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Example 10 - Chapter 11 Class 12 - Find angle between lines

Example 10 - Chapter 11 Class 12 Three Dimensional Geometry - Part 2
Example 10 - Chapter 11 Class 12 Three Dimensional Geometry - Part 3


Transcript

Example 10 Find the angle between the pair of lines (π‘₯ + 3)/3 = (𝑦 βˆ’ 1)/5 = (𝑧 + 3)/4 and (π‘₯ + 1)/1 = (𝑦 βˆ’ 4)/1 = (𝑧 βˆ’ 5)/2Angle between the pair of lines (π‘₯ βˆ’ π‘₯1)/π‘Ž1 = (𝑦 βˆ’ 𝑦1)/𝑏1 = (𝑧 βˆ’ 𝑧1)/𝑐1 and (π‘₯ βˆ’ π‘₯2)/π‘Ž2 = (𝑦 βˆ’ 𝑦2)/𝑏2 = (𝑧 βˆ’ 𝑧2)/𝑐2 is given by cos ΞΈ = |(𝒂_𝟏 𝒂_𝟐 + 𝒃_𝟏 𝒃_𝟐 +γ€– 𝒄〗_𝟏 𝒄_𝟐)/(√(〖𝒂_πŸγ€—^𝟐 + 〖𝒃_πŸγ€—^𝟐+ 〖𝒄_πŸγ€—^𝟐 ) √(〖𝒂_πŸγ€—^𝟐 +γ€–γ€– 𝒃〗_πŸγ€—^𝟐+ 〖𝒄_πŸγ€—^𝟐 ))| (𝒙 + πŸ‘)/πŸ‘ = (π’š βˆ’ 𝟏)/πŸ“ = (𝒛 + πŸ‘)/πŸ’ (π‘₯ βˆ’ (βˆ’3))/3 = (𝑦 βˆ’ 1)/5 = (𝑧 βˆ’ (βˆ’3))/4 Comparing with (π‘₯ βˆ’ π‘₯1)/π‘Ž1 = (𝑦 βˆ’ 𝑦1)/𝑏1 = (𝑧 βˆ’ 𝑧1)/𝑐1 x1 = βˆ’3, y1 = 1, z1 = –3 & π‘Ž1 = 3, b1 = 5, c1 = 4 (𝒙 + 𝟏)/𝟏 = (π’š βˆ’ πŸ’)/𝟏 = (𝒛 βˆ’ πŸ“)/𝟐 (π‘₯ βˆ’ (βˆ’1))/1 = (𝑦 βˆ’ 4)/1 = (𝑧 βˆ’ 5)/2 Comparing with (π‘₯ βˆ’ π‘₯2)/π‘Ž2 = (𝑦 βˆ’ 𝑦2)/𝑏2 = (𝑧 βˆ’ 𝑧2)/𝑐2 π‘₯2 = βˆ’1, y2 = 4, z2 = 5 & π‘Ž2 = 1, 𝑏2 = 1, 𝑐2 = 2 Now, cos ΞΈ = |(π‘Ž_1 π‘Ž_2 + 𝑏_1 𝑏_2 +γ€– 𝑐〗_1 𝑐_2)/(√(γ€–π‘Ž_1γ€—^2 + 〖𝑏_1γ€—^2+ 〖𝑐_1γ€—^2 ) √(γ€–π‘Ž_2γ€—^2 +γ€–γ€– 𝑏〗_2γ€—^2+ 〖𝑐_2γ€—^2 ))| = |((3 Γ— 1) + (5 Γ— 1) + (4 Γ— 2))/(√(3^2 + 5^2 + 4^2 ) Γ— √(1^2 + 1^2 + 2^2 ))| = |(3 + 5 + 8)/(√(9 + 25 + 16) √(1 + 1 + 4))| = |16/(√50 √6)| = |16/(5√2 Γ— √2 √3)| = |16/(5 Γ— 2 Γ— √3)| = 8/(5 √3) = 8/(5 √3) Γ— √3/√3 = (8√3)/(15 ) So, cos ΞΈ = (8√3)/(15 ) ∴ ΞΈ = cos-1((πŸ–βˆšπŸ‘)/(πŸπŸ“ )) Therefore, the angle between the given pair of line is cosβˆ’1 ((8√3)/(15 ))

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.