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Point $M$ is at distances of 5 and 4 from two parallel lines $m$ and $n$, and at a distance of 3 from the plane passing through these lines. Find the distance between the lines $m$ and $n$.
Let $M_1$ be the orthogonal projection of point $M$ onto the plane $\alpha$ passing through the parallel lines $m$ and $n$, and let $A$ and $B$ be the bases of the perpendiculars dropped from point $M$ to lines $m$ and $n$ respectively. Since $M_1A$ and $M_1B$ are the orthogonal projections of the oblique lines $MA$ and $MB$ onto the plane $\alpha$ and $MA \perp m$ and $MB \perp n$, by the theorem of three perpendiculars, $M_1A \perp m$ and $M_1B \perp n$. The lines $m$ and $n$ are parallel, so points $A, M_1$, and $B$ lie on the same line. Therefore, the distance between the lines $m$ and $n$ is equal to the length of the segment $AB$. From the right triangles $MAM_1$ and $MBM_1$, we find that

\[
\begin{aligned}
& AM_1 = \sqrt{AM^2 - MM_1^2} = \sqrt{25 - 9} = 4 \\
& BM_1 = \sqrt{BM^2 - MM_1^2} = \sqrt{16 - 9} = \sqrt{7}
\end{aligned}
\]

If point $M_1$ lies between points $A$ and $B$ (Fig. 1), then

\[
AB = AM_1 + M_1B = 4 + \sqrt{7}
\]

If point $M_1$ lies outside the segment $AB$ (Fig. 2), then

\[
AB = |AM_1 - M_1B| = |4 - \sqrt{7}| = 4 - \sqrt{7}
\]

## Answer

$4 + \sqrt{7}$ or $4 - \sqrt{7}$.

The line $l$ passes through a point lying on a circle with center $O$ and radius $r$. It is known that the orthogonal projection of the line $l$ onto the plane of the circle is a line tangent to this circle. Find the distance from point $O$ to the line $l$.

## Solution

Let $A$ be the point of tangency of the orthogonal projection $l_1$ of the given line $l$ with the given circle. Since $OA \perp l_1$, by the theorem of three perpendiculars, $OA \perp l$. Therefore, the distance from point $O$ to the line $l$ is equal to the length of the segment $OA$, i.e., $r$.

## Answer

$r$.