In the tetrahedron $A B C D$, $B D=D C=C B=\sqrt{2}, A C=\sqrt{3}, A B=A D=1, M$ is the midpoint of $C D$. Then the cosine of the angle formed by the skew lines $B M$ and $A C$ is 
(A) $\frac{\sqrt{2}}{3}$
(B) $\frac{\sqrt{2}}{2}$
(C) $\frac{\sqrt{3}}{3}$
(D) $\frac{\sqrt{6}}{3}$
4. (A).

Given $B A \perp A D$, take the midpoint $E$ of $A D$, then $M E / / A C$, so the angle formed by $A C$ and $B M$ is $\angle B M E$. From the given, $B M=\frac{\sqrt{3}}{2} B C=\frac{\sqrt{6}}{2}, M E=\frac{\sqrt{3}}{2}, B E=\sqrt{1+\left(\frac{1}{2}\right)^{2}}=\frac{\sqrt{5}}{2}$, thus $\cos \angle B M E$ $-\frac{B M^{2}+M E^{2}-B E^{\prime}}{2 B M \cdot M E}=\frac{\sqrt{2}}{3}$.