3. The points $A$ and $B$ move with constant velocities along the lines $a$ and $b$, and two corresponding positions $A_{1}, B_{1}$ and $A_{2}, B_{2}$ are known. Determine the position of the points $A$ and $B$ for which the length $A B$ is the smallest.
Solution. The vectors $\vec{a}=\overrightarrow{A_{1} A_{2}}$ and $\vec{b}=\overrightarrow{B_{1} B_{2}}$ are the velocity vectors of points $A$ and $B$, respectively. For an observer at point $A$, point $B$ moves with a velocity defined by the vector $\vec{b}-\vec{a}$. Let $Z$ and $C$ be points defined by

$$
\overrightarrow{B_{1} Z}=-\vec{a}, \overrightarrow{B_{1} C}=\vec{b}-\vec{a}
$$

The minimum distance between points $A$ and $B$ during their motion is equal to $A_{1} X$, where $X$ is the intersection of the line $B_{1} C$ with the normal to that line passing through point $A_{1}$, as shown in the diagram on the right. Let $\overrightarrow{B_{1} X}=\lambda \overrightarrow{B_{1} C}=\lambda(\vec{b}-\vec{a})$ and let $A_{0}$ and $B_{0}$ be the points defined by $\overrightarrow{A_{1} A_{0}}=\lambda \vec{a}, \overrightarrow{B_{1} B_{0}}=\lambda \vec{b}$. Clearly, points $A_{0}$ and $B_{0}$ are the respective positions of points $A$ and $B$ during the considered motion. Since

$$
\begin{aligned}
\overrightarrow{A_{0} B_{0}} & =\overrightarrow{A_{0} A_{1}}+\overrightarrow{A_{1} X}+\overrightarrow{X B_{1}}+\overrightarrow{B_{1} B_{0}} \\
& =-\lambda \vec{a}+\overrightarrow{A_{1} X}-\lambda(\vec{b}-\vec{a})+\lambda \vec{b}=\overrightarrow{A_{1} X}
\end{aligned}
$$

we conclude that $A_{0}$ and $B_{0}$ are the positions of points $A$ and $B$ at the moment when the distance between them is minimal.