11. (15 points) As shown in the figure, given that the skew lines $a, b$ 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 its endpoints $A, B$ moving on $a, b$ respectively. Find the equation of the trajectory of the midpoint $P$ of segment $A B$.


11. Solution: Clearly, point $P$ lies in the plane $\alpha$ passing through the midpoint $O$ of $E F$ and parallel to $a, b$. As shown in the figure, let the projections of $a, b$ in $\alpha$ be $a^{\prime}, b^{\prime}$, and the projections of $A, B$ in $\alpha$ be $A^{\prime}, B^{\prime}$. Connect $O A^{\prime}$, $O B^{\prime}$, $A^{\prime} B^{\prime}$. Then point $P$ is the midpoint of $A^{\prime} B^{\prime}$, and $\angle A^{\prime} O B^{\prime}=60^{\circ}$. From $A B=4,|E F|=2$, we get $A^{\prime} B^{\prime}=2 \sqrt{3}$. Establish a rectangular coordinate system with the angle bisector of $\angle A^{\prime} O B^{\prime}$ as the $x$-axis and $O$ as the origin. Let $\left|O A^{\prime}\right|=$ $m,\left|O B^{\prime}\right|=n$. Then, in $\triangle A^{\prime} O B^{\prime}$, by the cosine rule we have:
$$
\begin{array}{l}
m^{2}+n^{2}-m n=12 \text{. Let } P(x, y), \\
\text { then }\left\{\begin{array} { l } 
{ 2 x = \frac { \sqrt { 3 } } { 2 } ( m + n ) } \\
{ 2 y = \frac { 1 } { 2 } ( m - n ) }
\end{array} \text {, which gives } \left\{\begin{array}{l}
m=\frac{2}{\sqrt{3}} x+2 y \\
n=\frac{2}{\sqrt{3}} x-2 y
\end{array}\right.\right. \text {, substituting into the above equation }
\end{array}
$$

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 $\frac{x^{2}}{9}+y^{2}=1$.