If we take the position of ship ‘A’ as origin then position and velocities of both ships can be given as:

${\overrightarrow{\mathrm{v}}}_{\mathrm{A}}=(30\hat{\mathrm{i}}+50\hat{\mathrm{j}})\mathrm{km}/\mathrm{hr}$

${\overrightarrow{\mathrm{v}}}_{\mathrm{B}}=-10\hat{\mathrm{i}}\mathrm{km}/\mathrm{hr};{\overrightarrow{\mathrm{r}}}_{\mathrm{A}}=0\hat{\mathrm{i}}+0\hat{\mathrm{j}}$

${\overrightarrow{\mathrm{r}}}_{\mathrm{B}}=-(80\hat{\mathrm{i}}+150\hat{\mathrm{j}})\mathrm{km}$

Time after which distance between them will be minimum

$\mathrm{t}=-\frac{{\overrightarrow{\mathrm{r}}}_{\mathrm{BA}}\cdot {\overrightarrow{\mathrm{v}}}_{\mathrm{BA}}}{{\left|{\overrightarrow{\mathrm{v}}}_{\mathrm{BA}}\right|}^2};$

$\mathrm{Where}{\overrightarrow{\mathrm{r}}}_{\mathrm{BA}}=(80\hat{\mathrm{i}}+150\hat{\mathrm{j}})\mathrm{km}$

${\overrightarrow{\mathrm{V}}}_{\mathrm{BA}}=-10\hat{\mathrm{i}}-(30\hat{\mathrm{i}}+50\hat{\mathrm{j}})(-40\hat{\mathrm{i}}-50\hat{\mathrm{j}})\mathrm{km}/\mathrm{hr}$

$\therefore \mathrm{t}=\frac{(80\hat{\mathrm{i}}+150\hat{\mathrm{j}})\cdot (-40\hat{\mathrm{i}}-50\hat{\mathrm{h}})}{|-40\hat{\mathrm{i}}-50{\hat{\mathrm{h}}}^2|}$

$=\frac{3200+7500}{4100}\mathrm{hr}=\frac{10700}{4100}\mathrm{hr}=2.6\mathrm{hrs}$