Velocity of A, vA=4㎧ towards B
Velocity of B, vB=3㎧ perpendicular to AB
Distance between them, l=5m
Let they are closest after time t seconds
And A and B reach to A′ & B′respectively.
The distance traveled by A
=4×t
∴A′B=(5−4t)
And distance traveled by B=3×t
∴BB′=3×t
AAt this distance the distance between A and B is A′B′
In △A′BB′:(A′B′)2=(A′B)2+(BB′)2
=(5−4t)2+(3t)2
=25−40t+16t2+9t2
=25t2−40t+25
A′B′=√(5t−4)2+9
Now A′B′ is minimum when
(5t−4)2=0
⇒5t=4
t=45
At this time particles are nearest to each other