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

Transcript

Ex 11.2, 15 (Cartesian method) Find the shortest distance between the lines (๐‘ฅ + 1)/7 = (๐‘ฆ + 1)/( โˆ’ 6) = (๐‘ง + 1)/1 and (๐‘ฅ โˆ’ 3)/1 = (๐‘ฆ โˆ’ 5)/( โˆ’ 2) = (๐‘ง โˆ’ 7)/1 Shortest distance between two lines l1: (๐‘ฅ โˆ’ ๐‘ฅ_1)/๐‘Ž_1 = (๐‘ฆ โˆ’ ๐‘ฆ_1)/๐‘_1 = (๐‘ง โˆ’ ๐‘ง_1)/๐‘_1 l2: (๐‘ฅ โˆ’ ๐‘ฅ_2)/๐‘Ž_2 = (๐‘ฆ โˆ’ ๐‘ฆ_2)/๐‘_2 = (๐‘ง โˆ’ ๐‘ง_2)/๐‘_2 is ||โ– 8(๐’™_๐Ÿ โˆ’ ๐’™_๐Ÿ&๐’š_๐Ÿ โˆ’ ๐’š_๐Ÿ&๐’›_๐Ÿ โˆ’ ๐’›_๐Ÿ@๐’‚_๐Ÿ&๐’ƒ_๐Ÿ&๐’„_๐Ÿ@๐’‚_๐Ÿ&๐’ƒ_๐Ÿ&๐’„_๐Ÿ )|/โˆš((๐’‚_๐Ÿ ๐’ƒ_๐Ÿ โˆ’ ๐’‚_๐Ÿ ๐’ƒ_๐Ÿ )^๐Ÿ + (๐’ƒ_๐Ÿ ๐’„_(๐Ÿ )โˆ’ ๐’ƒ_๐Ÿ ๐’„_๐Ÿ )^๐Ÿ + (๐’„_๐Ÿ ๐’‚_๐Ÿ โˆ’ใ€– ๐’„ใ€—_๐Ÿ ๐’‚_๐Ÿ )^๐Ÿ )| (๐’™ + ๐Ÿ)/๐Ÿ• = (๐’š + ๐Ÿ)/( โˆ’ ๐Ÿ”) = (๐’› + ๐Ÿ)/๐Ÿ (๐‘ฅ โˆ’ (โˆ’1))/7 = (๐‘ฆ โˆ’ (โˆ’1))/( โˆ’6) = (๐‘ง โˆ’ (โˆ’1))/1 Comparing with l1: (๐‘ฅ โˆ’ ๐‘ฅ_1)/๐‘Ž_1 = (๐‘ฆ โˆ’ ๐‘ฆ_1)/๐‘_1 = (๐‘ง โˆ’ ๐‘ง_1)/๐‘_1 ๐‘ฅ_1 = โ€“1, ๐‘ฆ_1 = โ€“1, ๐‘ง_1 = โ€“1, & ๐‘Ž_1 = 7, ๐‘_1 = โ€“6, ๐‘_1 = 1, (๐’™ โˆ’ ๐Ÿ‘)/๐Ÿ = (๐’š โˆ’ ๐Ÿ“)/( โˆ’ ๐Ÿ) = (๐’› โˆ’ ๐Ÿ•)/๐Ÿ Comparing with l2: (๐‘ฅ โˆ’ ๐‘ฅ_2)/๐‘Ž_2 = (๐‘ฆ โˆ’ ๐‘ฆ_2)/๐‘_2 = (๐‘ง โˆ’ ๐‘ง_2)/๐‘_2 ๐‘ฅ_2 = 3, ๐‘ฆ_2 = 5, ๐‘ง_2 = 7, & ๐‘Ž_2 = 1, ๐‘_2 = โ€“2, ๐‘_2 = 1, d = ||โ– 8(๐‘ฅ_2โˆ’๐‘ฅ_1&๐‘ฆ_2 โˆ’ ๐‘ฆ_1&๐‘ง_2 โˆ’ ๐‘ง_1@๐‘Ž_1&๐‘_1&๐‘_1@๐‘Ž_2&๐‘_2&๐‘_2 )|/โˆš((๐‘Ž_1 ๐‘_2 โˆ’ ๐‘Ž_2 ๐‘_1 )^2 + (๐‘_1 ๐‘_(2 )โˆ’ ๐‘_2 ๐‘_1 )^2 + (๐‘_1 ๐‘Ž_2 โˆ’ใ€– ๐‘ใ€—_2 ๐‘Ž_1 )^2 )| d = ||โ– 8(3โˆ’(โˆ’1)&5โˆ’(โˆ’1)&7โˆ’(โˆ’1)@7&โˆ’6&1@1&โˆ’2&1)|/โˆš((7(โˆ’2) โˆ’1(โˆ’6))^2 + (โˆ’6(1)โˆ’(โˆ’2)1)^2 + (1(1) โˆ’1(7))^2 )| d = ||โ– 8(4&6&8@7&โˆ’6&1@1&โˆ’2&1)|/โˆš((โˆ’14 + 6)^2 + (โˆ’6 + 2)^2 + (1 โˆ’ 7)^2 )| d = ||โ– 8(4&6&8@7&โˆ’6&1@1&โˆ’2&1)|/โˆš((8)^2 + (โˆ’4)^2 + (โˆ’6)^2 )| d = ||โ– 8(4&6&8@7&โˆ’6&1@1&โˆ’2&1)|/โˆš116| d = |(4(โˆ’6(1) โˆ’ (โˆ’2)1) โˆ’ 6(7(1) โˆ’ 1(1)) + 8(7(โˆ’2) โˆ’ 1(โˆ’6)))/โˆš116| d = |(4(โˆ’6 + 2)โˆ’6(7 โˆ’ 1)+8(โˆ’14 + 6))/โˆš116| d = |(โˆ’16 โˆ’ 36 โˆ’ 64)/โˆš116| d = |(โˆ’116)/โˆš116| d = |โˆ’โˆš116| d = โˆš116 d = โˆš(4 ร— 29) d = ๐Ÿโˆš๐Ÿ๐Ÿ— Ex 11.2, 15 (Vector method) Find the shortest distance between the lines (๐‘ฅ + 1)/7 = (๐‘ฆ + 1)/( โˆ’ 6) = (๐‘ง + 1)/1 and (๐‘ฅ โˆ’ 3)/1 = (๐‘ฆ โˆ’ 5)/( โˆ’ 2) = (๐‘ง โˆ’ 7)/1 Shortest distance between two lines ๐‘Ÿ โƒ— = (๐‘Ž"1" ) โƒ— + ฮป(๐‘"1" ) โƒ— and ๐‘Ÿ โƒ— = (๐‘Ž"2" ) โƒ— + ฮผ(๐‘"2" ) โƒ— is |(((๐’ƒ๐Ÿ) โƒ— ร— (๐’ƒ๐Ÿ) โƒ— ).((๐’‚๐Ÿ) โƒ— ร— (๐’‚๐Ÿ) โƒ— ))/|(๐’ƒ๐Ÿ) โƒ— ร— (๐’ƒ๐Ÿ) โƒ— | | Ex 11.2, 15 (Vector method) Find the shortest distance between the lines (๐‘ฅ + 1)/7 = (๐‘ฆ + 1)/( โˆ’ 6) = (๐‘ง + 1)/1 and (๐‘ฅ โˆ’ 3)/1 = (๐‘ฆ โˆ’ 5)/( โˆ’ 2) = (๐‘ง โˆ’ 7)/1 Shortest distance between two lines ๐‘Ÿ โƒ— = (๐‘Ž"1" ) โƒ— + ฮป(๐‘"1" ) โƒ— and ๐‘Ÿ โƒ— = (๐‘Ž"2" ) โƒ— + ฮผ(๐‘"2" ) โƒ— is |(((๐’ƒ๐Ÿ) โƒ— ร— (๐’ƒ๐Ÿ) โƒ— ).((๐’‚๐Ÿ) โƒ— ร— (๐’‚๐Ÿ) โƒ— ))/|(๐’ƒ๐Ÿ) โƒ— ร— (๐’ƒ๐Ÿ) โƒ— | | (๐’™ + ๐Ÿ )/๐Ÿ• = (๐’š + ๐Ÿ )/(โˆ’๐Ÿ”) = (๐’› + ๐Ÿ )/๐Ÿ (๐‘ฅ โˆ’ (โˆ’1) )/7 = (๐‘ฆ โˆ’ (โˆ’1) )/(โˆ’6) = (๐‘ง โˆ’ (โˆ’1) )/1 Comparing with (๐‘ฅ โˆ’ ๐‘ฅ1 )/๐‘Ž1 = (๐‘ฆ โˆ’ ๐‘ฆ1 )/๐‘1 = (๐‘ง โˆ’ ๐‘ง1 )/๐‘1, ๐‘ฅ1 = โˆ’1, y1 = โˆ’1, ๐‘ง1= โˆ’1 ๐‘Ž1 = 7, b1 = โˆ’ 6, ๐‘1= 1 โˆด (๐‘Ž"1" ) โƒ— = ๐‘ฅ1๐‘– ฬ‚ + ๐‘ฆ1๐‘— ฬ‚ + ๐‘ง1๐‘˜ ฬ‚ = โˆ’1๐‘– ฬ‚ โˆ’ 1๐‘— ฬ‚ โˆ’ 1๐‘˜ ฬ‚ ("b1" ) โƒ— = ๐‘Ž1๐‘– ฬ‚ + ๐‘1๐‘— ฬ‚ + ๐‘1๐‘˜ ฬ‚ = 7๐‘– ฬ‚ โˆ’ 6๐‘— ฬ‚ +1๐‘˜ ฬ‚ (๐’™ โˆ’ ๐Ÿ‘ )/๐Ÿ = (๐’š โˆ’ ๐Ÿ“ )/( โˆ’ ๐Ÿ) = (๐’› โˆ’ ๐Ÿ•)/๐Ÿ Comparing with (๐‘ฅ โˆ’ ๐‘ฅ2 )/๐‘Ž2 = (๐‘ฆ โˆ’ ๐‘ฆ2 )/๐‘2 = (๐‘ง โˆ’ ๐‘ง2 )/๐‘2, ๐‘ฅ2 = 3, y2 = 5, ๐‘ง2= 7 ๐‘Ž2 = 1, b2 = โˆ’ 2, ๐‘2 = 1 โˆด (๐‘Ž"2" ) โƒ— = ๐‘ฅ2๐‘– ฬ‚ + ๐‘ฆ2๐‘— ฬ‚ + ๐‘ง2๐‘˜ ฬ‚ = 3๐‘– ฬ‚ + 5๐‘— ฬ‚ + 7๐‘˜ ฬ‚ ("b2" ) โƒ— = ๐‘Ž2๐‘– ฬ‚ + ๐‘2๐‘— ฬ‚ + ๐‘2๐‘˜ ฬ‚ = 1๐‘– ฬ‚ โˆ’ 2๐‘— ฬ‚ + 1๐‘˜ ฬ‚ Now, ((๐’‚"2" ) โƒ— โˆ’ (๐’‚"1" ) โƒ—) = (3๐‘– ฬ‚ + 5๐‘— + 7๐‘˜ ฬ‚) โˆ’ (โˆ’1๐‘– ฬ‚ โˆ’ 1๐‘— ฬ‚ โˆ’ 1๐‘˜ ฬ‚) = 3๐‘– ฬ‚ + 5๐‘— ฬ‚ + 7๐‘˜ ฬ‚ + 1๐‘– ฬ‚ + 1๐‘— ฬ‚ + 1๐‘˜ ฬ‚ = (3 + 1) ๐‘– ฬ‚ + (5 + 1)๐‘— ฬ‚ + (7 + 1)๐‘˜ ฬ‚ = 4๐’Š ฬ‚ + 6๐’‹ ฬ‚ + 8๐’Œ ฬ‚ (๐’ƒ"1" ) โƒ— ร— (๐’ƒ"2" ) โƒ— = |โ– 8(๐‘– ฬ‚&๐‘— ฬ‚&๐‘˜ ฬ‚@7& โˆ’6&1@1& โˆ’2&1)| = ๐‘– ฬ‚[(โˆ’6ร—1)โˆ’(โˆ’2ร—1)] โˆ’ ๐‘— ฬ‚[(โˆ’7ร—1)โˆ’(1ร—1)] + k[(7ร—โˆ’2)โˆ’(1ร—โˆ’6)] = ๐‘– ฬ‚[โˆ’6+2] โˆ’ ๐‘— ฬ‚ [(7โˆ’1)] + ๐‘˜ ฬ‚ [โˆ’14+6] = โˆ’4๐’Š ฬ‚ โˆ’ 6๐’‹ ฬ‚ โˆ’ 8๐’Œ ฬ‚ Magnitude of ((๐‘"1" ) โƒ—ร—(๐‘"2" ) โƒ— ) = โˆš((โˆ’4)2 + (โˆ’6)2 + (โˆ’8)2) |(๐’ƒ"1" ) โƒ—" " ร—" " (๐’ƒ"2" ) โƒ— | = โˆš116 = โˆš(4 ร— 29) = 2โˆš๐Ÿ๐Ÿ— Also, ((๐’ƒ"1" ) โƒ— ร—" " (๐’ƒ"2" ) โƒ—).((๐’‚"1" ) โƒ—" "โˆ’" " (๐’‚"2" ) โƒ—) = (โˆ’4๐‘– ฬ‚ โˆ’ 6๐‘— ฬ‚ โˆ’ 8๐‘˜ ฬ‚).(4๐‘– ฬ‚ + 6๐‘— ฬ‚ + 8๐‘˜ ฬ‚) = (โˆ’4 ร— 4) + (โˆ’6 ร— 6) + (โˆ’8 + 8) = โˆ’16 + (โˆ’36) + (โˆ’64) = โˆ’116 โˆด Shortest distance = |(((๐‘"1" ) โƒ— ร— (๐‘"2" ) โƒ— ).((๐‘Ž"2" ) โƒ— โˆ’ (๐‘Ž"1" ) โƒ—) )/|(๐‘"1" ) โƒ— ร— (๐‘"2" ) โƒ— | | = |(โˆ’116 )/(2โˆš29)| = |(โˆ’58 )/โˆš29| = |(โˆ’2 ร— 29 )/โˆš29| = ๐Ÿโˆš๐Ÿ๐Ÿ— Therefore, the shortest distance between the two given lines is 2โˆš29.

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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.