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Question
Find the shortest distance between the lines:
`vecr = (hati+2hatj+hatk) + lambda(hati-hatj+hatk)` and `vecr = 2hati - hatj - hatk + mu(2hati + hatj + 2hatk)`
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Solution
Given the equation of lines `vecr = (hati + 2hatj + hatk) + λ(hati - hatj + hatk)` .....(1)
and `vecr = 2hati - hatj - hatk + µ(2hati + hatj + 2hatk)` .....(2)
Comparing equation (1) with `vecr = vec(a_1) + λvec(b_1)` and equation (2) with `vecr = vec(a_2) + µvec(b_2)`,
`vec(a_1) = hati + 2hatj + hatk` and `vec(a_2) = 2hati - 2hatj - hatk`
`vec(b_1) = hati - hatj + hatk` and `vec(b_2) = 2hati + hatj + 2hatk`
Now, `vec(a_2) - vec(a_1) = hati - 3hatj - 2hatk`
and `vec(b_1) xx vec(b_2) = |(hati, hatj, hatk), (1, -1, 1), (2, 1, 2)|`
= `(-2 -1) hati - (2 - 2) hatj + (1 + 2) hatk`
= `-3hati - 0hatj + 3hatk`
Required shortest distance
`= |((vecb_1 xx vecb_2)* (veca_2 - veca_1))/|vecb_1 xx vecb_2||`
`= |((-3hati + 3hatk)* (hati - 3hatj - 2hatk))/|-3hati + 3hatk||`
`= |((-3) xx 1 + 0 xx (-3) + 3 xx (-2))/ sqrt((-3)^2 + 3^2)|`
`= 9/sqrt18`
`= 9/(3sqrt2)`
`= 3/sqrt2` units
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