The shortest distance between the lines
`vecr=(hati-hatj)+lamda(hati+2hatj-3hatk)`
and `vecr=(hati-hatj+2hatk)+mu(2hati+4hatj-5hatk)` is

Options

A. `6`
B. `6/(sqrt(5))`
C. `3/(sqrt(5))`
D. `2/(sqrt(5))`

Answer

B

Explanation

We know that the shortest distance between the lines `vecr=veca_(1)+lamdavecb_(1)` and `vecr=veca_(2)+muvecb_(2)` is given by
`d=|((veca_(2)-veca_(1)).(vecb_(1)xxvecb_(2)))/(|vecb_(1)xxvecb_(2)|)|`
Hence `veca_(1)=4hati-hatj,veca_(2)=hati-hatj+2hatk`,
`vecb_(1)=hati+2hatj-3hatk` and `vecb_(2)=2hati+4hatj-5hatk`
`:.veca_(2)-veca_(1)=-3hati+0hatj+2hatk`
and `vecb_(1)xxvecb_(2)=|(hati,hatj,hatk),(1,2,-3),(2,4,-5)|=2hati-hatj+0hatk`
`implies(veca_(2)-veca_(1)).(vecb_(1)xxvecb_(2))=(-3hati+0hatj+2hatk).(2hati-hatj+0hatk)=-6`
and `|vecb_(1)xxvecb_(2)|=sqrt(4+1+0)=sqrt(5)`
`:.d=|((veca_(2)-veca_(1)).(vecb_(1)xxvecb_(2)))/(|vecb_(1)xxvecb_(2)|)|=|(-6)/(sqrt(5))|=6/(sqrt(5))`