Misc 6 - Find a vector of magnitude 5 units, parallel to

Misc 6 - Chapter 10 Class 12 Vector Algebra - Part 2
Misc 6 - Chapter 10 Class 12 Vector Algebra - Part 3

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

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‚.Given ๐‘Ž โƒ— = 2๐‘– ฬ‚ + 3๐‘— ฬ‚ โˆ’ ๐‘˜ ฬ‚(, ๐‘) โƒ— = ๐‘– ฬ‚ โˆ’ 2๐‘— ฬ‚ + ๐‘˜ ฬ‚ Resultant of ๐’‚ โƒ— & ๐’ƒ โƒ— = ๐’‚ โƒ— + ๐’ƒ โƒ— (๐’‚ โƒ— + ๐’ƒ โƒ—) = (2 + 1)๐‘– ฬ‚ + (3 โˆ’ 2)๐‘— ฬ‚ + (โˆ’1 + 1)๐‘˜ ฬ‚ = 3๐’Š ฬ‚ + 1๐’‹ ฬ‚ + 0๐’Œ ฬ‚ Let ๐’„ โƒ— = (๐’‚ โƒ— + ๐’ƒ 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‚ + 5/โˆš10 ๐‘— ฬ‚ + 0๐‘˜ ฬ‚ = 15/โˆš10 ๐‘– ฬ‚ + 5/โˆš10 ๐‘— ฬ‚ Rationalizing = 15/โˆš10 ร— โˆš10/โˆš10 ๐‘– ฬ‚ + 5/โˆš10 "ร— " โˆš10/โˆš10 ๐‘— ฬ‚ = (15โˆš10)/10 ๐‘– ฬ‚ + (5โˆš10)/10 ๐‘— ฬ‚ = (๐Ÿ‘โˆš๐Ÿ๐ŸŽ)/๐Ÿ ๐’Š ฬ‚ + โˆš๐Ÿ๐ŸŽ/๐Ÿ 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 10 years. He provides courses for Maths and Science at Teachoo.