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Misc 7 If ๐‘Ž โƒ— = ๐‘– ฬ‚ + ๐‘— ฬ‚ + ๐‘˜ ฬ‚, ๐‘ โƒ— = 2๐‘– ฬ‚ โˆ’๐‘— ฬ‚ + 3๐‘˜ ฬ‚ and ๐‘ โƒ— = ๐‘– ฬ‚ โˆ’ 2๐‘— ฬ‚ + ๐‘˜ ฬ‚ , find a unit vector parallel to the vector 2๐‘Ž โƒ— โ€“ ๐‘ โƒ— + 3๐‘ โƒ— . Given ๐‘Ž โƒ— = ๐‘– ฬ‚ + ๐‘— ฬ‚ + ๐‘˜ ฬ‚ ๐‘ โƒ— = 2๐‘– ฬ‚ + ๐‘— ฬ‚ + 3๐‘˜ ฬ‚ ๐‘ โƒ— = ๐‘– ฬ‚ โˆ’ 2๐‘— ฬ‚ + ๐‘˜ ฬ‚ Let ๐’“ โƒ— = 2๐’‚ โƒ— โˆ’ ๐’ƒ โƒ— + 3๐’„ โƒ— = 2(๐‘– ฬ‚ + ๐‘— ฬ‚ + ๐‘˜ ฬ‚) โˆ’ (2๐‘– ฬ‚ โˆ’ ๐‘— ฬ‚ + 3๐‘˜ ฬ‚) + 3(๐‘– ฬ‚ โˆ’2๐‘— ฬ‚ + ๐‘˜ ฬ‚) = 2๐‘– ฬ‚ + 2๐‘— ฬ‚ + 2๐‘˜ ฬ‚ โˆ’ 2๐‘– ฬ‚ + 1๐‘— ฬ‚ โˆ’ 3๐‘˜ ฬ‚ + 3๐‘– ฬ‚ โˆ’ 6๐‘— ฬ‚ + 3๐‘˜ ฬ‚ = (2 โˆ’ 2 + 3) ๐‘– ฬ‚ + (2 + 1 โˆ’ 6) ๐‘— ฬ‚ + (2 โˆ’ 3 + 3) ๐‘˜ ฬ‚ = 3๐’Š ฬ‚ โ€“ 3๐’‹ ฬ‚ + 2๐’Œ ฬ‚ โˆด ๐’“ โƒ— = 3๐’Š ฬ‚ โ€“ 3๐’‹ ฬ‚ + 2๐’Œ ฬ‚ Magnitude of ๐‘Ÿ โƒ— = โˆš(32+(โˆ’3)2+22) |๐’“ โƒ— | = โˆš(9+9+4) = โˆš๐Ÿ๐Ÿ Unit vector in the direction of ๐‘Ÿ โƒ— = ๐Ÿ/|๐’“ โƒ— | x ๐’“ โƒ— = 1/โˆš22 ร— [3๐‘– ฬ‚ โˆ’3๐‘— ฬ‚+2๐‘˜ ฬ‚ ] = 3/โˆš22 ๐‘– ฬ‚ โ€“ 3/โˆš22 ๐‘— ฬ‚ + 2/โˆš22 ๐‘˜ ฬ‚ Hence the required vector is ๐Ÿ‘/โˆš๐Ÿ๐Ÿ ๐’Š ฬ‚ โ€“ ๐Ÿ‘/โˆš๐Ÿ๐Ÿ ๐’‹ ฬ‚ + ๐Ÿ/โˆš๐Ÿ๐Ÿ ๐’Œ ฬ‚

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

About the Author

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