18. For a regular triangular prism $A B C-A_{1} B_{1} C_{1}$ with a base edge length of $a$ and a lateral edge length of $\sqrt{2} a$, find the angle between $A C_{1}$ and the side face $A B B_{1} A_{1}$.
18. As shown in the figure, with $A$ as the origin, the line $A C$ as the $x$-axis, and the line $A A_{1}$ as the $z$-axis, a spatial rectangular coordinate system is established. According to the problem, the coordinates of the relevant points are: $A(0,0,0), B\left(\frac{a}{2},-\frac{\sqrt{3}}{2} a, 0\right)$, $A_{1}(0,0, \sqrt{2} a), C_{1}(a, 0, \sqrt{2} a)$.
Taking the midpoint $D$ of $A_{1} B_{1}$, and connecting $A D, C_{1} D$.
Since $C_{1} D \perp A_{1} B_{1}$,
it follows that $C_{1} D \perp$ plane $A B B_{1} A_{1}, A C_{1}$ and $A D$ form the angle,
which is the angle between $A C_{1}$ and the side face $A B B_{1} A_{1}$.
$$
\begin{array}{l}
\text { By } \overrightarrow{A D}=\left(\frac{a}{4},-\frac{\sqrt{3}}{4} a, \sqrt{2} a\right), \\
\overrightarrow{A C}_{1}=(a, 0, \sqrt{2} a),
\end{array}
$$

we get $\cos (\overrightarrow{A C}, \overrightarrow{A D})=\frac{\overrightarrow{A C_{1}} \cdot \overrightarrow{A D}}{\left|\overrightarrow{A C_{1}}\right||\overrightarrow{A D}|}=\frac{\sqrt{3}}{2}$.
Therefore, the angle formed by $\overrightarrow{A C}$ and $\overrightarrow{A D}$, which is the angle between $A C_{1}$ and the side face $A B B_{1} A_{1}$, is $30^{\circ}$.