1.6. Find the distance between two intersecting medians of the faces of a regular tetrahedron with edge 1. (Consider all possible arrangements of the medians.)

## § 2. Angles between lines and planes
1.6. Let $A B C D$ be a given regular tetrahedron, $K$ - the midpoint of $A B$, $M$ - the midpoint of $A C$. Consider the projection onto the plane $\mathrm{b}$, perpendicular to the face $A B C$ and passing through the line $A B$. Let $D_{1}$ be the projection of vertex $D$, $M_{1}$ - the projection of point $M$, i.e., the midpoint of segment $A K$. The distance between the lines $C K$ and $D M$ is equal to the distance from point $K$ to the line $D_{1} M_{1}$. In the right-angled triangle $D_{1} M_{1} K$, the leg $K M_{1}$ is equal to 1/4, and the leg $D_{1} M_{1}$ is equal to the height of the tetrahedron $A B C D$, i.e., it is equal to
$\sqrt{2 / 3}$. Therefore, the hypotenuse is

$\sqrt{35 / 48}$, and thus the desired distance is $\sqrt{2 / 35}$.

If $N$ is the midpoint of edge $C D$, then to find the distance between the medians $C K$ and $B N$, we can consider the projection onto the same plane as in the previous case. Let $N_{1}$ be the projection of point $N$, i.e., the midpoint of segment $D_{1} K$. In the right-angled triangle $B N_{1} K$, the leg $K B$ is equal to $1 / 2$, and the leg $K N_{1}$ is equal to $\sqrt{1 / 6}$. Therefore, the hypotenuse is $\sqrt{5 / 12}$, and the desired distance is $\sqrt{1 / 10}$.