## [ Coordinate method in space $\quad]$  [ Distance between skew lines ]

In a rectangular parallelepiped $A B C D A 1 B 1 C 1 D 1$, it is known that $A B=3, B C=2, C C 1=4$. On the edge $A B$, a point $M$ is taken such that $A M: M B=1: 2 ; K-$ is the point of intersection of the diagonals of the face $C C 1 D 1 D$. Find the angle and the distance between the lines $D 1 M$ and $B 1 K$.
Let's choose a coordinate system with the origin at point $D 1$. The $x$-axis is directed along the ray $D 1 C 1$, the $y$-axis along the ray $D 1 A 1$, and the $z$-axis along the ray $D 1 D$. Then the coordinates of the endpoints of segments $D 1 M$ and $B 1 K$ are:

$$
D 1(0 ; 0 ; 0), M(1 ; 2 ; 4), B 1(3 ; 2 ; 0), K\left(\frac{3}{2} ; 0 ; 2\right) .
$$

Let's find the coordinates of vectors $\overrightarrow{D_{1} M}$ and $\overrightarrow{B_{1} K}$:

$$
\overrightarrow{D_{1} M}=(1 ; 2 ; 4), \overrightarrow{B_{1} K}=\left(-\frac{3}{2} ;-2 ; 2\right)
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_0918ecac094e52bfe73bg-17.jpg?height=96&width=981&top_left_y=958&top_left_x=1)

$$
\cos \phi=\frac{\overrightarrow{D_{1} \vec{M}} \cdot \overrightarrow{B_{1} K}}{\left|\overrightarrow{D_{1} \mu A}\right| \cdot\left|\overrightarrow{B_{1} K}\right|}=\frac{-3-4+8}{\sqrt{(1+4+16)\left(9_{1}^{9}+4+4\right)}}=\frac{5}{\sqrt{21-41}}=\frac{5}{\sqrt{861}}
$$

If $\alpha$ is the angle between the lines $D 1 M$ and $B 1 K$, then

$$
\cos \alpha=|\cos \phi|=\frac{5}{\sqrt{861}}
$$

Let $\vec{n}=(a ; b ; c)$ be a vector perpendicular to the lines $D 1 M$ and $B 1 K$. Then $\vec{n} \cdot \overrightarrow{D_{1} M}=0$ and $\vec{n} \cdot \overrightarrow{B_{1} K}=0$, or

$$
\left\{\begin{array}{l}
a+2 b+4 c=0 \\
\frac{3}{2} a-2 b+2 c=0
\end{array}\right.
$$

By adding these equations term by term, we get $-\frac{1}{2} a+6 c=0$, or $a=12 c$. Let $c=1$. Then $a=12, b=\frac{1}{2}(-a-4 c)=-8$. Through point $D 1$, let's draw a plane perpendicular to the vector $\vec{n}=(12 ;-8 ; 1)$:

$$
12 x-8 y+z=0 .
$$

This plane passes through the line $D 1 M$ and is parallel to the line $B 1 K$, so the distance between the lines $D 1 M$ and $B 1 K$ is equal to the distance from any point on the line $B 1 K$ (for example, from point $B 1(3 ; 2 ; 0)$) to this plane.

If $\rho$ is the desired distance, then

$$
\rho=\frac{|12 \cdot 3-8 \cdot 2+0|}{\sqrt{12^{2}+8^{2}+1^{2}}}=\frac{20}{\sqrt{2008}}
$$

## Answer

$\arccos \frac{5}{\sqrt{861}}, \frac{20}{\sqrt{209}}$.