3. In rectangle $A B C D$, $A B=4, A D=3$, fold the rectangle along diagonal $A C$ so that the projection of point $D$ on plane $A B C$, point $E$, falls on line $A B$. At this moment, the dihedral angle $B-A C-D$ is $\qquad$
3. $\arccos \frac{9}{16}$.

As shown in Figure 2, draw $E G \perp A C$ 
at point $G$, and $B H \perp A C$ at point $H$, and connect $D G$.
By the theorem of three perpendiculars, we get
$D G \perp A C$.
Thus, $\angle E G D$ is the plane angle of the dihedral angle $B-A C-D$.

Since $B H / / E G$, the size of $\angle E G D$ is the same as the angle between vectors $H B$ and $G D$. Let $\angle D A C = \angle A C B = \alpha$. Because
$$
\begin{array}{l}
A C = \sqrt{3^{2} + 4^{2}} = 5, D G = B H = \frac{3 \times 4}{5} = \frac{12}{5}, \\
A G = C H = \sqrt{B C^{2} - B H^{2}} = \frac{9}{5},
\end{array}
$$

we have $\cos \alpha = \frac{A G}{A D} = \frac{3}{5}$.
Therefore, $\boldsymbol{H C} \cdot \boldsymbol{G A} = -|H C| \cdot |G A| = -\frac{81}{25}$,
$$
\begin{array}{l}
\boldsymbol{H C} \cdot \boldsymbol{A D} = |H C| \cdot |A D| \cos \alpha = \frac{9}{5} \times 3 \times \frac{3}{5} = \frac{81}{25}, \\
\boldsymbol{C B} \cdot \boldsymbol{G A} = |C B| \cdot |G A| \cos \alpha = 3 \times \frac{9}{5} \times \frac{3}{5} = \frac{81}{25}, \\
\boldsymbol{C B} \cdot \boldsymbol{A D} = \boldsymbol{C B} \cdot (\boldsymbol{A E} + \boldsymbol{E D}) = \boldsymbol{C B} \cdot \boldsymbol{A E} + \boldsymbol{C B} \cdot \boldsymbol{E D} \\
= 0 + 0 = 0 .
\end{array}
$$

Then, $\boldsymbol{H B} \cdot G D = (H C + C B) \cdot (G A + A D)$
$$
= \boldsymbol{H C} \cdot G A + H C \cdot A D + C B \cdot G A + C B \cdot A D = \frac{81}{25} \text{. }
$$

Thus, $\cos \angle E G D = \frac{H B \cdot G D}{|H B| \cdot |G D|} = \frac{9}{16}$.
Therefore, $\angle E G D = \arccos \frac{9}{16}$.