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Question
Find the shortest distance between lines `vecr = 6hati + 2hatj + 2hatk + lambda(hati - 2hatj + 2hatk)` and `vecr =-4hati - hatk + mu(3hati - 2hatj - 2hatk)`.
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Solution
If two points, a2 on the line and b1 on the line, and 22 is their direction, then the shortest distance between them is
If two points `vec(a_1), vec(a_2)` lie on a line and `vec(b_1)` and `vec(b_2`
their direction then minimum distance between them = `|((vec(a_1) - vec(a_2)). (vec(b_1) xx vec(b_2)))/|vec(b_1) xx vec(b_2)||`
Given lines, `vecr = 6hati + 2hatj + 2hatk + λ(hati - 2hatj+ 2hatk)`
And `vecr = -4hati - hatk + µ(3hati - 2hatj+ 2hatk)`
If `vec(a_1) = 6hati + 2hatj + 2hatk`, `vec(a_2) = -4hati - hatk`
`vec(b_1) = hati - 2hatj + 2hatk`, `vec(b_2) = 3hati - 2hatk -2hatk`
`vec(a_1) - vec(a_2) = (6hati + 2hatj + 2hatk) - (-4hati - hatk)`
= `10hati + 2hatj + 3hatk`
`vec(b_1) xx vec(b_2) = |(hati, hatj, hatk), (1, -2, 2), (3, -2, -2)|`
= `8hati + 8hatj + 4hatk`
`|vec(b_1) xx vec(b_2)|`
= `sqrt(8^2 + 8^2 + 4^2)`
= 12
∴ S.D. = `((vec(a_1) - vec(a_2)). (vec(b_1) xx vec(b_2)))/|vec(b_1) xx vec(b_2)|`
= `((10hati + 2hatj + 3hatk).(8hati + 8hatj + 4hatk))/12`
= `(80 + 16 + 12)/12`
= `108/12`
= 9
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