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Find the shortest distance between the following lines: λr→=3i^+5j^+7k^+λ(i^-2j^+k^) and μr→=(-i^-j^-k^)+μ(7i^-6j^+k^).

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Question

Find the shortest distance between the following lines:

`vecr = 3hati + 5hatj + 7hatk + λ(hati - 2hatj + hatk)` and `vecr = (-hati - hatj - hatk) + μ(7hati - 6hatj + hatk)`.

Sum
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Solution

Given lines are: `vecr = 3hati + 5hatj + 7hatk + λ(hati - 2hatj + hatk)`

and `vecr = (-hati - hatj - hatk) + μ(7hati - 6hatj + hatk)`

Let the given lines be `vecr = veca_1 + λvecb_2` and `vecr = veca_2 + λvecb_2`

Shortest distance between two lines

d = `|((veca_2 - veca_1).(vecb_1 xx vecb_2))/|vecb_1 xx vecb_2||`

∴ `veca_2 - veca_1 = (-hati - hatj - hatk) - (3hati + 5hatj + 7hatk)`

= `-4hati - 6hatj - 8hatk`

`vecb_1 xx vecb_2 = |(hati, hatj, hatk),(1, -2, 1),(7, -6, 1)|`

= `hati(-2 + 6) - hatj(1 - 7) + hatk(-6 + 14)`

= `4hati + 6hatj + 8hatk`

∴ `|vecb_1 xx vecb_2| = sqrt(4^2 + 6^2 + 8^2)`

= `sqrt(16 + 36 + 64)`

= `sqrt(116)`

Therefore, d = `|((-4hati - 6hatj - 8hatk).(4hati + 6hatj + 8hatk))/sqrt(116)|`

= `|(-16 - 36 - 64)/sqrt(116)|`

= `|(-116)/sqrt(116)|`

= `sqrt(116)` units

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2021-2022 (March) Term 2 - Outside Delhi Set 3

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