}=\frac{1}{2}(\overrightarrow{A B}+\overrightarrow{A C}), \overrightarrow{N B}=\frac{1}{2}(\overrightarrow{A B}+\overrightarrow{D B})$, so
$$
\overrightarrow{A M} \cdot \overrightarrow{N B}=\frac{1}{4}(\overrightarrow{A B}+\overrightarrow{A C}) \cdot(\overrightarrow{A B}+\overrightarrow{D B})=\frac{1}{2} \text {. }
$$

And $|\overrightarrow{A M}|=|\overrightarrow{N B}|=\frac{\sqrt{3}}{2}$, so the cosine value of the angle formed by $A M$ and $B N$ is
$$
\frac{\overrightarrow{A M} \cdot \overrightarrow{N B}}{|\overrightarrow{A M}||| \overrightarrow{N B} \mid}=\frac{2}{3} \text {. }
$$