Two particles P and Q are moving with velocity of `(hat(i)+hat(j))` and `(-hat(i)+2hat(j))` respectively. At time `t=0`, P is at origin and Q is at a point with positive vector `(2hat(i)+hat(j))`. Then the shortest distance between P & Q is :-

Options

A. `(2sqrt(5))/5`
B. `(4sqrt(5))/5`
C. `sqrt(5)`
D. `(3sqrt(5))/5`

Answer

B

Explanation

`vec(v)_(QP)=-hat(i)+2hat(j)-hat(i)-hat(j)=-2hat(i)+hat(j)`

So from sine rule `sqrt(5)/(sin 90^(@))=(x_("min"))/(sin theta)rArr x_(m)`
`=sqrt(5)xx2 "sin" theta/2 "cos" theta/2`
`=sqrt(5)xx2xx1/sqrt(5)xx2/sqrt(5)=4/sqrt(5)`