11.3 Given three pairwise perpendicular lines. The distance between any two of them is a. Find the area of the parallelogram, two vertices of which are located on one line, and the other two - on the other two lines.
Solution. Consider a cube $ABCD A_{1} B_{1} C_{1} D_{1} \mathrm{c}$ with edge $a$. We can assume that the given pairwise perpendicular skew lines are precisely the lines $AD, A_{1}B_{1}$, and $CC_{1}$ - see the figure. We can also assume that we are considering a parallelogram $KLMN$ with two vertices $K$ and $L$ lying on the line $CC_{1}$ (and $M$ and $N$ on the lines $AD$ and $A_{1}B_{1}$, respectively). Note that one such parallelogram is the rectangle $AA_{1}C_{1}C$. Its area is calculated trivially:

$$
S_{AA_{1}C_{1}C} = AA_{1} \cdot AC = a \cdot a \sqrt{2} = a^{2} \sqrt{2}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_d760a6f59d34e4511a30g-5.jpg?height=522&width=599&top_left_y=1304&top_left_x=1254)

To the solution of problem 11.3

We will show that any other parallelogram satisfying the condition has the same area.

The line $MN$ is parallel to the line $KL$ (opposite sides of a parallelogram are parallel), which passes through the edge $CC_{1}$. Therefore, the line $MN$ is perpendicular to both lines $AD$ and $A_{1}B_{1}$ and intersects both of them. Such a line is unique; the points of its intersection with the lines $AD$ and $A_{1}B_{1}$ are uniquely determined. Thus, $M = A$ and $N = A_{1}$.

The points $K$ and $L$ are not uniquely determined. For example, point $K$ can be placed at any point on the line $CC_{1}$. However, after this, point $L$ is constructed uniquely (the rays $KL$ and $NM$ are directed in the same direction, and segments $KL$ and $MN$ are equal. Essentially, we have shown that the segment $KL$ can be located at any position on the line $CC_{1}$.

It remains to note that in any such position of the segment, the length of the side $KL = MN$ does not change, nor does the height of the parallelogram dropped from point $M$ - this is the segment $AC$.

Answer: $a^{2} \sqrt{2}$.

| THERE IS IN THE SOLUTION | SCORE |
| :--- | :---: |
| Correct and justified answer | 7 points |
| In the correct solution, there are errors in formulas and/or calculations, possibly leading to an incorrect answer | -1 point for each error |
| The idea to consider a parallelepiped (cube) with edges lying on the given skew lines | 2 points |
| The problem is correctly solved in special cases (e.g., when the parallelogram is a rectangle) | 1 point |