Find the shortest distance between the lines whose vector equations are` -> r=(1-t) hat i+(t-2) hat j+(3-2t) hat k`and ` -> r=(s+1) hat i+(2s-1) hat j-(2s+1) hat k`

Explanation

Given equation of lines can be written as :
`vecr = (hati-2hatj+3hatk)+t(-hati+hatj-2hatk)`
and `vecr = (hati-hatj-hatk)+s(hati+2hatj-2hatk)`
Comparing these equation with `vecr = veca_(1)+tvecb_(1)` and `vecr = veca_(2)+ svecb_(2)`,
`veca_(1) = hati-2hatj+3hatk, vecb_(1)=-hati+hatj-2hatk`
and `veca_(2) = hati-hatj-hatk, vecb_(2)=hati+2hatj-2hatk`
Now, ` veca_(2)-veca_(1) = ( hati-hatj-hatk) - (hati-2hatj+3hatk)= hatj-4hatk`
and `vecb_(1)xxvecb_(2)=|{:(hati,hatj,hatk),(-1,1,-2),(1,2,2):}|`
`= (-2+4)hati-(2+2)hatj+(-2-1)hatk`
`= 2hati-4hatj-3hatk`
`rArr |vecb_(1)xxvecb_(2)|=sqrt((2)^(2)+(-4)^(2)+(-3)^(2))`
`= sqrt(4+16+9) = sqrt(29)`
`:.` Required shortest distance
`d=|{:(vecb_(1)xxvecb_(2)).(veca_(2)-veca_(1)))/(|vecb_(1)xxvecb_(2)|)|`
`=(|(2hati-4hatj-3hatk).(hatj-4hatk)|)/(sqrt(29))`
`= (|-4+12|)/(sqrt(29)) = (8)/(sqrt(29))`