6 For a cube $A B C D-A_{1} B_{1} C_{1} D_{1}$ with edge length 1, take point $M$ on the side diagonal $A_{1} D$ and point $N$ on $C D_{1}$, such that line segment $M N$ is parallel to the diagonal face $A_{1} A C C_{1}$. The minimum length of such $M N$ is ( ).
(A) $\frac{\sqrt{3}}{3}$
(B) $\frac{\sqrt{2}}{2}$
(C) 1
(D) $\sqrt{2}$
$6 \mathrm{~A}$ Hint: Draw $M M_{1} \perp A D$ at $M_{1}$, $N N_{1} \perp D C$ at $N_{1}$, it is easy to prove $M_{1} N_{1} / / A C$. Let $D M_{1}=D N_{1}=x$, then $M M_{1}=x, N N_{1}=$ $1-x$. Draw $M H \perp N N_{1}$ at $H, N H=$ $1-2 x, M_{1} N_{1}=\sqrt{2} x$. By the Pythagorean theorem
$$
\begin{aligned}
M N^{2} & =(\sqrt{2} x)^{2}+(1-2 x)^{2} \\
& =6\left(x-\frac{1}{3}\right)^{2}+\frac{1}{3} .
\end{aligned}
$$

When $x=\frac{1}{3}$, $M N$ reaches its minimum value $\frac{\sqrt{3}}{3}$.