(5) As shown in the figure, $A B C D-A_{1} B_{1} C_{1} D_{1}$ is a cube, and $P-A_{1} B_{1} C_{1} D_{1}$ is a regular quadrilateral pyramid. The distance from $P$ to the plane $A B C$ is $\frac{3}{2} A B$. Then, the cosine value of the angle between the skew lines $A_{1} P$ and $B C_{1}$ is ( ).
(A) $\frac{\sqrt{3}}{3}$
(B) $\frac{\sqrt{6}}{3}$
(C) $\frac{\sqrt{6}}{6}$
(D) $\frac{\sqrt{3}}{6}$
(5) B Hint: Connect $A C_{1}$, it is easy to prove that $A C_{1} / / A_{1} P$, thus $\angle B C_{1} A$ is the angle formed by $A_{1} P$ and $B C_{1}$. In the right triangle $\triangle A B C_{1}$, $\cos \angle B C_{1} A=\frac{B C_{1}}{A C_{1}}=$ $\frac{\sqrt{2}}{\sqrt{3}}=\frac{\sqrt{6}}{3}$. Therefore, the answer is B.
（Changqi No.2 High School, Liu Anning provided the problem）