3.79. Find the cosine of the angle between the non-intersecting diagonals of two adjacent lateral faces of a regular triangular prism, where the lateral edge is equal to the side of the base.
3.79. I m e t h o d. By the condition, $A B C A_{1} B_{1} C_{1}$ is a regular triangular prism, $A A_{1}=A B, A B_{1}$ and $B C_{1}$ are non-intersecting diagonals (Fig. 3.80); it is required to find $\angle (AB, BC_{1})$. Let $A B=A A_{1}=a$. Draw $B_{1} D \| B C_{1}, D \in (B C)$; then $\angle A B_{1} D = \varphi$ is the desired angle. From $\triangle A B D$ we find $A D^{2}=a^{2}+a^{2}-2 a^{2} \cos 120^{\circ}=3 a^{2}$. In $\triangle A B_{1} D$ we have $A D^{2}=2 A B_{1}^{2}-2 A B_{1}^{2} \cos \varphi$; since $A B_{1}^{2}=2 a^{2}$, then $3 a^{2}=4 a^{2}-4 a^{2} \cos \varphi$, from which $\cos \varphi=\frac{1}{4}$.

![](https://cdn.mathpix.com/cropped/2024_05_21_1e3d8d2f6605ac23e2a5g-550.jpg?height=558&width=375&top_left_y=921&top_left_x=778)

Fig. 3.80

II m e t h o d. Let $\overline{A B}=\bar{m}, \overline{B C}=\bar{n}, \overline{A A_{1}}=\bar{p}$; then $\overline{A B_{1}}=\bar{p}+\bar{m}, \overline{B C_{1}}=\bar{n}+\bar{p}$. Using the formula for finding the cosine of the angle between vectors, we get

$$
\cos \varphi=\frac{\overline{A B_{1}} \cdot \overline{B C_{1}}}{\left|\overline{A B_{1}}\right| \cdot\left|\overline{B C_{1}}\right|}=\frac{(\bar{p}+\bar{m})(\bar{n}+\bar{p})}{a \sqrt{2} \cdot a \sqrt{2}}=\frac{\bar{p} \bar{n}+\bar{p}^{2}+\bar{m} \bar{n}+\bar{m} \bar{p}}{2 a^{2}}=
$$

$$
=\frac{0+a^{2}+a^{2} \cos 120^{\circ}+0}{2 a^{2}}=\frac{1}{4} .
$$

![](https://cdn.mathpix.com/cropped/2024_05_21_1e3d8d2f6605ac23e2a5g-551.jpg?height=462&width=481&top_left_y=273&top_left_x=46)

Fig. 3.81