11. (15 points) Given that skew lines $a$ and $b$ form a $60^{\circ}$ angle, their common perpendicular segment is $E F$, and $|E F|=2$. The length of segment $A B$ is 4, with endpoints $A$ and $B$ moving on $a$ and $b$ respectively. Find the equation of the trajectory of the midpoint $P$ of segment $A B$.
11. Obviously, point $P$ lies in the plane $\alpha$ passing through the midpoint $O$ of $E F$ and parallel to $a$ and $b$.
As shown in Figure 2, let the projections of $a$ and $b$ in $\alpha$ be $a^{\prime}$ and $b^{\prime}$, and the projections of $A$ and $B$ in $\alpha$ be $A^{\prime}$ and $B^{\prime}$, respectively. Connect $O A^{\prime}$, $O B^{\prime}$, and $A^{\prime} B^{\prime}$. Then $P$ is the midpoint of $A^{\prime} B^{\prime}$, and $\angle A^{\prime} O B^{\prime}=60^{\circ}$.
From $A B=4$ and $|E F|=2$, we get $A^{\prime} B^{\prime}=2 \sqrt{3}$.
Establish a Cartesian coordinate system with the angle bisector of $\angle A^{\prime} O B^{\prime}$ as the $x$-axis and $O$ as the origin. Let $\left|O A^{\prime}\right|=m$ and $\left|O B^{\prime}\right|=n$.
In $\triangle A^{\prime} O B^{\prime}$, by the cosine rule we have
$$
m^{2}+n^{2}-m n=12 \text {. }
$$

Let $P(x, y)$. Then
$$
\left\{\begin{array}{l}
2 x = \frac{\sqrt{3}}{2} (m + n), \\
2 y = \frac{1}{2} (m - n)
\end{array} \Rightarrow \left\{\begin{array}{l}
m=\frac{2}{\sqrt{3}} x+2 y, \\
n=\frac{2}{\sqrt{3}} x-2 y .
\end{array}\right.\right.
$$

Substituting into equation (1) and simplifying, we get $\frac{x^{2}}{9}+y^{2}=1$.
This gives the arc of the ellipse within $\angle A^{\prime} O B^{\prime}$. Similarly, we can obtain the other three arcs.

In summary, the equation of the trajectory of point $P$ is $\frac{x^{2}}{9}+y^{2}=1$.