[ Coordinate method in space [ Distance between skew lines ]

The base of the pyramid $A B C S$ is an equilateral triangle $A B C$ with a side length of $4 \sqrt{2}$. The lateral edge $S C$ is perpendicular to the base plane and equals 2. Find the angle and the distance between the skew lines, one of which passes through point $S$ and the midpoint of edge $B C$, and the other passes through point $C$ and the midpoint of edge $A B$.
Let $M$ and $K$ be the midpoints of edges $B C$ and $A B$ respectively, and $P$ be the projection of point $M$ onto the line passing through vertex $C$ and parallel to $A B$. We choose a coordinate system with the origin at point $C$. The $x$-axis is directed along the ray $C P$, the $y$-axis along the ray $C K$, and the $z$-axis along the ray $C S$. Then the coordinates of the endpoints of segments $S M$ and $C K$ are:

$$
S(0 ; 0 ; 2), M(\sqrt{2} ; \sqrt{6} ; 0), C(0 ; 0 ; 0), K(0 ; 2 \sqrt{6} ; 0) .
$$

We find the coordinates of vectors $\overrightarrow{S M}$ and $\overrightarrow{C K}$:

$$
\overrightarrow{S M}=(\sqrt{2} ; \sqrt{6} ;-2), \overrightarrow{C K}=(0 ; 2 \sqrt{6} ; 0)
$$

Let $\phi$ be the angle between vectors $\overrightarrow{S M}$ and $\overrightarrow{C K}$. Then

$$
\left.\cos \phi=\frac{\overrightarrow{S M} \overrightarrow{C K}}{|\overrightarrow{S M}| \cdot|\cdot| C K} \right\rvert\,=\frac{12}{\sqrt{12 \cdot 24}}=\frac{1}{\sqrt{2}}
$$

If $\alpha$ is the angle between lines $S M$ and $C K$, then

$$
\cos \alpha=|\cos \phi|=\frac{1}{\sqrt{2}}
$$

Therefore, $\alpha=45^{\circ}$. Line $C K$ is parallel to line $M P$, so line $C K$ is parallel to the plane $S M P$. The equation of this plane is

$$
\frac{x}{C P}+\frac{z}{C S}=1
$$

(the equation of the plane in intercept form), or

$$
\frac{x}{\sqrt{2}}+\frac{z}{2}-1=0
$$

The distance between lines $S M$ and $C K$ is equal to the distance from any point on line $C K$ (for example, from point $C(0 ; 0 ; 0)$) to this plane. If $\rho$ is the desired distance, then

$$
\rho=\left|\sqrt{\left(\frac{1}{\sqrt{2}}\right)^{2}+\left(\frac{1}{2}\right)^{2}}\right|=\frac{2}{\sqrt{3}}
$$

## Answer

$45^{\circ}, \frac{2}{\sqrt{3}}$.