Two boat A and B 20 km apart as shown. Boat a streaming east and boat B in north by `30 km//hr` then find shortest approach and time taken for it.

Explanation

Let we find `" "vec(v)_(AB)`
`vec(v)_(BA)=vec(v)_(B)-vec(v)_(A)`
`30 hat(j)-30hat(i)=30sqrt(2) km//hr`
From `DeltaACB`
`(AC)/(AB)=sin 45^(@)`
`AC=10sqrt(2) km=BC`
time `" "t=(BC)/((vec(v)_(BA)))=(10sqrt(2))/(30sqrt(2))=1/3` hrs.
`t=20` min.

Alternative method:
After time t
`A'B'l=sqrt((20-30t)^(2)+(30t)^(2))`
For `l_("min") " "(dl)/(dt)=0`
`(dl)/(dt)=(1)/(2sqrt((20-30t)^(2)+(30t)^(2)))xx[2(20-30t)xx-30+1800t]=0`
`-(20-30t)+30t=0`
`60t-20`
`t=1/3` hrs.
Now `l_("min")=sqrt((20-30xx1/3)^(2)+(30xx1/3)^(2))=10sqrt(2) km`