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Find the shortest distance between lines r→=6i^+2j^+2k^+λ(i^-2j^+2k^) and r→=-4i^-k^+μ(3i^-2j^-2k^). - Mathematics

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प्रश्न

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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उत्तर

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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अध्याय 11: Three Dimensional Geometry - Exercise 11.4 [पृष्ठ ४९८]

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एनसीईआरटी Mathematics Part 1 and 2 [English] Class 12
अध्याय 11 Three Dimensional Geometry
Exercise 11.4 | Q 9 | पृष्ठ ४९८

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