** Given a fixed point $O$ and two fixed perpendicular lines $l_{1}, l_{2}$. Draw two moving lines through $O$ that are perpendicular to each other, one intersecting $l_{2}$ at $P$, the other intersecting $l_{1}$ at $Q$, and the projection of $O$ on $P Q$ is $R$. Find the locus of $R$.
With $O$ as the origin, and a line passing through $O$ and parallel to $l_{1}$ as the $x$-axis, establish a Cartesian coordinate system. Let the equation of $l_{1}$ be $y=k$, (1) and the equation of $l_{2}$ be $x=h$, (2) where $k, h$ are known numbers. Let the normal form equation of $P Q$ be $x \cos \alpha + y \sin \alpha - p = 0$, (3) then the coordinates of point $R$ are $x = p \cos \alpha, y = p \sin \alpha$. (4) From (1), (3) we get
the coordinates of point $Q$ as $\left(\frac{p - k \sin \alpha}{\cos \alpha}, k\right)$.
(5) From (2), (3) we get the coordinates of point $P$ as $\left(h, \frac{p - h \cos \alpha}{\sin \alpha}\right)$.
(6) From $O P \perp O Q$
and (5), (6) we get $\frac{h(p - k \sin \alpha)}{\cos \alpha} + \frac{k(p - h \cos \alpha)}{\sin \alpha} = 0$, which simplifies to
$h p \sin \alpha + k p \cos \alpha = k h$. (7) Substituting (4) into (7) yields the trajectory of point $R$ as $k x + h y = k h$. This is a straight line, with intercepts on the $x$-axis and $y$-axis being $h, k$ respectively. That is, if the intersection of $l_{2}$ with the $x$-axis is $R_{1}$, and the intersection of $l_{1}$ with the $y$-axis is $R_{2}$, then the line $R_{1} R_{2}$ is the required trajectory.