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  1. Chapter 11 Class 12 Three Dimensional Geometry
  2. Serial order wise

Transcript

Ex 11.2, 16 Find the shortest distance between the lines whose vector equations are ๐‘Ÿ โƒ— = (๐‘– ฬ‚ + 2๐‘— ฬ‚ + 3๐‘˜ ฬ‚) + ๐œ† (๐‘– ฬ‚ โ€“ 3๐‘— ฬ‚ + 2๐‘˜ ฬ‚) and ๐‘Ÿ โƒ— = (4๐‘– ฬ‚ + 5๐‘— ฬ‚ + 6๐‘˜ ฬ‚) + ๐œ‡ (2๐‘– ฬ‚ + 3๐‘— ฬ‚ + ๐‘˜ ฬ‚)Shortest distance between the lines with vector equations ๐‘Ÿ โƒ— = (๐‘Ž_1 ) โƒ— + ๐œ† (๐‘"1" ) โƒ— and ๐‘Ÿ โƒ— = (๐‘Ž"2" ) โƒ— + ๐œ‡(๐‘"2" ) โƒ— is |(((๐’ƒ๐Ÿ) โƒ— ร— (๐’ƒ๐Ÿ) โƒ— ).((๐’‚๐Ÿ) โƒ— โˆ’ (๐’‚๐Ÿ) โƒ— ))/|(๐’ƒ๐Ÿ) โƒ— ร— (๐’ƒ๐Ÿ) โƒ— | | Given, ๐’“ โƒ— = (๐’Š ฬ‚ + 2๐’‹ ฬ‚ + 3๐’Œ ฬ‚) + ๐œ† (๐’Š ฬ‚ โˆ’ 3๐’‹ ฬ‚ + 2๐’Œ ฬ‚) Comparing with ๐‘Ÿ โƒ— = (๐‘Ž1) โƒ— + ๐œ† (๐‘1) โƒ—, (๐‘Ž1) โƒ— = 1๐‘– ฬ‚ + 2๐‘— ฬ‚ + 3๐‘˜ ฬ‚ & (๐‘1) โƒ— = 1๐‘– ฬ‚ โˆ’ 3๐‘— ฬ‚ + 2๐‘˜ ฬ‚ ๐’“ โƒ— = (4๐’Š ฬ‚ + 5๐’‹ ฬ‚ + 6๐’Œ ฬ‚) + ๐(2๐’Š ฬ‚ + 3๐’‹ ฬ‚ + ๐’Œ ฬ‚) Comparing with ๐‘Ÿ โƒ— = (๐‘Ž2) โƒ— + ๐œ‡(๐‘2) โƒ—, (๐‘Ž2) โƒ— = 4๐‘– ฬ‚ + 5๐‘— ฬ‚ + 6๐‘˜ ฬ‚ & (๐‘2) โƒ— = 2๐‘– ฬ‚ + 3๐‘— ฬ‚ + 1๐‘˜ ฬ‚ Now, ((๐’‚๐Ÿ) โƒ— โˆ’ (๐’‚๐Ÿ) โƒ—) = (4๐‘– ฬ‚ + 5๐‘— ฬ‚ + 6๐‘˜ ฬ‚) โˆ’ (1๐‘– ฬ‚ + 2๐‘— ฬ‚ + 3๐‘˜ ฬ‚) = (4 โˆ’ 1)๐‘– ฬ‚ + (5 โˆ’ 2) ๐‘— ฬ‚ + (6 โˆ’ 3) ๐‘˜ = 3๐’Š ฬ‚ + 3๐’‹ ฬ‚ + 3๐’Œ ฬ‚ ((๐’ƒ๐Ÿ) โƒ— ร— (๐’ƒ๐Ÿ) โƒ—) = |โ– 8(๐‘– ฬ‚&๐‘— ฬ‚&๐‘˜ ฬ‚@1& โˆ’3&2@2&3&1)| = ๐‘– ฬ‚ [(โˆ’3ร—1)โˆ’(3ร—2)] โˆ’ ๐‘— ฬ‚ [(1ร—1)โˆ’(2ร—2)] + ๐‘˜ ฬ‚ [(1ร—3)โˆ’(2ร—โˆ’3)] = ๐‘– ฬ‚ [โˆ’3โˆ’6] โˆ’ ๐‘— ฬ‚ [1โˆ’4] + ๐‘˜ ฬ‚ [3+6] = ๐‘– ฬ‚ (โˆ’9) โˆ’ ๐‘— ฬ‚ (โˆ’3) + ๐‘˜ ฬ‚(9) = โˆ’ 9๐’Š ฬ‚ + 3๐’‹ ฬ‚ + 9๐’Œ ฬ‚ Magnitude of ((๐‘1) โƒ— ร— (๐‘2) โƒ—) = โˆš((โˆ’ 9)2+32+92) |(๐’ƒ๐Ÿ) โƒ—" ร— " (๐’ƒ๐Ÿ) โƒ— | = โˆš(81+9+81) = โˆš171 = โˆš(9ร—19 ) = 3โˆš๐Ÿ๐Ÿ— Also, ((๐’ƒ๐Ÿ) โƒ— ร— (๐’ƒ๐Ÿ) โƒ—) . ((๐’‚๐Ÿ) โƒ— โ€“ (๐’‚๐Ÿ) โƒ—) = (โˆ’9๐‘– ฬ‚ + 3๐‘— ฬ‚ + 9๐‘˜ ฬ‚).(3๐‘– ฬ‚ + 3๐‘— ฬ‚ + 3๐‘˜ ฬ‚) = (โˆ’9 ร— 3) + (3 ร— 3) + (9 ร— 3) = โˆ’27 + 9 + 27 = 9 So, shortest distance = |("(" (๐‘1) โƒ—ร—" " (๐‘2) โƒ—")" ."(" (๐‘Ž2) โƒ— โˆ’" " (๐‘Ž1) โƒ—")" )/|(๐‘1) โƒ—ร— (๐‘2) โƒ— | | = |9/(3โˆš19)| = ๐Ÿ‘/โˆš๐Ÿ๐Ÿ— Therefore, shortest distance between the given two lines is 3/โˆš19.

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.