3. (11th "Hope Cup" Invitational Competition Question) In the cube $A B C D-A_{1} B_{1} C_{1} D_{1}$, $E$ and $F$ are the midpoints of $A B$ and $B B_{1}$, respectively. The angle formed by $A_{1} E$ and $C_{1} F$ is ( ).
A. $\arcsin \frac{\sqrt{21}}{5}$
B. $\arccos \frac{3}{5}$
C. $\frac{\pi}{4}$
D. $\frac{\pi}{3}$
3. A. Reason: Take a point $G$ on $A_{1} B_{1}$ such that $B_{1} G=\frac{1}{4} A_{1} B_{1}$, then $G F / / A_{1} E$, $\angle G F C_{1}$ is the angle formed by $A_{1} E$ and $C_{1} F$, $\cos \angle G F C_{1}=\frac{2}{5}$, then $\angle G F C_{1}=\arccos \frac{2}{5}=\arcsin \frac{\sqrt{21}}{5}$.