6. The necessary and sufficient condition for the skew diagonals of two lateral faces of a regular triangular prism to be perpendicular to each other is: the ratio of the side length of its base to the side length of its lateral edge is ( ).
(A) $\sqrt{3}: 1$
(B) $1: \sqrt{3}$
(C) $\sqrt{2}: 1$
(D) $1: \sqrt{2}$
6.C.

As shown in Figure 4, let's assume the base edge length is $a>0$, and the side edge length is 1. According to the problem, we have
$$
\begin{array}{l}
B_{1} A \cdot B C_{1}=0 . \\
\text { Since } B_{1} A=B_{1} A_{1}+B_{1} B, \\
B C_{1}=B B_{1}+B_{1} C_{1}
\end{array}
$$

Then
$$
\begin{aligned}
& B_{1} \boldsymbol{A} \cdot B C_{1} \\
= & B_{1} A_{1} \cdot B B_{1}+ \\
& B_{1} B \cdot B B_{1}+B_{1} A_{1} \cdot B_{1} C_{1}+B_{1} B \cdot B_{1} C_{1} \\
= & -1+a^{2} \cos 60^{\circ}=-1+\frac{a^{2}}{2}=0 .
\end{aligned}
$$

Thus, $B_{1} A \perp B C_{1} \Leftrightarrow a=\sqrt{2}$.