Two particles start moving simultaneously with constant velocities `u_1 and u_2` as shown in (Fig. 5.194). First particle starts from `A` along `AO` and second starts from `O` along `OM`. Find the shortest distance between them during their motion.
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Answer

`(x_0 u_2 sin theta)/(sqrt(u_1^2 + u_2^2 + 2 u_1 u_2 cos theta))`.

Explanation

`vec u_1 = -u_1 hat j, vec u_2= u_2 cos theta hat j+u_2 sin theta hat i`
`rArr vec u_2 = vec u_1 u_2 sin theta hat i + (u_2 cos theta + u_1) hat j`
`tan prop = (u_2 sin theta)/(u_2 cos theta + u_1)`
Shortest distance,
`AB = x_0 sin prop =(x_0 u_2 sin theta)/(sqrt(u_1^2 + u_2^2 + 2 u_1 u_2 cos theta))`.
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