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Question
Find the distance between the parallel lines `x/2 = y/-1 = z/2` and `(x - 1)/2 = (y - 1)/-1 = (z - 1)/2`
Solution
Line `x/2 = y/-1 = z/2` passes through (0, 0, 0) and hos direction ratios 2, –1, 2
∴ Vector equation of the line is: `vecr = (0hati + 0hatj + 0hatk) + λ (2hati - hatj + 2hatk)`
i.e., `vecr = λ(2hati - hatj + 2hatk)`
Now line, `(x - 1)/2 = (y - 1)/-1 = (z - 1)/2`
Passes through (1, 1, 1) and hos direction ratios 2, –1, 2
∴ Vector equation of the line is: `vecr = λ(2hati - hatj + 2hatk)`
The distance between parallel is: `vecr = veca_1 + λvecb` and `vecr = veca_2 + λvecb` is `|(veca_2 - veca_1) xx hatb|`
Here, `veca_1 = 0, veca_2 = hati + hatj + hatk, vecb = 2hati - hatj + 2hatk`
∴ `hatb = vecb/|vecb| = (2hati - hatj + 2hatk)/sqrt(2^2 + (-1)^2 + (2)^2`
= `2/3hati - 1/3hatj + 2/3hatk`
∴ `(veca_2 - veca_1) xx vecb = |(hati, hatj, hatk),(1, 1, 1),(2/3, (-1)/3, 2/3)|`
= `hati (2/3 + 1/3) - hatj(2/3 - 2/3) + hatk(-1/3 - 2/3)`
= `hati - hatk`
Thus distance = `|(veca_2 - veca_1) xx hatb| = sqrt(1^2 + (-1)^2) = sqrt(2)` units.
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