Find the acute angle between the lines

(x - 4)/3 = (y + 3)/4 = (z + 1)/5 and (x - 1)/4 = (y + 1)/(-3) = (z + 10)/5

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  1. Class 12
  2. Solutions of Sample Papers and Past Year Papers - for Class 12 Boards

Transcript

Question 25 Find the acute angle between the lines (๐‘ฅ โˆ’ 4)/3 = (๐‘ฆ + 3)/4 = (๐‘ง + 1)/5 and (๐‘ฅ โˆ’ 1)/4 = (๐‘ฆ + 1)/(โˆ’3) = (๐‘ง + 10)/5 Angle between the pair of lines (๐‘ฅ โˆ’ ๐‘ฅ_1)/๐‘Ž_1 = (๐‘ฆ โˆ’ ๐‘ฆ_1)/๐‘_1 = (๐‘ง โˆ’ ๐‘ง_1)/๐‘_1 and (๐‘ฅ โˆ’ ๐‘ฅ_2)/๐‘Ž_2 = (๐‘ฆ โˆ’ ๐‘ฆ_2)/๐‘_2 = (๐‘ง โˆ’ ๐‘ง_2)/๐‘_2 is given by cos ฮธ = |(๐‘Ž_1 ๐‘Ž_2 + ๐‘_1 ๐‘_2 + ๐‘_1 ๐‘_2)/(โˆš(ใ€–๐‘Ž_1ใ€—^2 + ใ€–๐‘_1ใ€—^2 + ใ€–๐‘_1ใ€—^2 ) โˆš(ใ€–๐‘Ž_2ใ€—^2 + ใ€–๐‘_2ใ€—^2 + ใ€–๐‘_2ใ€—^2 ))| (๐’™ โˆ’ ๐Ÿ’)/๐Ÿ‘ = (๐’š + ๐Ÿ‘)/๐Ÿ’ = (๐’› + ๐Ÿ)/๐Ÿ“ Comparing with (๐‘ฅ โˆ’ ๐‘ฅ_1)/๐‘Ž_1 = (๐‘ฆ โˆ’ ๐‘ฆ_1)/๐‘_1 = (๐‘ง โˆ’ ๐‘ง_1)/๐‘_1 ๐‘Ž1 = 3, b1 = 4, c1 = 4 (๐’™ โˆ’ ๐Ÿ)/๐Ÿ’ = (๐’š + ๐Ÿ)/(โˆ’๐Ÿ‘) = (๐’› + ๐Ÿ๐ŸŽ)/๐Ÿ“ Comparing with (๐‘ฅ โˆ’ ๐‘ฅ_2)/๐‘Ž_2 = (๐‘ฆ โˆ’ ๐‘ฆ_2)/๐‘_2 = (๐‘ง โˆ’ ๐‘ง_2)/๐‘_2 ๐‘Ž2 = 4, ๐‘2 = โ€“3, ๐‘2 = 5 Now, cos ฮธ = |(๐‘Ž_1 ๐‘Ž_2 + ๐‘_1 ๐‘_2 + ๐‘_1 ๐‘_2)/(โˆš(ใ€–๐‘Ž_1ใ€—^2 + ใ€–๐‘_1ใ€—^2 + ใ€–๐‘_1ใ€—^2 ) โˆš(ใ€–๐‘Ž_2ใ€—^2 + ใ€–๐‘_2ใ€—^2 + ใ€–๐‘_2ใ€—^2 ))| = |(3 ร— 4 + 4 ร— (โˆ’3) + 5 ร— 5)/(โˆš(3^2 + 4^2 + 5^2 ) โˆš(4^2 +(โˆ’3)^2 + 5^2 ))| = |(12 โˆ’ 12 + 25)/(โˆš(9 + 16 + 25) โˆš(16 + 9 + 25))| = |25/(โˆš50 โˆš50)| = |25/50| = |1/2| = 1/2 So, cos ฮธ = 1/2 โˆด ฮธ = 60ยฐ = ๐…/๐Ÿ‘ Therefore, required angle is ๐…/๐Ÿ‘ Note: Please write angle in radians and not degree

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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.