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

Transcript

Example 10 Find the angle between the pair of lines (๐‘ฅ + 3)/3 = (๐‘ฆ โˆ’ 1)/5 = (๐‘ง + 3)/4 and (๐‘ฅ + 1)/1 = (๐‘ฆ โˆ’ 4)/1 = (๐‘ง โˆ’ 5)/2Angle 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Ÿ‘ = (๐’š โˆ’ ๐Ÿ)/๐Ÿ“ = (๐’› + ๐Ÿ‘)/๐Ÿ’ (๐‘ฅ โˆ’ (โˆ’3))/3 = (๐‘ฆ โˆ’ 1)/5 = (๐‘ง โˆ’ (โˆ’3))/4 Comparing with (๐‘ฅ โˆ’ ๐‘ฅ1)/๐‘Ž1 = (๐‘ฆ โˆ’ ๐‘ฆ1)/๐‘1 = (๐‘ง โˆ’ ๐‘ง1)/๐‘1 x1 = โˆ’3, y1 = 1, z1 = โ€“3 & ๐‘Ž1 = 3, b1 = 5, c1 = 4 (๐’™ + ๐Ÿ)/๐Ÿ = (๐’š โˆ’ ๐Ÿ’)/๐Ÿ = (๐’› โˆ’ ๐Ÿ“)/๐Ÿ (๐‘ฅ โˆ’ (โˆ’1))/1 = (๐‘ฆ โˆ’ 4)/1 = (๐‘ง โˆ’ 5)/2 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’1, y2 = 4, z2 = 5 & ๐‘Ž2 = 1, ๐‘2 = 1, ๐‘2 = 2 Now, cos ฮธ = |(๐‘Ž_1 ๐‘Ž_2 + ๐‘_1 ๐‘_2 +ใ€– ๐‘ใ€—_1 ๐‘_2)/(โˆš(ใ€–๐‘Ž_1ใ€—^2 + ใ€–๐‘_1ใ€—^2+ ใ€–๐‘_1ใ€—^2 ) โˆš(ใ€–๐‘Ž_2ใ€—^2 +ใ€–ใ€– ๐‘ใ€—_2ใ€—^2+ ใ€–๐‘_2ใ€—^2 ))| = |((3 ร— 1) + (5 ร— 1) + (4 ร— 2))/(โˆš(3^2 + 5^2 + 4^2 ) ร— โˆš(1^2 + 1^2 + 2^2 ))| = |(3 + 5 + 8)/(โˆš(9 + 25 + 16) โˆš(1 + 1 + 4))| = |16/(โˆš50 โˆš6)| = |16/(5โˆš2 ร— โˆš2 โˆš3)| = |16/(5 ร— 2 ร— โˆš3)| = 8/(5 โˆš3) = 8/(5 โˆš3) ร— โˆš3/โˆš3 = (8โˆš3)/(15 ) So, cos ฮธ = (8โˆš3)/(15 ) โˆด ฮธ = cos-1((๐Ÿ–โˆš๐Ÿ‘)/(๐Ÿ๐Ÿ“ )) Therefore, the angle between the given pair of line is cosโˆ’1 ((8โˆš3)/(15 ))

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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.