3. As shown in the figure, in quadrilateral $A B C D$, it is known that $\cos \angle B A D=\frac{3}{4}, \angle B A C=\angle D A C, A D<A B$, and $A B=5, A C=B D=\sqrt{14}$. If $\overrightarrow{A C}=\lambda \overrightarrow{A B}+\mu \overrightarrow{A D}(\lambda, \mu \in \mathbf{R})$, then $\lambda+\mu=$ $\qquad$
In $\triangle A B D$, $14=x^{2}+25-$
$2 x \cdot 5 \cdot \frac{3}{4} \Rightarrow 2 x^{2}-15 x+22=0$
$\Rightarrow x=2$ or $\frac{11}{2}$ (discard), then in $\triangle A C D$,
$y^{2}=4+14-2 \cdot 2 \cdot \sqrt{14} \cdot \frac{\sqrt{14}}{4}=4 \Rightarrow y=2$, thus $D C / / A B \Rightarrow A E C D$ is a rhombus $\Rightarrow \overrightarrow{A C}=\overrightarrow{A E}+\overrightarrow{A D} \Rightarrow \lambda=\frac{2}{5}, \mu=1 \Rightarrow \lambda+\mu=\frac{7}{5}$.