Two ships are 10 km apart on the west-east line at some instant, the first one streaming to  the east at 30 km/h and the second one to north at 40 km/h. what is the distance of closest  approach between them in km?

${\overrightarrow{v}}_{21}\left(\text{relative\hspace{0.17em}velocity\hspace{0.17em}of\hspace{0.17em}2\hspace{0.17em}in\hspace{0.17em}realtive\hspace{0.17em}to\hspace{0.17em}1}\right),$ 
${\overrightarrow{v}}_{21}={\overrightarrow{v}}_2-{\overrightarrow{v}}_1=40\widehat{\text{j}}-30\widehat{\text{i}}$ 
 
From the concept of relative velocity 1 may be assumed to be fixed and 2 to be moving with the velocity $(-30\widehat{\text{i}}+40\widehat{\text{j}}).$ This is a vector of magnitude 50 and inclined at $\theta ={\tan }^{-1}(4/3).$ The shortest distance is obviously the perpendicular distance of the relative velocity vector from the fixed position of ship 1.
$\therefore S_{\text{shortest}}=s_{12}\sin \theta \text{\hspace{0.17em}where\hspace{0.17em}}{s}_{12}\text{\hspace{0.17em}is\hspace{0.17em}the\hspace{0.17em}initial\hspace{0.17em}distance}$ between 1 and 2.   
$s_{12}=10\times (4/5)=8\text{km}$