3. $A B C D-A_{1} B_{1} C_{1} D_{1}$ is a unit cube, two ants, one black and one white, both start from point $A$, and crawl forward along the edges. Each edge they walk is called "one segment". The white ant's route is $A A_{1} \rightarrow A_{1} D_{1} \rightarrow \cdots$, and the black ant's route is $A B \rightarrow B B_{1} \rightarrow \cdots$. They both follow the rule that the line of the $(i+2)$-th segment must be a skew line to the line of the $i$-th segment (where $i \in \mathbf{N}$). After both ants have completed the 2008th segment and stop at some vertex of the cube, the distance between the black and white ants is
A. 1
B. $\sqrt{2}$
C. $\sqrt{3}$
D. 0
3. B According to the conditions, list the route of the termite: $A A_{1} \rightarrow A_{1} D_{1} \rightarrow D_{1} C_{1} \rightarrow C_{1} C \rightarrow C B \rightarrow B A \rightarrow$ $A A_{1} \rightarrow \cdots$, it can be immediately observed that the termite returns to point $A$ after walking 6 segments. It can be verified that the black termite must return to the starting point after walking 6 segments, which means 6 segments form a cycle.
$2008=6 \times 334+4$, thus the problem is reduced to considering the position of the black termite after walking 4 segments. It is not difficult to calculate that the black termite is at point $D_{1}$ after walking 4 segments, and the termite is at point $C$. Therefore, the required distance is $\sqrt{2}$. Choose $\mathrm{B}$.