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Question
Show that the lines `vec"r" = (6hat"i" + hat"j" + 2hat"k") + "s"(hat"i" + 2hat"j" - 3hat"k")` and `vec"r" = (3hat"i" + 2hat"j" - 2hat"k") + "t"(2hat"i" + 4hat"j" - 5hat"k")` are skew lines and hence find the shortest distance between them
Solution
`vec"r" = vec"a" + "s"vec"b"`, `vec"r" = vec"c" + "s"vec"d"`
`vec"a" = 6hat"i" + hat"j" + 2hat"k"`, `vec"b" = hat"i" + 2hat"j" - 3hat"k"`
`vec"c" = 3hat"i" + 2hat"j" - 2hat"k"`, `vec"d" = 2hat"i" + 4hat"j" - 5hat"k"`
`vec"b"` is not a scalar multiple of `vec"d"`
∴ They are not parallel.
∴ The given lines are skew lines.
The shortest distance δ = `(|(vec"c" - vec"a")*(vec"b" xx vec"d")|)/(|vec"b" xx vec"d"|)`
`vec"b" xx vec"d" = |(hat"i", hat"j", hat"k"),(1, 2, -3),(2, 4, 5)|`
= `"i"(- 10 + 12) - hat"j"(- 5 + 6) + hat"k"(4 - 4)`
= `2hat"i" - hat"j"`
`|vec"b" xx vec"d"| = sqrt(2^2 + 1^2) = sqrt(5)`
`vec"c" - vec"a" = 3hat"i" + 2hat"j" - 2hat"k" - 6hat"i" - hat"j" - 2hat"k"`
= `3hat"i" + hat"j" - 4hat"k"`
`("c" - "a")*(vec"b" xx vec"d") = (-3hat"i" + hat"j" - 4hat"k")*(2hat"i" - hat"j")`
= – 6 – 1 + 0
= – 7
δ = `(|(vec"c" - vec"a")*(vec"b" xx vec"d")|)/(|vec"b" xx vec"d"|)`
= `|-7|/sqrt(5)`
= `7/sqrt(5)` units
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