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An insect is crawling along the line λr¯=6i^+2j^+2k^+λ(i^-2j^+2k^) and another insect is crawling along the line μr¯=-4i^-k^+μ(3i^-2j^-2k^). - Mathematics

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Question

An insect is crawling along the line `barr = 6hati + 2hatj + 2hatk + λ(hati - 2hatj + 2hatk)` and another insect is crawling along the line `barr = - 4hati - hatk + μ(3hati - 2hatj - 2hatk)`. At what points on the lines should they reach so that the distance between them s the shortest? Find the shortest possible distance between them.

Sum

Solution

The given lines are non-parallel lines. There is a unique line segment PQ (P lying on one and Q on the other) at right angles to both lines. PQ is the shortest distance between the lines. Hence, the shortest possible distance between the insects = PQ

The position vector of P lying on the line

`barr = 6hati + 2hatj + 2hatk + λ(hati - 2hatj + 2hatk)` is `(6 + λ)hati + (2 - 2λ)hatj + (2 + 2λ)hatk` for some λ

The position vector of Q lying on the line

`vecr = - 4hati - hatk + μ(3hati - 2hatj - 2hatk)` is `(-4 + 3μ)hati + (-2μ)hatj + (-1 - 2μ)hatk` for some μ

`vec(PQ) = (- 10 + 3μ - λ)hati + (-2μ - 2 + 2λ)hatj + (-3 - 2μ - 2λ)hatk`

Since PQ is perpendicular to both lines

`(-10 + 3μ - λ) + (-2μ - 2 + 2λ)(-2) + (-3 - 2μ - 2λ)2` = 0,

i.e., μ – 3λ = 4  ...(i)

And (–10 + 3μ – λ)3 + (–2μ –2 + 2λ)(–2) + (–3 – 2μ – 2λ)(–2) = 0,

i.e., 17μ – 3λ = 20  ...(ii) 

Solving (i) and (ii) for λ and μ, we get = 1, 1 = –1.

The position vector of the points, at which they should be so that the distance between them is the shortest, is `5hati + 4hatj` and `-hati - 2hatj - 3hatk`

`vec(PQ) = - 6hati - 6hatj - 3hatk`

The shortest distance = `|vec(PQ)|`

= `sqrt(6^2 + 6^2 + 3^2)`

= 9

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