6. In a regular tetrahedron $ABCD$, the line segment connecting the midpoint of edge $AD$ and the centroid of face $BCD$ is $MN$, and the line segment connecting the midpoint of edge $CD$ and the centroid of face $ABC$ is $PQ$. Then the angle formed by $MN$ and $PQ$ is $\qquad$
6. $\arccos \frac{1}{18}$.

Let the edge length of a regular tetrahedron be $a$. On the line segment $AM$, intercept the line segment $ME = \frac{a}{3}$, and construct the height $DF$ from plane $BCD$ and the height $AF$ from plane $ABC$, as shown in Figure 1. Then, $FN = \frac{1}{3} FD$, $FQ = \frac{1}{3} FA$, $NQ \parallel AD$, and $NQ = \frac{1}{3} AD = \frac{a}{3}$.
Thus, $M N Q E$ is a parallelogram.
Therefore, $EQ \parallel MN$.
Also, $MN = PQ = \frac{a}{2}$. In $\triangle DPE$, $DP = \frac{a}{2}$, $DE = \frac{5}{6} a$, and $\angle PDE = 60^\circ$, so $EP^2 = \frac{19}{36} a^2$.
In $\triangle PQE$, by the cosine rule, we get
$$
\cos \angle PQE = -\frac{1}{18}.
$$

Hence, the angle between $PQ$ and $MN$ is $\arccos \frac{1}{18}$.