Two particles, 1 and 2, move with constant velocities `v_1` and `v_2` along two mutually perpendicular straight lines toward the intersection point O. At the moment `t=0` the particles were located at the distances `l_1` and `l_2` from the point O. How soon will the distance between the particles become the smallest? What is it equal to?

Explanation

Let particles (A) and (B) reach at (A') and (B') after time (t).`
thus ` OA'= v_1 t ` and ` BB' =v_2 t`
So, ` AA' = (d_1 -v_1 t) ` and ` OB' = (d_2 -v_2 t),
` If `` A' B' = L, ` then ` L_^2 (d_1 -v_1)^2 + (d_2 -v_2 t)^2 ` …(i)
Dfferentiating it w.r.t. time get
` 2 L (dL)/(dt) = 2 (d_1 -v_1 t) (- v_1 ) + 2 (d_2 -v_2 t) (- v_2)`
For (L) to be minimum, ` (dL)/(dt0 =0`
:. 0=- (d_1 v_1 -v_1^2 t) - 2 9d_2 v_2 - v_2^2 t0`
or ` (v_1^2 _ v_2^2 ) t=d_1 v-1 + d_2 v_2 or ` t = (d_1 v_1 + d_2 v_2 )/( (v_1^2 + v_2^2))`
Putting this value of (t) in (i). and simplifyng it, we get,
` L_(min0 = ( d_1 v_2 - d_2 v_1)/( sqrt (v-1^2 + v_2^2)`
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