The $A B C A_{1} B_{1} C_{1}$ regular triangular prism has its side edge $A A_{1}$ perpendicular to the base $A B C$ and twice as long as the height of the base triangle. Determine the angle between the lines $A B_{1}$ and $B C_{1}$.
The lines $A B_{1}$ and $B C_{1}$ are skew. Their angle can be better visualized and the situation made suitable for calculation if we draw a parallel through $B_{1}$ to $C_{1} B$. In other words, this means that we translate $C_{1} B$ in the plane of the side $C_{1} C B$ by the vector $\overrightarrow{C_{1} B_{1}}$ to the position $B_{1} B^{\prime}$.

![](https://cdn.mathpix.com/cropped/2024_05_02_22ed75ff4d17b4d07323g-1.jpg?height=1011&width=659&top_left_y=308&top_left_x=717)

Then $A B^{\prime}$ is clearly twice the height of the base triangle, i.e., $a \sqrt{3}$ - which is the length of the edge $B B_{1}$ - and $C B^{\prime} B_{1}$ is an equilateral triangle with side length $2a$. From the triangle $A B_{1} B^{\prime}$, using the cosine rule,

$$
\cos \angle A B_{1} B^{\prime} = \frac{B_{1} A^{2} + B_{1} B^{\prime 2} - A B^{\prime 2}}{2 B_{1} A \cdot B_{1} B^{\prime}} = \frac{5}{8}
$$

the sought angle is $51^{\circ} 19^{\prime}$.

Remark. In principle, equivalent to the above is the planar construction of the angle if we translate $B C_{1}$ by the vector $\overrightarrow{B B_{1}}$, but the calculation is somewhat longer.