11. The ellipse $\frac{x^{2}}{2002^{2}}+\frac{y^{2}}{1949^{2}}=1$ intersects with two lines passing through the origin at points $A, B, C, D$. Among these, the slope of line $AB$ is $k$, and the slope of line $CD$ is $-\frac{1949^{2}}{2002^{2} \cdot k}$. If there exists a $\triangle PQR$ such that $PQ=OA, PR=OC, \angle QPR=\left|\angle AOC-\frac{\pi}{2}\right|$, then $QR-$ $\qquad$ .
11. 53

Since $AB$ and $CD$ are a pair of conjugate diameters, according to Apollonius's theorem for the ellipse, we have $OA^{2} + OC^{2} = 1949^{2} + 2002^{2}, OA \cdot OC \cdot \sin \angle AOC = 1949 \times 2002$, thus $QR^{2} = OA^{2} + OC^{2} - 2 \cdot OA \cdot OC \cdot \cos \left(\angle AOC - \frac{\pi}{2}\right)$
$$
-1949^{2} + 2002^{2} - 2 \times 1949 \times 2002 = 53^{2}
$$

Therefore, $QR = 53$