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Ex 10.3, 10 If ๐‘Ž โƒ— = 2๐‘– ฬ‚ + 2๐‘— ฬ‚ + 3๐‘˜ ฬ‚, ๐‘ โƒ— = โˆ’๐‘– ฬ‚ + 2๐‘— ฬ‚ + ๐‘˜ ฬ‚ and ๐‘ โƒ— = 3๐‘– ฬ‚ + ๐‘— ฬ‚ are such that ๐‘Ž โƒ— +๐œ†๐‘ โƒ— is perpendicular to ๐‘ โƒ— , then find the value of ๐œ†.๐‘Ž โƒ— = 2๐‘– ฬ‚ + 2๐‘— ฬ‚ + 3๐‘˜ ฬ‚ ๐‘ โƒ— = โˆ’๐‘– ฬ‚ + 2๐‘— ฬ‚ + ๐‘˜ ฬ‚ = โˆ’1๐‘– ฬ‚ + 2๐‘— ฬ‚ + 1๐‘˜ ฬ‚ ๐‘ โƒ— = 3๐‘– ฬ‚ + ๐‘— ฬ‚ = 3๐‘– ฬ‚ + 1๐‘— ฬ‚ + 0๐‘˜ ฬ‚ Now, (๐‘Ž โƒ— + ๐œ†๐‘ โƒ—) = (2๐‘– ฬ‚ + 2๐‘— ฬ‚ + 3๐‘˜ ฬ‚) + ๐œ† (-1๐‘– ฬ‚ + 2๐‘— ฬ‚ + 1๐‘˜ ฬ‚) = 2๐‘– ฬ‚ + 2๐‘— ฬ‚ + 3๐‘˜ ฬ‚ โˆ’ ๐œ†๐‘– ฬ‚ + 2๐œ†๐‘— ฬ‚ + ๐œ†๐‘˜ ฬ‚ = (2 โˆ’ ๐œ†) ๐‘– ฬ‚ + (2 + 2๐œ†) ๐‘— ฬ‚ + (3 + ๐œ†) ๐‘˜ ฬ‚ Since (๐‘Ž โƒ— + ๐œ†๐‘ โƒ—) is perpendicular to ๐‘ โƒ— (๐‘Ž โƒ— + ๐œ†๐‘ โƒ—). ๐‘ โƒ— = 0 [(2โˆ’๐œ†) ๐‘– ฬ‚+(2+2๐œ†) ๐‘— ฬ‚+(3+๐œ†)๐‘˜ ฬ‚ ] . (3๐‘– ฬ‚ + 1๐‘— ฬ‚ + 0๐‘˜ ฬ‚) = 0 (2 โˆ’ ๐œ†).3 + (2 + 2๐œ†).1 + (3 + ๐œ† ).0 = 0 3.2 โˆ’ 3๐œ† + 2 + 2๐œ† + 0 = 0 6 โ€“ 3๐œ† + 2 + 2๐œ† = 0 8 โˆ’ ๐œ† = 0 ๐œ† = 8 โˆด ๐œ† = 8 (Dot product of perpendicular vectors is 0)

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

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo