11. In tetrahedron $ABCD$, $AB=CD=a, BC=AD=b, CA=BD=c$. If the angle between the skew lines $AB$ and $CD$ is $\theta$, then, $\cos \theta=$ $\qquad$ .
11. $\frac{\left|b^{2}-c^{2}\right|}{a^{2}}$.

Solution 1: Let $D C=a, D A=b, D B=c$. From the given conditions, we have
$$
|\boldsymbol{a}-\boldsymbol{b}|=|\boldsymbol{c}|, a^{2}+b^{2}-2 \boldsymbol{a} \cdot \boldsymbol{b}=c^{2} .
$$

Thus, $\boldsymbol{a} \cdot \boldsymbol{b}=\frac{a^{2}+b^{2}-c^{2}}{2}$.
Similarly, $a \cdot c=\frac{a^{2}+c^{2}-b^{2}}{2}$.
Therefore, $\cos \theta=\frac{|\boldsymbol{a} \cdot(\boldsymbol{b}-\boldsymbol{c})|}{|\boldsymbol{a}| \cdot|\boldsymbol{b}-\boldsymbol{c}|}=\frac{\left|b^{2}-c^{2}\right|}{a^{2}}$.
Solution 2: The edges of the tetrahedron are the face diagonals of a rectangular prism. Let the three sides of the rectangular prism be $x, y, z$, then
$$
a^{2}=y^{2}+z^{2}, b^{2}=z^{2}+x^{2}, c^{2}=x^{2}+y^{2} \text {. }
$$

Let $\theta$ be the angle between the diagonals of the $y, z$ rectangle, hence
$$
\cos \theta=\frac{\left|\left(\frac{a}{2}\right)^{2}+\left(\frac{a}{2}\right)^{2}-y^{2}\right|}{2 \cdot \frac{a}{2} \cdot \frac{a}{2}}=\frac{\left|b^{2}-c^{2}\right|}{a^{2}} .
$$