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  1. Chapter 11 Class 12 Three Dimensional Geometry
  2. Serial order wise

Transcript

Ex 11.2, 10 Find the angle between the following pairs of lines: (i) ๐‘Ÿ โƒ— = 2๐‘– ฬ‚โˆ’ 5๐‘— ฬ‚ + ๐‘˜ ฬ‚ + ๐œ† (3๐‘– ฬ‚ + 2๐‘— ฬ‚ + 6๐‘˜ ฬ‚) and ๐‘Ÿ โƒ— = 7๐‘– ฬ‚ โ€“ 6๐‘˜ ฬ‚ + ๐œ‡(๐‘– ฬ‚ + 2๐‘— ฬ‚ + 2๐‘˜ ฬ‚) Ex 11.2, 10 Find the angle between the following pairs of lines: (i) ๐‘Ÿ โƒ— = 2๐‘– ฬ‚โˆ’ 5๐‘— ฬ‚ + ๐‘˜ ฬ‚ + ๐œ† (3๐‘– ฬ‚ + 2๐‘— ฬ‚ + 6๐‘˜ ฬ‚) and ๐‘Ÿ โƒ— = 7๐‘– ฬ‚ โ€“ 6๐‘˜ ฬ‚ + ๐œ‡(๐‘– ฬ‚ + 2๐‘— ฬ‚ + 2๐‘˜ ฬ‚) Angle between two vectors ๐‘Ÿ โƒ— = (๐‘Ž1) โƒ— + ๐œ† (๐‘1) โƒ— & ๐‘Ÿ โƒ— = (๐‘Ž2) โƒ— + ๐œ‡ (๐‘2) โƒ— is given by cos ฮธ = |((๐’ƒ๐Ÿ) โƒ— . (๐’ƒ๐Ÿ) โƒ—)/|(๐’ƒ๐Ÿ) โƒ— ||(๐’ƒ๐Ÿ) โƒ— | | Given, the pair of lines is ๐’“ โƒ— = (2๐’Š ฬ‚ โˆ’ 5๐’‹ ฬ‚ + ๐’Œ ฬ‚) + ๐œ† (3๐’Š ฬ‚ + 2๐’‹ ฬ‚ + 6๐’Œ ฬ‚) So, (๐‘Ž1) โƒ— = 2๐‘– ฬ‚ โˆ’ 5๐‘— ฬ‚ + 1๐‘˜ ฬ‚ (๐‘1) โƒ— = 3๐‘– ฬ‚ + 2๐‘— ฬ‚ + 6๐‘˜ ฬ‚ ๐’“ โƒ— = (7๐’Š ฬ‚ โˆ’ 6๐’Œ ฬ‚) + ๐ (๐’Š ฬ‚ + 2๐’‹ ฬ‚ + 2๐’Œ ฬ‚) So, (๐‘Ž2) โƒ— = 7๐‘– ฬ‚ + 0๐‘— ฬ‚ โˆ’ 6๐‘˜ ฬ‚ (๐‘2) โƒ— = 1๐‘– ฬ‚ + 2๐‘— ฬ‚ + 2๐‘˜ ฬ‚ Now, (๐‘1) โƒ—.(๐‘2) โƒ— = (3๐‘– ฬ‚ + 2๐‘— ฬ‚ + 6๐‘˜ ฬ‚) . (1๐‘– ฬ‚ + 2๐‘— ฬ‚ + 2๐‘˜ ฬ‚) = (3 ร— 1) + (2 ร— 2) + (6 ร— 2) = 3 + 4 + 12 = 19 Magnitude of (๐‘1) โƒ— = โˆš(32 + 22 + 62) |(๐‘1) โƒ— | = โˆš(9 + 4 + 36) = โˆš49 = 7 Magnitude of (๐‘2) โƒ— = โˆš(12+22+22) |(๐‘2) โƒ— | = โˆš(1+4+4) = โˆš9 = 3 Now, cos ฮธ = |((๐‘1) โƒ—.(๐‘2) โƒ—)/|(๐‘1) โƒ— ||(๐‘2) โƒ— | | cos ฮธ = |19/(7 ร— 3 )| cos ฮธ = 19/(21 ) โˆด ฮธ = cosโˆ’1 (๐Ÿ๐Ÿ—/(๐Ÿ๐Ÿ )) Therefore, the angle between the given vectors is cos โˆ’1(19/(21 )) Ex 11.2, 10 Find the angle between the following pairs of lines: (ii) ๐‘Ÿ โƒ— = (3๐‘– ฬ‚ + ๐‘— ฬ‚ โˆ’ 2๐‘˜ ฬ‚) + ๐œ† (๐‘– ฬ‚ โˆ’ ๐‘— ฬ‚ โˆ’ 2๐‘˜ ฬ‚) and ๐‘Ÿ โƒ— = (2๐‘– ฬ‚ โˆ’ ๐‘— ฬ‚ โˆ’ 56๐‘˜ ฬ‚) + ๐œ‡ (3๐‘– ฬ‚ โ€“ 5๐‘— ฬ‚ โˆ’ 4๐‘˜ ฬ‚) Angle between two vectors ๐‘Ÿ โƒ— = (๐‘Ž1) โƒ— + ๐œ† (๐‘1) โƒ— & ๐‘Ÿ โƒ— = (๐‘Ž2) โƒ— + ๐œ‡ (๐‘2) โƒ— is given by cos ฮธ = |((๐’ƒ๐Ÿ) โƒ— . (๐’ƒ๐Ÿ) โƒ—)/|(๐’ƒ๐Ÿ) โƒ— ||(๐’ƒ๐Ÿ) โƒ— | | Given, the pair of lines is ๐’“ โƒ— = (3๐’Š ฬ‚ + ๐’‹ ฬ‚ โˆ’ 2๐’Œ ฬ‚) + ๐œ† (๐’Š ฬ‚ โˆ’ ๐’‹ ฬ‚ โˆ’ 2๐’Œ ฬ‚) So , (๐‘Ž1) โƒ—= 3๐‘– ฬ‚ + 1๐‘— ฬ‚ โˆ’ 2๐‘˜ ฬ‚ (๐‘1) โƒ— = 1๐‘– ฬ‚ โˆ’ 1๐‘— ฬ‚ โˆ’ 2๐‘˜ ฬ‚ ๐’“ โƒ— = (2๐’Š ฬ‚ โˆ’ ๐’‹ ฬ‚ โˆ’ 56๐’Œ ฬ‚) + ๐ (๐Ÿ‘๐’Š ฬ‚ โ€“ 5๐’‹ ฬ‚ โˆ’ 4๐’Œ ฬ‚) So, (๐‘Ž2) โƒ— = 2๐‘– ฬ‚ โˆ’ 1๐‘— ฬ‚ โˆ’ 56๐‘˜ ฬ‚ (๐‘2) โƒ— = 3๐‘– ฬ‚ โˆ’ 5๐‘— ฬ‚ โˆ’ 4๐‘˜ ฬ‚ Now, (๐‘1) โƒ—. (๐‘2) โƒ— = (1๐‘– ฬ‚ โˆ’ 1๐‘— ฬ‚ โˆ’ 2๐‘˜ ฬ‚).(3๐‘– ฬ‚ โˆ’ 5๐‘— ฬ‚ โˆ’ 4๐‘˜ ฬ‚) = (1 ร— 3) + ( โˆ’1 ร— โˆ’5) + ( โˆ’2 ร— โ€“4) = 3 + 5 + 8 = 16 Magnitude of (๐‘1) โƒ— = โˆš(1^2+(โˆ’1)^2+(โˆ’2)^2 ) |(๐‘1) โƒ— | = โˆš(1+1+4) = โˆš6 Magnitude of (๐‘2) โƒ— = โˆš(3^2+(โˆ’5)^2+( โˆ’4)2) |(๐‘2) โƒ— | = โˆš(9+25+16) = โˆš50 = โˆš(25ร—2) = 5โˆš2 Now, cos ฮธ = |((๐‘1) โƒ—.(๐‘2) โƒ—)/|(๐‘1) โƒ— ||(๐‘2) โƒ— | | = |16/(โˆš6 ร— 5โˆš2)| = |16/(โˆš3 ร— โˆš2 ร— 5 ร— โˆš2)| = |16/(โˆš3 ร— 2 ร— 5 )| = 8/(5โˆš3 ) โˆด ฮธ = cos-1 (8/(5โˆš3 )) Therefore, the angle between the given vectors is cos โˆ’ 1(8/(5โˆš3 ))

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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.