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

Transcript

Ex 10.4, 3 If a unit vector ๐‘Ž โƒ— makes angles ๐œ‹/3 with ๐‘– ฬ‚, ๐œ‹/4 , with ๐‘— ฬ‚ and an acute angle ฮธ with ๐‘˜ ฬ‚ , then find ฮธ and hence, the components of ๐‘Ž โƒ— . Let us take a unit vector ๐‘Ž โƒ— = ๐‘ฅ๐‘– ฬ‚ + y๐‘— ฬ‚ + z๐‘˜ ฬ‚ So, magnitude of ๐‘Ž โƒ— = |๐‘Ž โƒ— | = 1 Angle of ๐’‚ โƒ— with ๐’Š ฬ‚ = ๐…/๐Ÿ‘ ๐‘Ž โƒ— . ๐‘– ฬ‚ = |๐‘Ž โƒ— ||๐‘– ฬ‚ | cos ๐œ‹/3 (x๐‘– ฬ‚ + y๐‘— ฬ‚ + z๐‘˜ ฬ‚). ๐‘– ฬ‚ = 1 ร— 1 ร— 1/2 (x๐‘– ฬ‚ + y๐‘— ฬ‚ + z๐‘˜ ฬ‚). (1๐‘– ฬ‚ + 0๐‘— ฬ‚ + 0๐‘˜ ฬ‚) = 1/2 (x ร— 1) + (y ร— 0) + (z ร— 0) = 1/2 x + 0 + 0 = 1/2 x = ๐Ÿ/๐Ÿ Angle of ๐’‚ โƒ— with ๐’‹ ฬ‚ = ๐…/๐Ÿ’ ๐‘Ž โƒ— . ๐‘— ฬ‚ = |๐‘Ž โƒ— ||๐‘— ฬ‚ | cos ๐œ‹/4 (x๐‘– ฬ‚ + y๐‘— ฬ‚ + z๐‘˜ ฬ‚). ๐‘— ฬ‚ = 1 ร— 1 ร— 1/โˆš2 (x๐‘– ฬ‚ + y๐‘— ฬ‚ + z๐‘˜ ฬ‚). (0๐‘– ฬ‚ + 1๐‘— ฬ‚ + 0๐‘˜ ฬ‚) = 1/โˆš2 (x ร— 0) + (y ร— 1) + (z ร— 0) = 1/โˆš2 0 + y + 0 = 1/โˆš2 y = ๐Ÿ/โˆš๐Ÿ Also, Angle of ๐‘Ž โƒ— with ๐‘˜ ฬ‚ = ฮธ ๐‘Ž โƒ—. ๐‘˜ ฬ‚ = |๐‘Ž โƒ— ||๐‘˜ ฬ‚ |ร—cosโก"ฮธ" (x๐‘– ฬ‚ + y๐‘— ฬ‚ + z๐‘˜ ฬ‚). (0๐‘– ฬ‚ + .0๐‘— ฬ‚ + 1๐‘˜ ฬ‚) = 1 ร— 1 ร— cos ฮธ (x ร— 0) + (y ร— 0) + (z ร— 1) = cosฮธ 0 + 0 + z = cos ฮธ z = cos ฮธ Now, Magnitude of ๐‘Ž โƒ— = โˆš(๐‘ฅ^2+๐‘ฆ2+๐‘ง2) 1 = โˆš((1/2)^2+(1/โˆš2)^2+๐‘๐‘œ๐‘ 2"ฮธ" ) 1 = โˆš(1/4+1/2+๐‘๐‘œ๐‘ 2"ฮธ" ) 1 = โˆš(3/4+๐‘๐‘œ๐‘ 2"ฮธ" ) โˆš(3/4+๐‘๐‘œ๐‘ 2"ฮธ" ) = 1 (โˆš(3/4+๐‘๐‘œ๐‘ 2"ฮธ" ))^2 = 12 3/4 + ๐‘๐‘œ๐‘ 2" ฮธ" = 1 ๐‘๐‘œ๐‘ 2 "ฮธ" = 1 โˆ’ 3/4 ๐‘๐‘œ๐‘ 2" ฮธ" = 1/4 cosโก"ฮธ" = ยฑ โˆš(1/4) cosโก"ฮธ" = ยฑ 1/2 Since ฮธ is given an acute angle So, ฮธ < 90ยฐ โˆด ฮธ is in 1st quadrant And, cos ฮธ is positive in 1st quadrant= So, cos ฮธ = 1/2 โˆด ฮธ = 60ยฐ = ๐…/๐Ÿ‘ Also, z = cos ฮธ = cos 60ยฐ = ๐Ÿ/๐Ÿ Hence x = 1/2 , y = 1/โˆš2 & z = 1/2 The required vector ๐‘Ž โƒ— is 1/2 ๐‘– ฬ‚ + 1/โˆš2 ๐‘— ฬ‚ + 1/2 ๐‘˜ ฬ‚ So, components of ๐‘Ž โƒ— are ๐Ÿ/๐Ÿ , ๐Ÿ/โˆš๐Ÿ & ๐Ÿ/๐Ÿ The required vector ๐‘Ž โƒ— is 1/2 ๐‘– ฬ‚ + 1/โˆš2 ๐‘— ฬ‚ + 1/2 ๐‘˜ ฬ‚ So, components of ๐‘Ž โƒ— are ๐Ÿ/๐Ÿ , ๐Ÿ/โˆš๐Ÿ & ๐Ÿ/๐Ÿ

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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.