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Question
Find the shortest distance between the lines `vecr = (4hati - hatj) + lambda(hati+2hatj-3hatk)` and `vecr = (hati - hatj + 2hatk) + mu(2hati + 4hatj - 5hatk)`
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Solution
Shortest distance between two lines = `|((A_2-A_1).(B_1xxB_2))/|B_1xxB_2||``
`A_2 - A_1 = (hati - hatj + 2hatk) - (4hati - hatj) = -3hati + 2hatk`
`B_1 xx B_2 = |(hati,hatj,hatk),(1,2,-3),(2,4,-5)| = hati(-10+12) - hatj(-5+6) + hatk (4-4) = 2hati - hatj`
`(A_2 - A_1).(B_1xxB_2) = (-3hati + 2hatk).(2hati - hatj) = 6`
`|B_1xxB_2| = sqrt(2^2 + (-1)^2) = sqrt5`
∴ Shortest distance between two lines = `|(-6)/(sqrt5)| = 6/sqrt5 = (6sqrt5)/5` units
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