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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 is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.