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  1. Class 12
  2. Important Question for exams Class 12

Transcript

Misc 6 Find a vector of magnitude 5 units, and parallel to the resultant of the vectors ๐‘Ž โƒ— = 2๐‘– ฬ‚ + 3๐‘— ฬ‚ โˆ’ ๐‘˜ ฬ‚ and ๐‘ โƒ— = ๐‘– ฬ‚ โˆ’ 2๐‘— ฬ‚ + ๐‘˜ ฬ‚. ๐‘Ž โƒ— = 2๐‘– ฬ‚ + 3๐‘— ฬ‚ โˆ’ ๐‘˜ ฬ‚ ๐‘ โƒ— = ๐‘– ฬ‚ โˆ’ 2๐‘— ฬ‚ + ๐‘˜ ฬ‚ (๐‘Ž โƒ— + ๐‘ โƒ—) = (2 + 1)๐‘– ฬ‚ + (3 โˆ’ 2)๐‘— ฬ‚ + (โˆ’1 + 1)๐‘˜ ฬ‚ = 3๐‘– ฬ‚ + 1๐‘— ฬ‚ + 0๐‘˜ ฬ‚ Let ๐‘ โƒ— = (๐‘Ž โƒ— + ๐‘ โƒ—) โˆด ๐‘ โƒ— = 3๐‘– ฬ‚ + 1๐‘— ฬ‚ + 0๐‘˜ ฬ‚ Magnitude of ๐‘ โƒ— = โˆš(32+12+02) |๐‘ โƒ— | = โˆš(9+1) = โˆš10 Unit vector in direction of ๐‘ โƒ— = 1/|๐‘ โƒ— | ร— ๐‘ โƒ— ๐‘ ฬ‚ = 1/โˆš10 ร— [3๐‘– ฬ‚+1๐‘— ฬ‚+0๐‘˜ ฬ‚ ] ๐‘ ฬ‚ = ๐Ÿ‘/โˆš๐Ÿ๐ŸŽ ๐’Š ฬ‚ + ๐Ÿ/โˆš๐Ÿ๐ŸŽ ๐’‹ ฬ‚ + 0๐’Œ ฬ‚ Vector with magnitude 1 = 3/โˆš10 ๐‘– ฬ‚ + 1/โˆš10 ๐‘— ฬ‚ + 0๐‘˜ ฬ‚ Vector with magnitude 5 = 5 ร— [3/โˆš10 " " ๐‘– ฬ‚" + " 1/โˆš10 ๐‘— ฬ‚" + 0" ๐‘˜ ฬ‚ ] = 15/โˆš10 ๐‘– ฬ‚ + 2/โˆš10 ๐‘— ฬ‚ + 0๐‘˜ ฬ‚ = 15/โˆš10 ๐‘– ฬ‚ + 2/โˆš10 ๐‘— ฬ‚ Rationalizing = 15/โˆš10 ร— โˆš10/โˆš10 ๐‘– ฬ‚ + 2/โˆš10 "ร— " โˆš10/โˆš10 ๐‘— ฬ‚ = (15โˆš10)/10 ๐‘– ฬ‚ + (2โˆš10)/10 ๐‘— ฬ‚ = (3โˆš10)/2 ๐‘– ฬ‚ + โˆš10/5 ๐‘— ฬ‚ Hence the required vector is (๐Ÿ‘โˆš๐Ÿ๐ŸŽ)/๐Ÿ ๐’Š ฬ‚ + โˆš๐Ÿ๐ŸŽ/๐Ÿ“ ๐’‹ 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.