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Misc 17 Let ๐‘Ž โƒ— and ๐‘ โƒ— be two unit vectors and ฮธ is the angle between them. Then ๐‘Ž โƒ— + ๐‘ โƒ— is a unit vector if (A) ฮธ = ๐œ‹/4 (B) ฮธ = ๐œ‹/3 (C) ฮธ = ๐œ‹/2 (D) ฮธ = 2๐œ‹/3 Given ๐‘Ž โƒ— & ๐‘ โƒ— are unit vectors, So, |๐’‚ โƒ— | = 1 & |๐’ƒ โƒ— | = 1 We need to find ฮธ if ๐’‚ โƒ— + ๐’ƒ โƒ— is a unit vector Assuming ๐‘Ž โƒ— + ๐‘ โƒ— is a unit vector Magnitude of ๐’‚ โƒ— + ๐’ƒ โƒ— = 1 |๐‘Ž โƒ—+๐‘ โƒ— |=1 |๐’‚ โƒ—+๐’ƒ โƒ— |^๐Ÿ=1^2 (๐’‚ โƒ—+๐’ƒ โƒ— ).(๐’‚ โƒ—+๐’ƒ โƒ— ) = 1 ๐‘Ž โƒ—. (๐‘Ž โƒ—+๐‘ โƒ— ) + ๐‘ โƒ—. (๐‘Ž โƒ—+๐‘ โƒ— ) = 1 ๐’‚ โƒ— . ๐’‚ โƒ— + ๐‘Ž โƒ— . ๐‘ โƒ— + ๐‘ โƒ—.๐‘Ž โƒ— + ๐’ƒ โƒ—.๐’ƒ โƒ— = 1 |๐’‚ โƒ— |^๐Ÿ + ๐‘Ž โƒ—.๐‘ โƒ— + ๐‘ โƒ—.๐‘Ž โƒ— + |๐’ƒ โƒ— |^๐Ÿ=1 12 + ๐‘Ž โƒ—. ๐‘ โƒ— + ๐‘ โƒ—.๐‘Ž โƒ— + 12 = 1 2 + ๐‘Ž โƒ—. ๐‘ โƒ— + ๐’ƒ โƒ—.๐’‚ โƒ— = 1 2 + ๐‘Ž โƒ—. ๐‘ โƒ— + ๐’‚ โƒ—. ๐’ƒ โƒ— = 1 2 + 2 ๐‘Ž โƒ—. ๐‘ โƒ— = 1 2๐‘Ž โƒ—. ๐‘ โƒ— = 1 โˆ’ 2 ๐‘Ž โƒ—. ๐‘ โƒ— = (โˆ’1)/2 |๐‘Ž โƒ— ||๐‘ โƒ— | cosโก ฮธ = (โˆ’1)/2 1 ร— 1 ร— cos ฮธ = (โˆ’1)/2 cos ฮธ = (โˆ’๐Ÿ)/๐Ÿ So, ฮธ = ๐Ÿ๐…/๐Ÿ‘ Hence, option (D) is correct Rough We know that cos 60ยฐ = 1/2 & cos is negative in 2nd quadrant, So, ฮธ = 180 โˆ’ 60ยฐ ฮธ = 120ยฐ ฮธ = 120 ร— ๐œ‹/180 ฮธ = 2๐œ‹/3

  1. Chapter 10 Class 12 Vector Algebra
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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