11. Given $\boldsymbol{a}=(\cos \alpha, \sin \alpha), \boldsymbol{b}=(\cos \beta, \sin \beta), \boldsymbol{a}$ and $\boldsymbol{b}$ satisfy the relation $|k \boldsymbol{a}+\boldsymbol{b}|=\sqrt{3}|\boldsymbol{a} - k \boldsymbol{b}|$, where $k>$ 0. Then the minimum value of $\boldsymbol{a} \cdot \boldsymbol{b}$ is $\qquad$ .
11. $\frac{1}{2}$.

From $|k \boldsymbol{a}+\boldsymbol{b}|^{2}=(\sqrt{3}|\boldsymbol{a}-k \boldsymbol{b}|)^{2}$ we get
$8 k \boldsymbol{a} \cdot \boldsymbol{b}=\left(3-k^{2}\right) \boldsymbol{a}^{2}+\left(3 k^{2}-1\right) \boldsymbol{b}^{2}$.
Therefore, $\boldsymbol{a} \cdot \boldsymbol{b}=\frac{\left(3-k^{2}\right) \boldsymbol{a}^{2}+\left(3 k^{2}-1\right) \boldsymbol{b}^{2}}{8 k}$.
Since $\boldsymbol{a}=(\cos \alpha, \sin \alpha), \boldsymbol{b}=(\cos \beta, \sin \beta)$, we have,
$$
a^{2}=1, b^{2}=1, a \cdot b=\frac{k^{2}+1}{4 k} \text {. }
$$

Since $k>0, k^{2}+1 \geqslant 2 k$, then $\frac{k^{2}+1}{4 k} \geqslant \frac{2 k}{4 k}=\frac{1}{2}$.
Therefore, the minimum value of $\boldsymbol{a} \cdot \boldsymbol{b}$ is $\frac{1}{2}$.