[ Coordinate method in space ] [ Rectangular parallelepipeds ]

The non-intersecting diagonals of two adjacent faces of a rectangular parallelepiped are inclined to the base plane at angles $\alpha$ and $\beta$. Find the angle between these diagonals.
Consider a rectangular parallelepiped $A B C D A 1 B 1 C 1 D 1$. Let $A D 1$ and $D C 1$ be the specified diagonals. Then $\angle A D 1 A 1=\alpha, \angle D C D 1=\beta$. Choose a coordinate system with the origin at point $A$. The $x$-axis is directed along the ray $A B$, the $y$-axis along the ray $A D$, and the $z$-axis along the ray $A A 1$. Let $A B=1$. Then

$$
A A 1=D D 1=D 1 C 1 \operatorname{tg} \beta=\operatorname{tg} \beta, A D=A 1 D 1=A A 1 \operatorname{ctg} \alpha=\operatorname{tg} \beta \operatorname{ctg} \alpha .
$$

Then the coordinates of the endpoints of segments $A D 1$ and $D C 1$ are:

$$
A(0 ; 0 ; 0), D 1(0 ; \operatorname{tg} \beta \operatorname{ctg} \alpha ; \operatorname{tg} \beta)
$$

$$
D(0 ; \operatorname{tg} \beta \operatorname{ctg} \alpha ; 0), C 1(1 ; \operatorname{tg} \beta \operatorname{ctg} \alpha ; \operatorname{tg} \beta)
$$

Let $\phi$ be the angle between the vectors $\overrightarrow{A D_{\mathbf{1}}}=(0 ; \operatorname{tg} \beta \operatorname{ctg} \alpha ; \operatorname{tg} \beta)$ and $\overrightarrow{D C_{\mathbf{1}}}=(1 ; 0 ; \operatorname{tg} \beta)$. Then

![](https://cdn.mathpix.com/cropped/2024_05_06_dbba6af20ebd45cbbb54g-27.jpg?height=128&width=832&top_left_y=1517&top_left_x=610)

![](https://cdn.mathpix.com/cropped/2024_05_06_dbba6af20ebd45cbbb54g-27.jpg?height=137&width=729&top_left_y=1709&top_left_x=670)

If $\gamma$ is the angle between the specified lines, then

$$
\cos \gamma=|\cos \phi|=\sin \alpha \sin \beta
$$

## Answer

$\arccos (\sin \alpha \sin \beta)$.