8,9\end{aligned}
$$

Given a rectangular parallelepiped $A B C D A 1 B 1 C 1 D 1$, where $A B=4, A D=2, A A 1=6$. Point $N$ is the midpoint of edge $C D$, point $M$ is located on edge $C C 1$, such that $C 1 M: C M=1: 2$, and $K$ is the point of intersection of the diagonals of the face $A A 1 D 1 D$. Find the angle between the lines $K M$ and $A 1 N$.
Let $A D=x, A B=y, A A 1=z, \overrightarrow{A D}=\vec{x}, \overrightarrow{A B}=\vec{y}, \overrightarrow{A A_{1}}=\vec{z}, x=2, y=4, z=6$. Then

$$
\overrightarrow{K M}=\frac{1}{2} \vec{x}+\vec{y}+\frac{1}{6} \vec{z}, \overrightarrow{A_{1} N}=\vec{x}+\frac{1}{2} \vec{y}-\vec{z}
$$

If $\alpha$ is the angle between the vectors $\overrightarrow{K M}$ and $\overrightarrow{A_{1} N}$, then

![](https://cdn.mathpix.com/cropped/2024_05_06_cdc67dfeb844af03d94fg-11.jpg?height=137&width=806&top_left_y=500&top_left_x=632)

$$
\begin{aligned}
& =\frac{\frac{1}{2} \vec{x} \cdot \vec{x}+\frac{1}{2} \vec{y} \cdot \vec{y}-\frac{1}{6} \cdot \vec{z} \cdot \vec{z}}{\sqrt{\left(\frac{x^{2}}{1}+y^{2}+\frac{z^{2}}{3}\right)\left(x^{2}+\frac{x^{2}}{1}+z^{2}\right)}}= \\
& =\frac{2+5-6}{\sqrt{(1+16+1)(4+4+36)}}=\frac{4}{\sqrt{18 \cdot 44}}=\frac{\sqrt{22}}{33}
\end{aligned}
$$

Let $\phi$ be the angle between the lines $K M$ and $A 1 N$. Then

$$
\cos \phi=|\cos \alpha|=\frac{\sqrt{22}}{33}
$$

Therefore, $\alpha=\arccos \frac{\sqrt{22}}{33}$.

## Answer

$\arccos \frac{\sqrt{22}}{33}$