6. Given $\overrightarrow{O A}$ and $\overrightarrow{O B}$ are non-zero and non-collinear vectors. Let $\overrightarrow{O C}=\frac{1}{1+r} \overrightarrow{O A}+\frac{r}{1+r} \overrightarrow{O B}$. Define the set of points
$$
M=\left\{K \left\lvert\, \frac{\mid \overrightarrow{K A} \cdot \overrightarrow{K C}}{|\overrightarrow{K A}|}=\frac{\overrightarrow{K B} \cdot \overrightarrow{K C}}{|\overrightarrow{K B}|}\right.\right\} .
$$

When $K_{1} 、 K_{2} \in M$, if for any $r \geqslant 2$, the inequality $\left|\overrightarrow{K_{1} K_{2}}\right| \leqslant c|\overrightarrow{A B}|$ always holds, then the minimum value of the real number $c$ is $\qquad$
6. $\frac{4}{3}$.
From the given, we know that $K C$ bisects $\angle A K B$.
Thus, $\frac{K A}{K B}=\frac{A C}{B C}=r$.
Assume without loss of generality that $A B=2$. Then the locus of point $K$ is a circle with center $\left(\frac{r^{2}+1}{r^{2}-1}, 0\right)$ and radius $\frac{2 r}{r^{2}-1}$.
Therefore, $c \geqslant \frac{\left|\overrightarrow{K_{1} K_{2}}\right|}{2}$.
And $\frac{\left|\overrightarrow{K_{1} K_{2}}\right|}{2} \leqslant \frac{2 r}{r^{2}-1} \leqslant \frac{4}{3}$, hence $c_{\min }=\frac{4}{3}$.