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

Transcript

Example 30 If with reference to the right handed system of mutually perpendicular unit vectors ๐‘– ฬ‚, ๐‘— ฬ‚ and ๐‘˜ ฬ‚, "ฮฑ" โƒ— = 3๐‘– ฬ‚ โˆ’ ๐‘— ฬ‚, "ฮฒ" โƒ— = 2๐‘– ฬ‚ + ๐‘— ฬ‚ โ€“ 3๐‘˜ ฬ‚, then express "ฮฒ" ฬ‚ in the form "ฮฒ" โƒ— = "ฮฒ" โƒ—1 + "ฮฒ" โƒ—2, where "ฮฒ" โƒ—1 is parallel to "ฮฑ" โƒ— and "ฮฒ" โƒ—2 is perpendicular to "ฮฑ" โƒ—.๐›ผ โƒ— = 3๐‘– ฬ‚ โˆ’ ๐‘— ฬ‚ = 3๐‘– ฬ‚ โˆ’ ๐‘— ฬ‚ + 0๐‘˜ ฬ‚ "ฮฒ" โƒ— = 2๐‘– ฬ‚ + ๐‘— ฬ‚ โˆ’ 3๐‘˜ ฬ‚ = 2๐‘– ฬ‚ + 1๐‘— ฬ‚ โˆ’ 3๐‘˜ ฬ‚ To show: "ฮฒ" โƒ— = "ฮฒ" โƒ—1 + "ฮฒ" โƒ—2 Given, "ฮฒ" โƒ—1 is parallel to ๐›ผ โƒ— & "ฮฒ" โƒ—2 is perpendicular to ๐›ผ โƒ— Let "ฮฒ" โƒ—1 = ๐€๐œถ โƒ— , ๐œ† being a scalar. "ฮฒ" โƒ—1 = ๐œ† (3๐‘– ฬ‚ โˆ’ 1๐‘— ฬ‚ + 0๐‘˜ ฬ‚) = 3๐œ† ๐‘– ฬ‚ โˆ’ ๐œ†๐‘— ฬ‚ + 0๐‘˜ ฬ‚ Now, "ฮฒ" โƒ—2 = "ฮฒ" โƒ— โˆ’ "ฮฒ" โƒ—1 = ["2" ๐‘– ฬ‚" + 1" ๐‘— ฬ‚" โˆ’ 3" ๐‘˜ ฬ‚ ] โˆ’ ["3" ๐œ†๐‘– ฬ‚" โˆ’ ๐œ†" ๐‘— ฬ‚" + 0" ๐‘˜ ฬ‚ ] = 2๐‘– ฬ‚ + 1๐‘— ฬ‚ โˆ’ 3๐‘˜ ฬ‚ โˆ’ 3๐œ†๐‘– ฬ‚ + ๐œ†๐‘— ฬ‚ + 0๐‘˜ ฬ‚ = (2 โˆ’ 3๐œ†) ๐‘– ฬ‚ + (1 + ๐œ†) ๐‘— ฬ‚ โˆ’ 3๐‘˜ ฬ‚ Also, since "ฮฒ" โƒ—2 is perpendicular to ๐›ผ โƒ— So, "ฮฒ" โƒ—2 . ๐œถ โƒ— = 0 ["(2 โˆ’ 3๐œ†) " ๐‘– ฬ‚" + (1 + ๐œ†) " ๐‘— ฬ‚" โˆ’ 3" ๐‘˜ ฬ‚ ]. (3๐‘– ฬ‚ โˆ’ 1๐‘— ฬ‚ + 0๐‘˜ ฬ‚) = 0 (2 โˆ’ "3๐œ†") ร— 3 + (1 + ๐œ†) ร— โˆ’1 + (โˆ’3) ร— 0 = 0 6 โˆ’ 9"๐œ†" โˆ’ 1 โˆ’ ๐œ† = 0 5 โˆ’ 10๐œ† = 0 10๐œ† = 5 ๐œ† = 5/10 ๐œ† = ๐Ÿ/๐Ÿ Putting value of ๐œ† in "ฮฒ" โƒ—1 and "ฮฒ" โƒ—2 , "ฮฒ" โƒ—1 = 3๐œ†๐‘– ฬ‚ โˆ’ ๐œ†๐‘— ฬ‚ + 0๐‘˜ ฬ‚ = 3. 1/2 ๐‘– ฬ‚ โˆ’ 1/2 ๐‘— ฬ‚ + 0 ๐‘˜ ฬ‚ = ๐Ÿ‘/๐Ÿ ๐’Š ฬ‚ โˆ’ ๐Ÿ/๐Ÿ ๐’‹ ฬ‚ "ฮฒ" โƒ—2 = (2 โˆ’ 3๐œ†) ๐‘– ฬ‚ + (1 + ๐œ†) ๐‘— ฬ‚ โˆ’ 3๐‘˜ ฬ‚ = ("2 โˆ’ 3. " 1/2) ๐‘– ฬ‚ + (1+1/2) ๐‘— ฬ‚ โˆ’ 3๐‘˜ ฬ‚ = ๐Ÿ/๐Ÿ ๐’Š ฬ‚ + ๐Ÿ‘/๐Ÿ ๐’‹ ฬ‚โˆ’ 3๐’Œ ฬ‚

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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.