Two particles A and B move with constant velocity $\overrightarrow{v_1}$ and $\overrightarrow{v_2}$ along two mutually perpendicular straight lines towards intersection point O as shown in figure. At moment t = 0 particles were located at distance $l_1$ and $l_2$ respectively from O. Then minimum distance between the particles and time taken are respectively

Let the separation between the particles be minimum at time t, Then

Since  $\mathrm{OB}={\mathrm{l}}_2-{\mathrm{v}}_2\mathrm{t}\text{ and }\mathrm{OA}={\mathrm{l}}_1-{\mathrm{v}}_1\mathrm{t}\text{ and }$

$A{B}^2=O{B}^2+O{A}^2\Rightarrow s^2={\left(l_1-v_{1}t\right)}^2+{\left(l_2-v_{2}t\right)}^2$

For s to be minimum  $\frac{ds}{dt}=0\text{ or }\frac{d}{dt}\left(s^2\right)=0$

$$\begin{gathered} \Rightarrow 2s\frac{ds}{dt}=2\left(l_1-v_{1}t\right)\left(-v_1\right)+2\left(l_2-v_{2}t\right)\left(-v_2\right)=0 \\ -l_1{v}_1+v_1^{2}t-l_2{v}_2+v_2^{2}t=0\Rightarrow t=\frac{l_1{v}_1+l_2{v}_2}{v_1^2+v_2^2} \end{gathered}$$

$S_{\min }^2={\left[l_1-\left(\frac{l_1{v}_1+l_2{v}_2}{v_1^2+v_2^2}\right)v_1\right]}^2+{\left[l_2-v_2\left(\frac{l_1{v}_1+l_2{v}_2}{v_1^2+v_2^2}\right)\right]}^2$

$\Rightarrow S_{min}^2=\frac{{\left(l_1{v}_2-l_2{v}_1\right)}^2}{v_1^2+v_2^2}\Rightarrow S_{\min }=\frac{\left|l_1{v}_2-l_2{v}_1\right|}{\sqrt{v_1^2+v_2^2}}$