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

Transcript

Angle between the pair of lines (๐‘ฅ โˆ’ ๐‘ฅ1)/๐‘Ž1 = (๐‘ฆ โˆ’ ๐‘ฆ1)/๐‘1 = (๐‘ง โˆ’ ๐‘ง1)/๐‘1 and (๐‘ฅ โˆ’ ๐‘ฅ2)/๐‘Ž2 = (๐‘ฆ โˆ’ ๐‘ฆ2)/๐‘2 = (๐‘ง โˆ’ ๐‘ง2)/๐‘2 is given by cos ฮธ = |(๐’‚_๐Ÿ ๐’‚_๐Ÿ + ๐’ƒ_๐Ÿ ๐’ƒ_๐Ÿ +ใ€– ๐’„ใ€—_๐Ÿ ๐’„_๐Ÿ)/(โˆš(ใ€–๐’‚_๐Ÿใ€—^๐Ÿ + ใ€–๐’ƒ_๐Ÿใ€—^๐Ÿ+ ใ€–๐’„_๐Ÿใ€—^๐Ÿ ) โˆš(ใ€–๐’‚_๐Ÿใ€—^๐Ÿ +ใ€–ใ€– ๐’ƒใ€—_๐Ÿใ€—^๐Ÿ+ ใ€–๐’„_๐Ÿใ€—^๐Ÿ ))| Angle between the pair of lines (๐‘ฅ โˆ’ ๐‘ฅ1)/๐‘Ž1 = (๐‘ฆ โˆ’ ๐‘ฆ1)/๐‘1 = (๐‘ง โˆ’ ๐‘ง1)/๐‘1 and (๐‘ฅ โˆ’ ๐‘ฅ2)/๐‘Ž2 = (๐‘ฆ โˆ’ ๐‘ฆ2)/๐‘2 = (๐‘ง โˆ’ ๐‘ง2)/๐‘2 is given by cos ฮธ = |(๐’‚_๐Ÿ ๐’‚_๐Ÿ + ๐’ƒ_๐Ÿ ๐’ƒ_๐Ÿ +ใ€– ๐’„ใ€—_๐Ÿ ๐’„_๐Ÿ)/(โˆš(ใ€–๐’‚_๐Ÿใ€—^๐Ÿ + ใ€–๐’ƒ_๐Ÿใ€—^๐Ÿ+ ใ€–๐’„_๐Ÿใ€—^๐Ÿ ) โˆš(ใ€–๐’‚_๐Ÿใ€—^๐Ÿ +ใ€–ใ€– ๐’ƒใ€—_๐Ÿใ€—^๐Ÿ+ ใ€–๐’„_๐Ÿใ€—^๐Ÿ ))| (๐’™ โˆ’ ๐Ÿ)/๐Ÿ = (๐’š โˆ’ ๐Ÿ)/๐Ÿ“ = (๐’› + ๐Ÿ‘)/( โˆ’ ๐Ÿ‘) (๐‘ฅ โˆ’ 2)/2 = (๐‘ฆ โˆ’ 1)/5 = (๐‘ง โˆ’ (โˆ’3))/( โˆ’ 3) Comparing with (๐‘ฅ โˆ’ ๐‘ฅ1)/๐‘Ž1 = (๐‘ฆ โˆ’ ๐‘ฆ1)/๐‘1 = (๐‘ง โˆ’ ๐‘ง1)/๐‘1 x1 = 2, y1 = 1, z1 = โ€“3 & ๐‘Ž1 = 2, b1 = 5, c1 = โ€“3 (๐’™ + ๐Ÿ)/( โˆ’ ๐Ÿ) = (๐’š โˆ’ ๐Ÿ’)/๐Ÿ– = (๐’› โˆ’ ๐Ÿ“)/๐Ÿ’ (๐‘ฅ โˆ’ (โˆ’ 2))/( โˆ’ 1) = (๐‘ฆ โˆ’ 4)/8 = (๐‘ง โˆ’ 5)/4 Comparing with (๐‘ฅ โˆ’ ๐‘ฅ2)/๐‘Ž2 = (๐‘ฆ โˆ’ ๐‘ฆ2)/๐‘2 = (๐‘ง โˆ’ ๐‘ง2)/๐‘2 ๐‘ฅ2 = โˆ’ 2, y2 = 4, z2 = 5 & ๐‘Ž2 = โ€“1, ๐‘2 = 8, ๐‘2 = 4 Now, cos ฮธ = |(๐’‚_๐Ÿ ๐’‚_๐Ÿ + ๐’ƒ_๐Ÿ ๐’ƒ_๐Ÿ +ใ€– ๐’„ใ€—_๐Ÿ ๐’„_๐Ÿ)/(โˆš(ใ€–๐’‚_๐Ÿใ€—^๐Ÿ + ใ€–๐’ƒ_๐Ÿใ€—^๐Ÿ+ ใ€–๐’„_๐Ÿใ€—^๐Ÿ ) โˆš(ใ€–๐’‚_๐Ÿใ€—^๐Ÿ +ใ€–ใ€– ๐’ƒใ€—_๐Ÿใ€—^๐Ÿ+ ใ€–๐’„_๐Ÿใ€—^๐Ÿ ))| = |((2 ร— โˆ’1) + (5 ร— 8) + ( โˆ’ 3 ร— 4) )/(โˆš(2^2 + 5^2 + ใ€–(โˆ’3)ใ€—^2 ) โˆš(ใ€–(โˆ’1)ใ€—^2 + 8^2 + 4^2 ))| = |( โˆ’2 + 40 + (โˆ’12) )/(โˆš(4 + 25 + 9) โˆš(1 + 64 + 16))| = |26/(โˆš38 โˆš81)| = |26/(โˆš38 ร— 9)| = 26/(9โˆš38 ) So, cos ฮธ = 26/(9โˆš38 ) โˆด ฮธ = cosโˆ’1 (๐Ÿ๐Ÿ”/(๐Ÿ—โˆš๐Ÿ‘๐Ÿ– )) Therefore, the angle between the given lines is cos-1 (26/(9โˆš38 )). = |( โˆ’2 + 40 + (โˆ’12) )/(โˆš(4 + 25 + 9) โˆš(1 + 64 + 16))| = |26/(โˆš38 โˆš81)| = |26/(โˆš38 ร— 9)| = 26/(9โˆš38 ) So, cos ฮธ = 26/(9โˆš38 ) โˆด ฮธ = cosโˆ’1 (๐Ÿ๐Ÿ”/(๐Ÿ—โˆš๐Ÿ‘๐Ÿ– )) Therefore, the angle between the given lines is cos-1 (26/(9โˆš38 )). Angle between the pair of lines (๐‘ฅ โˆ’ ๐‘ฅ1)/๐‘Ž1 = (๐‘ฆ โˆ’ ๐‘ฆ1)/๐‘1 = (๐‘ง โˆ’ ๐‘ง1)/๐‘1 and (๐‘ฅ โˆ’ ๐‘ฅ2)/๐‘Ž2 = (๐‘ฆ โˆ’ ๐‘ฆ2)/๐‘2 = (๐‘ง โˆ’ ๐‘ง2)/๐‘2 is given by cos ฮธ = |(๐’‚_๐Ÿ ๐’‚_๐Ÿ + ๐’ƒ_๐Ÿ ๐’ƒ_๐Ÿ +ใ€– ๐’„ใ€—_๐Ÿ ๐’„_๐Ÿ)/(โˆš(ใ€–๐’‚_๐Ÿใ€—^๐Ÿ + ใ€–๐’ƒ_๐Ÿใ€—^๐Ÿ+ ใ€–๐’„_๐Ÿใ€—^๐Ÿ ) โˆš(ใ€–๐’‚_๐Ÿใ€—^๐Ÿ +ใ€–ใ€– ๐’ƒใ€—_๐Ÿใ€—^๐Ÿ+ ใ€–๐’„_๐Ÿใ€—^๐Ÿ ))| ๐’™/๐Ÿ = ๐’š/๐Ÿ = ๐’›/๐Ÿ (๐‘ฅ โˆ’ 0)/2 = (๐‘ฆ โˆ’ 0)/2 = (๐‘ง โˆ’ 0)/1 Comparing with (๐‘ฅ โˆ’ ๐‘ฅ1)/๐‘Ž1 = (๐‘ฆ โˆ’ ๐‘ฆ1)/๐‘1 = (๐‘ง โˆ’ ๐‘ง1)/๐‘1 x1 = 0, y1 = 0, z1 = 0 & ๐‘Ž1 = 2, b1 = 2, c1 = 1 (๐’™ โˆ’ ๐Ÿ“)/๐Ÿ’ = (๐’š โˆ’ ๐Ÿ)/๐Ÿ = (๐’› โˆ’ ๐Ÿ“)/๐Ÿ– Comparing with (๐‘ฅ โˆ’ ๐‘ฅ2)/๐‘Ž2 = (๐‘ฆ โˆ’ ๐‘ฆ2)/๐‘2 = (๐‘ง โˆ’ ๐‘ง2)/๐‘2 ๐‘ฅ2 = 5, y2 = 2, z2 = 5 & ๐‘Ž2 = 4, ๐‘2 = 1, ๐‘2 = 8 Now, cos ฮธ = |(๐’‚_๐Ÿ ๐’‚_๐Ÿ + ๐’ƒ_๐Ÿ ๐’ƒ_๐Ÿ +ใ€– ๐’„ใ€—_๐Ÿ ๐’„_๐Ÿ)/(โˆš(ใ€–๐’‚_๐Ÿใ€—^๐Ÿ + ใ€–๐’ƒ_๐Ÿใ€—^๐Ÿ+ ใ€–๐’„_๐Ÿใ€—^๐Ÿ ) โˆš(ใ€–๐’‚_๐Ÿใ€—^๐Ÿ +ใ€–ใ€– ๐’ƒใ€—_๐Ÿใ€—^๐Ÿ+ ใ€–๐’„_๐Ÿใ€—^๐Ÿ ))| = |((2 ร— 4) + (2 ร— 1) + (1 ร— 8) )/(โˆš(22 + 22 + 12) ร— โˆš(42 + 12 + 82))| = |(8 + 2 + 8 )/(โˆš(4 + 4 + 1) โˆš(16 + 1 + 64))| = |18/(โˆš9 ร— โˆš81)| = 18/(3 ร— 9) = 2/3 So, cos ฮธ = 2/3 โˆด ฮธ = cos-1 (๐Ÿ/๐Ÿ‘) Therefore, the angle between the given lines is cosโˆ’1(๐Ÿ/๐Ÿ‘).

About the Author

Davneet Singh's photo - Teacher, Computer Engineer, Marketer
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.