Given a line and a point $1 \mathrm{~cm}$ away from it. What is the volume of the cube, one of whose vertices is the point, and one of whose space diagonals is the line?
Any space diagonal of a cube connects two vertices, each of which is connected to another vertex of the cube by an edge. The plane defined by the space diagonal and an edge starting from one of its endpoints intersects the cube in a rectangle (diagram). Therefore, we can assume that the point is the $D$ point shown in the diagram, and the line is the $H B$ line.

![](https://cdn.mathpix.com/cropped/2024_05_02_5d8722e3e13ee4fe1e75g-1.jpg?height=667&width=639&top_left_y=301&top_left_x=732)

Let the edge length of the cube be $a$; then $D B = \sqrt{2} \cdot a, H B = \sqrt{3} \cdot a$. The altitude to the hypotenuse of the right triangle $B D H$ is exactly the distance from point $D$ to line $H B$, which is $1 \text{ cm}$. The area of triangle $D B H$ can be written in two ways:

$$
\frac{B H \cdot 1}{2} = \frac{D B \cdot D H}{2}, \quad \text{that is} \quad \frac{\sqrt{3} a \cdot 1}{2} = \frac{\sqrt{2} a \cdot a}{2}.
$$

From this: $a = \frac{\sqrt{6}}{2} \text{ cm}$, so the volume of the cube is $a^3 = \frac{3 \sqrt{6}}{4} \text{ cm}^3 \left(\approx 1.84 \text{ cm}^3\right)$.