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Question
Find the shortest distance between the lines whose vector equations are `vecr = (hati + 2hatj + 3hatk) + lambda(hati - 3hatj + 2hatk)` and `vecr = 4hati + 5hatj + 6hatk + mu(2hati + 3hatj + hatk)`.
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Solution
Compare the given equations
`vecr = vec(a_1) + λvec(b_1)` and `vecr = vec(a_2) + µvec(b_2)` respectively, we have `vec(a_1) = hati + 2hatj + 3hatk, vec(b_1) = hati - 3hatj + 2hatk`
`vec(a_2) = 4hati + 5hatj + 6hatk` and `vec(b_2) = 2hati + 3hatj + hatk`
Now, `vec(a_2) - vec(a_1) = 3hati + 3hatj + 3hatk`
And `vec(b_1) xx vec(b_2) = |(hati, hatj, hatk), (1, -3, 2), (2, 3, 1)|`
= `-9hati + 3hatj + 9hatk`
`|vec(b_1) xx vec(b_2)| = 3sqrt19`
∴ `(vec(a_2) - vec(a_1)). (vec(b_1) xx vec(b_2)) = 9`
d = `|((vec(a_2) - vec(a_1)). (vec(b_1) xx vec(b_2)))/|vec(b_1) xx vec(b_2)||`
`= |(-9 xx 3 + 3 xx 3 + 9 xx3)/sqrt((-9)^2 + 3^2 + 9^2)|`
`= |9/ (sqrt(3^2) sqrt(3^2 + 1 + 3^2))|`
= `9/(3sqrt19)`
= `3/sqrt19`
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