Find the distance from the center of a face of a unit cube to the vertices of the opposite face.

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Let $M$ be the center of the face $A_{1} B_{1} C_{1} D_{1}$ of the unit cube $A B C D A_{1} B_{1} C_{1} D_{1}$. From the right triangle $A M A_{1}$, using the Pythagorean theorem, we find that

![](https://cdn.mathpix.com/cropped/2024_05_06_d69082500b56170eb8ddg-12.jpg?height=1108&width=948&top_left_y=1258&top_left_x=1118)

$M A=\sqrt{A A_{1}^{2} + M A_{1}^{2}} = \sqrt{1 + \left(\frac{\sqrt{2}}{2}\right)^{2}} = \sqrt{\frac{3}{2}} = \frac{\sqrt{6}}{2}$.

It is clear that

$M D = M C = M B = M A = \frac{\sqrt{6}}{2}$.

The orthogonal projections of a segment onto three mutually perpendicular lines are 1, 2, and 3. Find the length of this segment.

## Solution

Let $O M$ be the given segment. Through point $O$, draw lines $a, b$, and $c$, respectively parallel to the three given mutually perpendicular lines. Then the orthogonal projections of segment $O M$ onto lines $a, b$, and $c$ are also 1, 2, and 3, and segment $O M$ is the diagonal of a rectangular parallelepiped with dimensions 1, 2, and 3. Therefore,

$$
O M = \sqrt{1^{2} + 2^{2} + 3^{2}} = \sqrt{14}
$$

## Answer

$\sqrt{14}$