−−→VAB=−→VA−−→VB=2(→i−→j)⇒|−−→VAB|=2√2.
Assuming B to be at rest, A will move with velocity −−→VAB in the direction shown in figure. The distance between them will first decrease from A to C and then increase beyond C.
Minimum distance between them is BC which is equal to 4√2 or 2√2 and the time after which they are at closest distance is :
t=AC−−→VAB=2√22√2=1second.