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Question
Show that the straight lines x + 1 = 2y = – 12z and x = y + 2 = 6z – 6 are skew and hence find the shortest distance between them
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Solution
x + 1 = 2y = – 12z
x + 1 = `y/(1/2) = z/((-1)/12)`
x = y + 2 = 6z – 6
x = y + 2 = `(z - 1)/(1/6)`
(x1, y1, z1) = (- 1, 0, 0) and (x2, y2, z2)
= (0, - 2, 1)
(b1, b2, b3) = `(1, 1/2, (-1)/12)` and (d1, d2, d3) = `(1, 1, 1/6)`
Condition for skew lines
`|(x_2 - x_1, y_2 - y_1, z_2 - z_1),("b"_1, "b"_2, "b"_3),("d"_1, "d"_2, "d"_3)| ≠ 0`
OR
`[(vec"c" - vec"a") vec"b" vec"d"] ≠ 0`
`[(vec"c" - vec"a") vec"b" vec"d"] = |(1, -2, 1),(1, 1/2, (-1)/12),(1, 1, 1/6)|`
= `1[1/12 + 1/12] + 2[1/6 + 1/12] + 1[1 - 1/2]`
= `1/6 + 2/6 + 2/12 + 1 - 1/2`
= `1/6 + 2/6 + 1/6 + 1 - 1/2`
= `2/3 + 1/2 = 7/6 ≠ 0``
Given Lines are skew lines
`vec"b" xx vec"d" = |(vec"i", vec"j", vec"k"),(1, 1/2, - 1/12),(1, 1, 1/6)|`
= `vec"i"(1/6) - vec"j"(3/12) + vec"k"(1/2)`
= `1/6 vec"i" - 1/4 vec"j" + 1/2 vec"k"`
`|vec"b" xx vec"d"| = sqrt(1/36 + 1/16 + 1/4)`
= `sqrt((16 + 36 + 144)/576`
= `sqrt(196/576)`
= `14/24`
= `7/12`
Shortest distance between skew lines
= `(|[vec"c" - vec"a" vec"b" vec"d"]|)/|vec"b" xx vec"d"|`
= `(7/6)/(7/12)`
= `7/6 xx 12/7`
= 2 units
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