Two particles A and B move with constant velocities `v_(1)` and `v_(2)` along two mutually perpendicular strainght lines. Towards the intersection point O. At moment `t = 0` the particle were located at distance `l_(1)` and `l_(2)` from O respectively. Find the time when they are nearest and also this shortest distance

Explanation

As shown in Fig. 4.54, in time t, A will move a distance `v_(1) t` while `B, v_(2)t`, so after time `t` the distnace of A and B from O will be `(l_(1) - v_(1)t)` and `(l_(2) - v_(2) t)` respectively. So the distance between them L at time t will be given by
`L^(2) = (l_(1) - v_(1)t)^(2) + (l_(2) - v_(2)t)^(2)`.....(i)
Differentiating Eqn. (i) with respect to time,
`2L (dL)/(dt) = 2 (l_(1) - v_(1) t) + 2 (l_(2) - v_(2)t) (-v_(2))`
Now for L to be minimum `(dL//dt) = 0`
So, `(v_(1)^(2) + v_(2)^(2)) t = l_(1) v_(1) + l_(2) v_(2)`
i.e., `t = (l_(1) v_(1) + l_(2) v_(2))/((v_(1)^(2) + v_(2)^(2)))`....(ii)
Substituting this value of `t` from Eqn. (ii) in (i) and simplifying, we get
`L_("min") = (l_(1) v_(2) - l_(2) v_(1))/(sqrt((v_(1)^(2) + v_(2)^(2))))`.....(iii)
Eqn. (ii) and (iii) are the desired results.