Two particles, 1 and 2, move with constant velocities v1 and v2 along two mutually perpendicular straight lines toward the intersection point O. At the moment t=0 the particles were located at the distances l1 and l2 from the point O. At time tm=xv1l1+v2l2v21+v22, the distance between the particles become the smallest. Value of x will be
Let the particle 1 and 2 be at points B and A at t=0 at the distances l1 and l2 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 →v12=→v1−→v2, 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 to the straight line BP (figure shown below).
From the figure shown below, v12=√v21+v22, and tanθ=v1v2 (1)
The shortest distance
AP=AMsinθ=(OA−OM)sinθ=(l2−l1cotθ)sinθ
or AP=(l2−l1v2v1)v1√v21+v22=v1l2−v2l1√v21+v22 (using 1)
The sought time can be obtained directly from the condition that (l1−v1t)2+(l2−v2t)2 is minimum. This gives t=l1v1+l2v2v21+v22.