Question

Find the shortest distance between the lines:

`(x + 1)/2 = (y - 1)/1 = (z - 9)/3 and  (x - 3)/2 = (y + 15)/(-7) = (z - 9)/5`

Answer 1

`(x + 1)/2 = (y - 1)/1 = (z - 9)/3`

`(x - 3)/2 = (y + 15)/(-7) = (z - 9)/5`

Here, `vec(a_1) = -hati + hatj + 9hatk`

`vec(b_1) = 2hati + hatj - 3hatk`

`vec(a_2) = 3hati - 15hatj + 9hatk`

`vec(b_2) = 2hati - 7hatj + 5hatk`

`vec(a_1) - vec(a_2) = (3hati - 15hatj + 9hatk) - (-hati + hatj + 9hatk)`

= `4hati - 16hatj + 0hatk`

`vec(a_1) - vec(a_2) = (3hati - 15hatj + 9hatk) - (-hati + hatj + 9hatk)`

= `4hati - 16hatj + 0hatk`

`vecb_1 xx vecb_2 = |(hati, hatj, hatk), (2, 1, -3), (2, -7, 5)|`

= `hati(5 - 21) - hatj(10 + 6) + hatk(-14 - 2)`

= `hati(-16) - hatj(16) + hatk(-16)`

= `-16hati - 16hatj - 16hatk`

`|vecb_1 xx vecb_2| = sqrt(16^2 + 16^2 + 16^2)`

= `16sqrt3`

Shortest distance = `|((veca_2 - veca_1) . (vecb_1 xx vecb_2))/(|vecb_1 xx vecb_2|)|`

= `|((4hati - 16hatj) . (-16hati - 16hatj - 16hatk))/(|16sqrt3|)|`

= `|(4(-16)hati + (-16)(-16)hatj + 0(-16)hatk)/(|16sqrt3|)|`

= `|(-64 + 256)/(16sqrt3)|`

= `192/(16sqrt3)`

= `(12sqrt3)/3`

= `4sqrt3` units