Question

Two particles, 1 and 2, move with constant velocities $v_1$ and $v_2$ along two mutually perpendicular straight lines toward the intersection point $O$. At the moment $t=0$ the particles were located at the distances $l_1$ and $l_2$ from the point $O$. At time $t_m=x\frac{v_1{l}_1+v_2{l}_2}{v_1^2+v_2^2}$, the distance between the particles become the smallest. Value of $x$ will be

Answer 1

Let the particle 1 and 2 be at points $B$ and $A$ at $t=0$ at the distances $l_1$ and $l_2$ from intersection point $O$.
Let us fix the inertial frame with the particle 2. Now the particle 1 moves in relative to this reference frame with a relative velocity $\overrightarrow{v_{12}}=\overrightarrow{v_1}-\overrightarrow{v_2}$, and its trajectory is the straight line $BP$. Obviously, the minimum distance between the particles is equal to the length of the perpendicular $AP$ dropped from point $A$ on to the straight line $BP$ (figure shown below).
From the figure shown below, $v_{12}=\sqrt{v_1^2+v_2^2}$, and $\tan \theta =\frac{v_1}{v_2}$ (1)
The shortest distance
$AP=AM\sin \theta =(OA-OM)\sin \theta =(l_2-l_1\cot \theta )\sin \theta$
or $AP=(l_2-l_1\frac{v_2}{v_1})\frac{v_1}{\sqrt{v_1^2+v_2^2}}=\frac{v_1{l}_2-v_2{l}_1}{\sqrt{v_1^2+v_2^2}}$ (using 1)
The sought time can be obtained directly from the condition that ${(l_1-v_{1}t)}^2+{(l_2-v_{2}t)}^2$ is minimum. This gives $t=\frac{l_1{v}_1+l_2{v}_2}{v_1^2+v_2^2}$.