Misc 12 - Find vector d, perpendicular to a, b and c.d = 15

Misc 12 - Chapter 10 Class 12 Vector Algebra - Part 2
Misc 12 - Chapter 10 Class 12 Vector Algebra - Part 3
Misc 12 - Chapter 10 Class 12 Vector Algebra - Part 4
Misc 12 - Chapter 10 Class 12 Vector Algebra - Part 5

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

Transcript

Misc 12 Let ๐‘Ž โƒ— = ๐‘– ฬ‚ + 4๐‘— ฬ‚ + 2๐‘˜ ฬ‚, ๐‘ โƒ— = 3๐‘– ฬ‚ โˆ’ 2๐‘— ฬ‚ + 7๐‘˜ ฬ‚ and ๐‘ โƒ— = 2๐‘– ฬ‚ โˆ’ ๐‘— ฬ‚ + 4๐‘˜ ฬ‚ . Find a vector ๐‘‘ โƒ— which is perpendicular to both ๐‘Ž โƒ— and ๐‘ โƒ— and ๐‘ โƒ— โ‹… ๐‘‘ โƒ— = 15 . Given ๐’‚ โƒ— = ๐‘– ฬ‚ + 4๐‘— ฬ‚ + 2๐‘˜ ฬ‚ ๐’ƒ โƒ— = 3๐‘– ฬ‚ + 2๐‘— ฬ‚ + 7๐‘˜ ฬ‚ ๐’„ โƒ— = 2๐‘– ฬ‚ + ๐‘— ฬ‚ + 4๐‘˜ ฬ‚ Let ๐’… โƒ— = x๐’Š ฬ‚ + y๐’‹ ฬ‚ + z๐’Œ ฬ‚ Since ๐’… โƒ— is perpendicular to ๐‘Ž โƒ— and ๐‘ โƒ— ๐‘‘ โƒ— . ๐‘Ž โƒ— = 0 & ๐‘‘ โƒ— . ๐‘ โƒ— = 0 ๐’… โƒ— . ๐’‚ โƒ— = 0 (x๐‘– ฬ‚ + y๐‘— ฬ‚ + z๐‘˜ ฬ‚). (1๐‘– ฬ‚ + 4๐‘— ฬ‚ + 2๐‘˜ ฬ‚) = 0 (x ร— 1) + (y ร— 4) + (z ร— 2) = 0 x + 4y + 2z = 0 ๐’… โƒ— . ๐’ƒ โƒ— = 0 (x๐‘– ฬ‚ + y๐‘— ฬ‚ + z๐‘˜ ฬ‚). (3๐‘– ฬ‚ โˆ’ 2๐‘— ฬ‚ + 7๐‘˜ ฬ‚) = 0 (x ร— 3) + (y ร— -2) + (z ร— 7) = 0 3x โˆ’ 2y + 7z = 0 Also, ๐‘ โƒ— . ๐‘‘ โƒ— = 15 (2๐‘– ฬ‚ โ€“ 1๐‘— ฬ‚ + 4๐‘˜ ฬ‚). (x๐‘– ฬ‚ + y๐‘— ฬ‚ + z๐‘˜ ฬ‚) = 15 (2 ร— x) + (โˆ’1 ร— y) + (4 ร— 2) = 15 2x โˆ’ y + 4z = 15 Now, we need to solve equations x + 4y + 2z = 0 3x โˆ’ 2y + 7z = 0 & 2x โˆ’ y + 4z = 15 Solving x + 4y + 2z = 0 3x โ€“ 2y + 7z = 0 ๐‘ฅ/(28 โˆ’ (โˆ’4) ) = ๐‘ฆ/(6 โˆ’ 7 ) = ๐‘ง/(โˆ’2 โˆ’ 12 ) ๐’™/(๐Ÿ‘๐Ÿ ) = ๐’š/(โˆ’๐Ÿ ) = ๐’›/(โˆ’๐Ÿ๐Ÿ’ ) Writing x & y in terms of z โˆด x = 32๐‘ง/(โˆ’14) = (โˆ’๐Ÿ๐Ÿ”๐’›)/๐Ÿ• & y = (โˆ’1๐‘ง)/(โˆ’14) = ๐’›/๐Ÿ๐Ÿ’ Putting values of x and y in (3) 2x โ€“ y + 4z = 15 2 ((โˆ’16๐‘ง)/7 ) โˆ’ (๐‘ง/14 ) + 4z = 5 (โˆ’32)/7 z โˆ’ 1/14 z + 4/1 z = 15 ((โˆ’64 โˆ’ 1 + 56))/14 z = 15 (โˆ’9)/14 z = 15 z = 15 ร— 14/(โˆ’9) z = (โˆ’๐Ÿ•๐ŸŽ)/๐Ÿ‘ Putting value of z in x & y, x = (โˆ’16๐‘ง)/7 = (โˆ’16)/7 ร— (โˆ’70)/3 = ๐Ÿ๐Ÿ”๐ŸŽ/๐Ÿ‘ y = ๐‘ง/14 = 1/14 ร— (โˆ’70)/3 = (โˆ’๐Ÿ“)/๐Ÿ‘ Therefore, the required vector ๐‘‘ โƒ— = x๐‘– ฬ‚ + y๐‘— ฬ‚ + z๐‘˜ ฬ‚ = 160/3 ๐‘– ฬ‚ โˆ’ 5/3 ๐‘— ฬ‚ โ€“ 70/3 ๐‘˜ ฬ‚ = ๐Ÿ/๐Ÿ‘ (160๐’Š ฬ‚ โˆ’ 5๐’‹ ฬ‚ โ€“ 70๐’Œ ฬ‚) Note: Answer given in the book is incorrect If we have made any mistake, please email at admin@teachoo.com

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.