1. For a regular tetrahedron $ABCD$ with edge length $1$, $G$ is the centroid of the base $ABC$, and point $M$ lies on the segment $DG$ such that $\angle AMB=90^{\circ}$. Then the length of $DM$ is $\qquad$.
1. $\frac{\sqrt{6}}{6}$.

As shown in Figure 2, take the midpoint $N$ of $AB$, then $MN=\frac{1}{2} AB=\frac{1}{2}, GN=\frac{1}{3} \times \frac{\sqrt{3}}{2}=\frac{\sqrt{3}}{6}$. In the right triangle $\triangle MGN$, we have
$$
MG=\sqrt{MN^{2}-GN^{2}}=\frac{\sqrt{6}}{6} .
$$

Thus, $DM=DG-MG=\frac{\sqrt{6}}{6}$.