7. Given a regular quadrilateral pyramid $V-A B C D$ where the height equals half the length of $A B$, $M$ is the midpoint of the lateral edge $V B$, and $N$ is a point on the lateral edge $V D$ such that $D N=2 V N$, then the cosine value of the angle formed by the skew lines $A M, B N$ is $\qquad$ .
Answer: $\frac{\sqrt{11}}{11}$.
Solution: As shown in the figure, take the center $O$ of the base $ABCD$ as the origin of coordinates, and the directions of $\overline{AB}, \overline{BC}, \overline{OV}$ as the positive directions of the $x, y, z$ axes, respectively, to establish a spatial rectangular coordinate system. Without loss of generality, let $AB=2$, at this time the height $VO=1$, thus $A(-1,-1,0), B(1,-1,0), D(-1,1,0), V(0,0,1)$. According to the conditions, $M\left(\frac{1}{2},-\frac{1}{2}, \frac{1}{2}\right), N\left(-\frac{1}{3}, \frac{1}{3}, \frac{2}{3}\right)$, therefore
$$
\overline{AM}=\left(\frac{3}{2}, \frac{1}{2}, \frac{1}{2}\right), \overline{BN}=\left(-\frac{4}{3}, \frac{4}{3}, \frac{2}{3}\right) \text {. }
$$

Let the angle formed by the skew lines $AM, BN$ be $\theta$, then
$$
\cos \theta=\frac{|\overrightarrow{AM} \cdot \overline{BN}|}{|\overline{AM}| \cdot|\overline{BN}|}=\frac{|-1|}{\frac{\sqrt{11}}{2} \times 2}=\frac{\sqrt{11}}{11} .
$$