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Find the shortest distance between the following lines: r→=(i^+j^-k^)+s(2i^+j^+k^) r→=(i^+j^-2k^)+t(4i^+2j^+2k^) - Mathematics

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Question

Find the shortest distance between the following lines:

`vecr = (hati + hatj - hatk) + s(2hati + hatj + hatk)`

`vecr = (hati + hatj - 2hatk) + t(4hati + 2hatj + 2hatk)`

Sum
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Solution

Here, the lines are parallel.

The shortest distance = `(|(veca_2 - veca_1) xx vecb|)/|vecb|`

= `(|(3hatk) xx (2hati + hatj + hatk)|)/sqrt(4 + 1 + 1)`

`(3hatk) xx (2hati + hatj + hatk) = |(hati, hatj, hatk),(0, 0, 3),(2, 1, 1)| = -3hati + 6hatj`

Hence, the required shortest distance = `(3sqrt(5))/sqrt(6)` units.

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2021-2022 (March) Term 2 Sample

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