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  1. Chapter 10 Class 12 Vector Algebra
  2. Serial order wise

Transcript

Misc 10 The two adjacent sides of a parallelogram are 2๐‘– ฬ‚ โˆ’ 4๐‘— ฬ‚ + 5๐‘˜ ฬ‚ and ๐‘– ฬ‚ โˆ’ 2๐‘— ฬ‚ โˆ’ 3๐‘˜ ฬ‚ Find the unit vector parallel to its diagonal. Also, find its area. Let ๐‘Ž โƒ— and ๐‘ โƒ— are adjacent side of a parallelogram, where ๐‘Ž โƒ— = 2๐‘– ฬ‚ โˆ’ 4๐‘— ฬ‚ + 5๐‘˜ ฬ‚ ๐‘ โƒ— = ๐‘– ฬ‚ โˆ’ 2๐‘— ฬ‚ โˆ’ 3๐‘˜ ฬ‚ Let diagonal be ๐‘ โƒ— Hence, ๐’„ โƒ— = ๐’‚ โƒ— + ๐’ƒ โƒ— = (2๐‘– ฬ‚ โˆ’ 4๐‘— ฬ‚ + 5๐‘˜ ฬ‚) + (1๐‘– ฬ‚ โˆ’ 2๐‘— ฬ‚ โˆ’ 3๐‘˜ ฬ‚) = (2 + 1) ๐‘– ฬ‚ โˆ’ (4 + 2) ๐‘— ฬ‚ + (5 โˆ’ 3)๐‘˜ ฬ‚ = 3๐’Š ฬ‚ โˆ’ 6๐’‹ ฬ‚ + 2๐’Œ ฬ‚ Magnitude of โŒˆ๐‘ โƒ— โŒ‰ = โˆš((3)^2+(โˆ’6)^2+(2)^2 ) = โˆš(9+36+4) = โˆš49 = 7 Unit vector in direction of ๐‘ โƒ— = 1/(๐‘š๐‘Ž๐‘”๐‘›๐‘–๐‘ก๐‘ข๐‘‘๐‘’ ๐‘œ๐‘“ ๐‘ โƒ— ) ร— ๐‘ โƒ— ๐’„ ฬ‚ = ๐Ÿ/๐Ÿ• ("3" ๐’Š ฬ‚" โˆ’ 6" ๐’‹ ฬ‚" + 2" ๐’Œ ฬ‚ ) Area of parallelogram = |๐‘Ž โƒ— ร— ๐‘ โƒ— | So ๐‘Ž โƒ—ร—๐‘ โƒ— = |โ– 8(๐‘– ฬ‚&๐‘— ฬ‚&๐‘˜ ฬ‚@2&โˆ’4&5@1&โˆ’2&โˆ’3)| = ๐‘– ฬ‚ [(โˆ’3ร—โˆ’4)โˆ’(โˆ’2ร—5)] โˆ’ ๐‘— ฬ‚ [(2ร—โˆ’3)โˆ’(1ร—5)] + ๐‘˜ ฬ‚ [(โˆ’2ร—2)โˆ’(โˆ’4ร—1)] = ๐‘– ฬ‚ (12 โˆ’ (โˆ’10) โˆ’ ๐‘— ฬ‚ (โˆ’6 โˆ’5) + ๐‘˜ ฬ‚ (โˆ’4 โˆ’ (โˆ’4)) = ๐‘– ฬ‚ (12 + 10) โˆ’ ๐‘— ฬ‚ (โˆ’11) + ๐‘˜ ฬ‚ (โˆ’4 + 4) = 22๐‘– ฬ‚ + 11๐‘— ฬ‚ + 0๐‘˜ ฬ‚ = 22๐‘– ฬ‚ + 11๐‘— ฬ‚ So ๐’‚ โƒ— ร— ๐’ƒ โƒ— = 22๐’Š ฬ‚ + 11๐’‹ ฬ‚ |๐‘Ž โƒ— ร— ๐‘ โƒ— | = โˆš(ใ€–22ใ€—^2+ใ€–11ใ€—^2 ) = โˆš(2^2 (ใ€–11)ใ€—^2+ใ€–11ใ€—^2 ) = โˆš( ใ€–11ใ€—^2 (2^2+1)) = 11โˆš5 Hence, Area of parallelogram = |๐‘Ž โƒ—ร—๐‘ โƒ— |= ๐Ÿ๐Ÿโˆš๐Ÿ“

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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.