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

Transcript

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)

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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.