Find the shortest distance between the following lines: `(x+1)/7=(y+1)/(-6)=(z+1)/1;(3-x)/(-1)=(y-5)/(-2)=(z-7)/1`

Explanation

Given lines are ltbr `(x+1)/(7) = (y+1)/(-6) = (z+1)/(1)`
`(x1)/(7) = (y+1)/(-6)=(z+1)/(1)`
and `(x-3)/(1) = (y-5)/(-2) = (z-7)/(1)`
The direction ratios of first line are `(7,-6,1)` and it passes through the point `(-1,-1,-1)`. Therefore, the vector equation of the given line is
`rArr vecr_(1) = -hati-hatj-hatk+lambda(7hati-6hatj+hatk)`
Similarly the vector equati on of second line is :
`vecr_(2)=3hati+5hatj+7hatk+mu(hati-2hatj+hatk)`
which are in the form `vecr_(1) = veca_(1)+lambdavecb_(1)`
and `vecr_(2)= veca_(2)+muvecb_(2)`
where, `veca_(1) = -hati-hatj-hatk, vecb_(1) = vecb_(1) =7hati-6hatj+hatk`
and `veca_(2) = 3hati+5hatj+7hatk,vecb_(2)=hati-2hatj+hatk`
Now, `veca_(2)-veca_(1) = (3hati+5hatj+7hatk) - (-hati-hatj-hatk)`
`= 4hati+6hatk+8hatk`,
and `vecb_(1)xxvecb_(2) = 4hati+6hatj+8hatk`,
and `vecb_(1)+vecb_(2) = |{:(hati,hatj,hatk),(7,-6,1),(1,-2,1):}|`
`= hati(-6+2)-hatj(7-1)+hatk(-14+6)`
`= -4hati-6hatj-8hatk`
`|vecb_(1)xxvecb_(2)| = sqrt((-4)^(2)+(-6)^(2)+(-8)^(2))`
`=sqrt(166+36+64) = sqrt(116) = 2sqrt(29)`
`:.` Shortest distance betweent the given lines
`d=|((vecb_(1)xxvecb_(2)).(veca_(2)-veca_(1)))/(|vecb_(1)xxvecb_(2)|)|`
`=(|(-4hati-6hatj-8hatk).(4hati+6hatj+8hatk)|)/(2sqrt(29))`
`= (|-16-36-64|)/(2sqrt(29))`
`= (116)/(2sqrt(29)) = (58)/(sqrt(29)) = 2sqrt(29)` unit