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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

Find the acute angle between lines  (x - 4)/3 = (y + 3)/4 = (z + 1)/5

Question 25 - CBSE Class 12 Sample Paper for 2020 Boards - Part 2
Question 25 - CBSE Class 12 Sample Paper for 2020 Boards - Part 3

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Question 25 Find the acute angle between the lines (π‘₯ βˆ’ 4)/3 = (𝑦 + 3)/4 = (𝑧 + 1)/5 and (π‘₯ βˆ’ 1)/4 = (𝑦 + 1)/(βˆ’3) = (𝑧 + 10)/5 Angle between the pair of lines (π‘₯ βˆ’ π‘₯_1)/π‘Ž_1 = (𝑦 βˆ’ 𝑦_1)/𝑏_1 = (𝑧 βˆ’ 𝑧_1)/𝑐_1 and (π‘₯ βˆ’ π‘₯_2)/π‘Ž_2 = (𝑦 βˆ’ 𝑦_2)/𝑏_2 = (𝑧 βˆ’ 𝑧_2)/𝑐_2 is given by cos ΞΈ = |(π‘Ž_1 π‘Ž_2 + 𝑏_1 𝑏_2 + 𝑐_1 𝑐_2)/(√(γ€–π‘Ž_1γ€—^2 + 〖𝑏_1γ€—^2 + 〖𝑐_1γ€—^2 ) √(γ€–π‘Ž_2γ€—^2 + 〖𝑏_2γ€—^2 + 〖𝑐_2γ€—^2 ))| (𝒙 βˆ’ πŸ’)/πŸ‘ = (π’š + πŸ‘)/πŸ’ = (𝒛 + 𝟏)/πŸ“ Comparing with (π‘₯ βˆ’ π‘₯_1)/π‘Ž_1 = (𝑦 βˆ’ 𝑦_1)/𝑏_1 = (𝑧 βˆ’ 𝑧_1)/𝑐_1 π‘Ž1 = 3, b1 = 4, c1 = 4 (𝒙 βˆ’ 𝟏)/πŸ’ = (π’š + 𝟏)/(βˆ’πŸ‘) = (𝒛 + 𝟏𝟎)/πŸ“ Comparing with (π‘₯ βˆ’ π‘₯_2)/π‘Ž_2 = (𝑦 βˆ’ 𝑦_2)/𝑏_2 = (𝑧 βˆ’ 𝑧_2)/𝑐_2 π‘Ž2 = 4, 𝑏2 = –3, 𝑐2 = 5 Now, cos ΞΈ = |(π‘Ž_1 π‘Ž_2 + 𝑏_1 𝑏_2 + 𝑐_1 𝑐_2)/(√(γ€–π‘Ž_1γ€—^2 + 〖𝑏_1γ€—^2 + 〖𝑐_1γ€—^2 ) √(γ€–π‘Ž_2γ€—^2 + 〖𝑏_2γ€—^2 + 〖𝑐_2γ€—^2 ))| = |(3 Γ— 4 + 4 Γ— (βˆ’3) + 5 Γ— 5)/(√(3^2 + 4^2 + 5^2 ) √(4^2 +(βˆ’3)^2 + 5^2 ))| = |(12 βˆ’ 12 + 25)/(√(9 + 16 + 25) √(16 + 9 + 25))| = |25/(√50 √50)| = |25/50| = |1/2| = 1/2 So, cos ΞΈ = 1/2 ∴ ΞΈ = 60Β° = 𝝅/πŸ‘ Therefore, required angle is 𝝅/πŸ‘ Note: Please write angle in radians and not degree

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.