We know that the shortest distance between the lines, `vecr=veca_(1)+lamda vecb_(1)and vecr=veca_(2)+muvecb_(2)` is given as,
`=|((veca_(2)-veca_(1))*(vecb_(1)xx vecb_(2)))/(|vecb_(1)xxvecb_(2)|)|`
Here, `veca_(1)=4hati-hatj, veca_(2)=hati-hatj+2hatk,`
`vecb_(1)=hati+2hatj-3hatk`
and `vecb_(2)=hati+4hatj-5hatj`
Now, `veca_(2)-veca_(1)=(hati-hatj+2hatk)-(4hati-hatj)`
`=-3hati+0hatj+2hatk`
`=-3hati+2hatk`
and `vecb_(1)xxvecb_(2)=|{:(hati,hatj, hatk),(1,2,-3),(1,4,-5):}|`
`=hati(-10+12)-hatj(-5+3)+hatk(4-2)`
`=2hati+2hatj+2hatk`
`therefore(veca_(2)-veca_(1))*(vecb_(1)xxvecb_(2))=(-3hati+2hatk)*(2hati+2hatj+2hatk)`
`=-6+4=-2`
and `|vecb_(1)xxvecb_(2)|=sqrt((2)^(2)+(2)^(2)+(2)^(2))=sqrt12`
`therefore` Shortest distance, `d=|(-2)/(sqrt12)|`
`=(2)/(2sqrt3)=(1)/(sqrt3)` units.
