3. As shown in Figure 1, in the spatial quadrilateral $ABCD$, $E$ and $F$ are points on $AB$ and $CD$ respectively, such that $\frac{AE}{EB}=\frac{DF}{FC}=2007$. Then $\boldsymbol{EF}=(\quad)$.
3.A.

Let $\frac{A E}{E B}=2007=a, \frac{F C}{D F}=\frac{1}{2007}=b$, then $a b=1$. Therefore, $\frac{1}{a+1}+\frac{1}{b+1}=\frac{1}{a+1}+\frac{1}{\frac{1}{a}+1}=\frac{1}{a+1}+\frac{a}{a+1}=1$.
Thus, $E F=E B+B C+C F$
$$
\begin{array}{l}
=\frac{A B}{a+1}+B C+\frac{C D}{b+1} \\
=\frac{A B}{a+1}+\left(\frac{B C}{a+1}+\frac{B C}{b+1}\right)+\frac{C D}{b+1} \\
=\frac{A B+B C}{a+1}+\frac{B C+C D}{b+1} \\
=\frac{A C}{a+1}+\frac{B D}{b+1}=\frac{A C}{a+1}+\frac{B D}{\frac{1}{a}+1} \\
=\frac{A C}{a+1}+\frac{a B D}{a+1}=\frac{A C+a B D}{a+1} \\
=\frac{A C+2007 B D}{2008} .
\end{array}
$$