Find the shortest distance `vecr=hati+2hatj+3hatk+lambda(hati-3hatj+2hatk)and
vecr=
4hati+5hatj+6hatk+mu(2hati+3hatj+hatk)`.

Explanation

Given lines are
`vecr = hati+2hatj+3hatk+lambda(hati-3hatj+2hatk)`
and `vecr = 4hati+5hatj+6hatk+mu(2hati+3hatj+hatk)`
Comparing the given equatiions with `vecr = veca_(1) + lambdavecb_(1)` and `vecr = veca_(2) + muvecb_(2)`,
`vecb_(1)xxvecb_(2)=|{:(hati,hatj,hatk),(1,-3,2),(2,3,1):}|`
`= veci(-3-6)-hatj(1-4)+hatk(3+6)`
`= - 9hati+3hatj+9hatk`
`rArr |vecb_(1) xx vecb_(2)|= sqrt((-9)^(2)+(3)^(2)+(9)^(2))`
`= sqrt(81+9+81)`
`= sqrt(171) = 3sqrt(19)`
`:.` Required shorted distance,
`d=|{:(vecb_(1)xxvecb_(2)).(veca_(2)-veca_(1)))/(|vecb_(1)xxvecb_(2)|)|`
`=(|(-9hati+3hatj+9hatk).(3hati+3hatj+3hatk)|)/(3sqrt(19))`
`= (|-9xx3+3xx3+9xx3|)/(3sqrt(19))`
` = (-27+9+27)/(3sqrt(19))= (9)/(3sqrt(19))= 3/(sqrt(19))`
Therefore, shortest distance between the given two lines is `(3)/(sqrt(19))` units.