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?

Answer

`t_m=(v_1l_1+v_2l_2)/(v_1^2+v_2^2)`, `l_(min)=(|l_1v_2-l_2v_1|)/(sqrt(v_1^2+v_2^2))`.

Explanation

Let the particle 1 and 2 be at points B and A at `t=0` at the distances `l_1` and `l_2` from intersection point O.
Let us fix the inertial frame with the particle 2. Now the particle 1 moves in relative to this reference frame with a relative velocity `vec(v_(12))=vec(v_1)-vec(v_2)`, and its trajectory is the straight line BP. Obviously, the minimum distance between the particles is equal to the length of the perpendicular AP dropped from point A on the straight line BP(figure).


From Figure (b), `v_(12)=sqrt(v_1^2+v_2^2)`, and `tan theta=v_1/v_2` (1)
The shortest distance
`AP=AM sin theta=(OA-OM)sin theta=(l_2-l_1cot theta) sin theta`
or `AP=(l_2-l_1v_2/v_1)(v_1)/(sqrt(v_1^2+v_2^2))=(v_1l_2-v_2l_1)/(sqrt(v_1^2+v_2^2))` (using 1)
The sought time can be obtained directly from the condition that `(l_1-v_1t)^2+(l_2-v_2t)^2` is minimum. This gives `t=(l_1v_1+l_2v_2)/(v_1^2+v_2^2)`.