6. Given point $P$ is in the plane of Rt $\triangle A B C$, $\angle B A C=90^{\circ}, \angle C A P$ is an acute angle, and
$|\overrightarrow{A P}|=2, \overrightarrow{A P} \cdot \overrightarrow{A C}=2, \overrightarrow{A P} \cdot \overrightarrow{A B}=1$.
When $| \overrightarrow{A B}+\overrightarrow{A C}+\overrightarrow{A P} |$ is minimized,
$\tan \angle C A P=$ $\qquad$ .
6. $\frac{\sqrt{2}}{2}$.

Let $\angle C A P=\alpha$.
From the problem, we know $\angle B A P=\frac{\pi}{2}-\alpha$.
$$
\begin{aligned}
\text { Given } & |\overrightarrow{A P}|=2, \overrightarrow{A P} \cdot \overrightarrow{A C}=2, \overrightarrow{A P} \cdot \overrightarrow{A B}=1 \\
\Rightarrow & |\overrightarrow{A C}|=\frac{1}{\cos \alpha},|\overrightarrow{A B}|=\frac{1}{2 \sin \alpha} \\
\Rightarrow & |\overrightarrow{A B}+\overrightarrow{A C}+\overrightarrow{A P}|^{2} \\
& =|\overrightarrow{A B}|^{2}+|\overrightarrow{A C}|^{2}+|\overrightarrow{A P}|^{2}+2 \overrightarrow{A B} \cdot \overrightarrow{A C}+ \\
& 2 \overrightarrow{A B} \cdot \overrightarrow{A P}+2 \overrightarrow{A C} \cdot \overrightarrow{A P} \\
= & \frac{\sin ^{2} \alpha+\cos ^{2} \alpha}{4 \sin ^{2} \alpha}+\frac{\sin ^{2} \alpha+\cos ^{2} \alpha}{\cos ^{2} \alpha}+10 \\
= & \frac{\cos ^{2} \alpha}{4 \sin ^{2} \alpha}+\frac{\sin ^{2} \alpha}{\cos ^{2} \alpha}+\frac{45}{4} \\
\geqslant & \geqslant \sqrt{\frac{\cos ^{2} \alpha}{4 \sin ^{2} \alpha} \cdot \frac{\sin ^{2} \alpha}{\cos ^{2} \alpha}}+\frac{45}{4}=\frac{49}{4},
\end{aligned}
$$

When and only when $\frac{\cos ^{2} \alpha}{4 \sin ^{2} \alpha}=\frac{\sin ^{2} \alpha}{\cos ^{2} \alpha}$, i.e., $\tan \alpha=\frac{\sqrt{2}}{2}$, $|\overrightarrow{A B}+\overrightarrow{A C}+\overrightarrow{A P}|$ achieves its minimum value $\frac{7}{2}$.