6. (8th "Hope Cup" Invitational Competition Question) In the dihedral angle $\alpha-l-\beta$, there are lines $a$ and $b$ in the planes $\alpha$ and $\beta$ respectively, and they are not perpendicular to the edge $l$. Then ( ).
A. When the dihedral angle is a right dihedral angle, it is possible that $a / / b$, and it is also possible that $a \perp b$
B. When the dihedral angle is a right dihedral angle, it is possible that $a / / b$, but it is impossible that $a \perp b$
C. When the dihedral angle is not a right dihedral angle, it is possible that $a / / b$, but it is impossible that $a \perp b$
D. When the dihedral angle is not a right dihedral angle, it is impossible for $a / / b$, and it is also impossible for $a \perp b$
6. B. Reason: Clearly, $\mathrm{C}$ and $\mathrm{D}$ are incorrect. Suppose $\alpha-l-\beta$ is a right dihedral angle, take point $O$ on $l$, draw $O A / / a$ in $\alpha$, and draw $O B / / b$ in $\beta$. If $a \perp b$, then $O A \perp O B$. In $\beta$, draw $O B^{\prime} \perp l$, according to the problem, $O B$ and $O B^{\prime}$ do not coincide, thus, by $O B^{\prime} \perp \alpha$, we have $O B^{\prime} \perp O A$, hence $O A \perp \beta$, which implies $O A \perp l$, i.e., $a \perp l$, contradicting the given condition, thus $\mathrm{A}$ is incorrect.