3. As shown in Figure 1, in the cube $A B C D-A_{1} B_{1} C_{1} D_{1}$, it is known that $M$ is the midpoint of $B B_{1}$. Then the cosine value of the dihedral angle $M-C D_{1}-A$ is ( ).
(A) $\frac{\sqrt{3}}{6}$
(B) $\frac{1}{2}$
(C) $\frac{\sqrt{3}}{3}$
(D) $\frac{\sqrt{6}}{3}$
3. C.

Taking $D$ as the origin, the lines $D A$, $D C$, and $D D_{1}$ as the $x$-axis, $y$-axis, and $z$-axis respectively to establish a spatial rectangular coordinate system. Then
$$
\begin{array}{l}
D(0,0,0), A(1,0,0), C(0,1,0), \\
D_{1}(0,0,1), M\left(1,1, \frac{1}{2}\right),
\end{array}
$$

and the normal vector of plane $A C D_{1}$ is $\boldsymbol{n}_{1}=(1,1,1)$, and the normal vector of plane $M C D_{1}$ is $\boldsymbol{n}_{2}=(-1,2,2)$.

Thus, $\cos \left\langle\boldsymbol{n}_{1}, \boldsymbol{n}_{2}\right\rangle=\frac{\sqrt{3}}{3}$, which means the cosine value of the dihedral angle $M-C D_{1}-A$ is $\frac{\sqrt{3}}{3}$.