6. 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 $A B B_{1} A_{1}$.
6. As shown in Figure $20-18$, with $A$ as the origin, $A C$ as the $x$-axis, and $A A_{1}$ as the $z$-axis, establish a spatial rectangular coordinate system. According to the problem, the coordinates of the relevant points are: $A(0,0,0), B\left(\frac{a}{2},-\frac{\sqrt{3}}{2} a, 0\right)$, $A_{1}(0,0, \sqrt{2} a), C_{1}(a, 0, \sqrt{2} a)$; take the midpoint $D$ of $A_{1} B_{1}$, and connect $A D, C_{1} D$.
Since $C_{1} D \perp A_{1} B_{1}$, it follows that $C_{1} D \perp$ plane $A B B_{1} A_{1}$.
The angle formed by $A C_{1}$ and $A D$ is the angle formed by $A C_{1}$ and the side face $A B B_{1} A_{1}$.
Given $\overrightarrow{A D}=\left(\frac{a}{4},-\frac{\sqrt{3}}{4} a, \sqrt{2} a\right), \overrightarrow{A C}_{1}=(a, 0, \sqrt{2} a)$,
we get $\cos \left\langle\overrightarrow{A C}_{1}, \overrightarrow{A D}\right\rangle=\frac{\left|\overrightarrow{A C_{1}} \cdot \overrightarrow{A D}\right|}{\left|\overrightarrow{A C_{1}}\right||\overrightarrow{A D}|}=\frac{\sqrt{3}}{2}$,
thus the angle formed by $\overrightarrow{A C_{1}}$ and $\overrightarrow{A D}$, which is the angle formed by $A C_{1}$ and the side face $A B B_{1} A_{1}$, is $30^{\circ}$.