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Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class


Transcript

Example 8 Find the angle between the pair of lines (𝑥 + 3)/3 = (𝑦 − 1)/5 = (𝑧 + 3)/4 and (𝑥 + 1)/1 = (𝑦 − 4)/1 = (𝑧 − 5)/2 Angle between the pair of lines (𝑥 − 𝑥1)/𝑎1 = (𝑦 − 𝑦1)/𝑏1 = (𝑧 − 𝑧1)/𝑐1 and (𝑥 − 𝑥2)/𝑎2 = (𝑦 − 𝑦2)/𝑏2 = (𝑧 − 𝑧2)/𝑐2 is given by cos θ = |(𝒂_𝟏 𝒂_𝟐 + 𝒃_𝟏 𝒃_𝟐 +〖 𝒄〗_𝟏 𝒄_𝟐)/(√(〖𝒂_𝟏〗^𝟐 + 〖𝒃_𝟏〗^𝟐+ 〖𝒄_𝟏〗^𝟐 ) √(〖𝒂_𝟐〗^𝟐 +〖〖 𝒃〗_𝟐〗^𝟐+ 〖𝒄_𝟐〗^𝟐 ))| (𝒙 + 𝟑)/𝟑 = (𝒚 − 𝟏)/𝟓 = (𝒛 + 𝟑)/𝟒 (𝑥 − (−3))/3 = (𝑦 − 1)/5 = (𝑧 − (−3))/4 Comparing with (𝑥 − 𝑥1)/𝑎1 = (𝑦 − 𝑦1)/𝑏1 = (𝑧 − 𝑧1)/𝑐1 x1 = −3, y1 = 1, z1 = –3 & 𝒂1 = 3, b1 = 5, c1 = 4 (𝒙 + 𝟏)/𝟏 = (𝒚 − 𝟒)/𝟏 = (𝒛 − 𝟓)/𝟐 (𝑥 − (−1))/1 = (𝑦 − 4)/1 = (𝑧 − 5)/2 Comparing with (𝑥 − 𝑥2)/𝑎2 = (𝑦 − 𝑦2)/𝑏2 = (𝑧 − 𝑧2)/𝑐2 𝑥2 = −1, y2 = 4, z2 = 5 & 𝒂2 = 1, 𝒃2 = 1, 𝒄2 = 2 Now, cos θ = |(𝑎_1 𝑎_2 + 𝑏_1 𝑏_2 +〖 𝑐〗_1 𝑐_2)/(√(〖𝑎_1〗^2 + 〖𝑏_1〗^2+ 〖𝑐_1〗^2 ) √(〖𝑎_2〗^2 +〖〖 𝑏〗_2〗^2+ 〖𝑐_2〗^2 ))| = |((𝟑 × 𝟏) + (𝟓 × 𝟏) + (𝟒 × 𝟐))/(√(𝟑^𝟐 + 𝟓^𝟐 + 𝟒^𝟐 ) × √(𝟏^𝟐 + 𝟏^𝟐 + 𝟐^𝟐 ))| = |(3 + 5 + 8)/(√(9 + 25 + 16) √(1 + 1 + 4))| = |16/(√50 √6)| = |16/(5√2 × √2 √3)| = |16/(5 × 2 × √3)| = 𝟖/(𝟓 √𝟑) = 8/(5 √3) × √3/√3 = (𝟖√𝟑)/(𝟏𝟓 ) So, cos θ = (8√3)/(15 ) ∴ θ = cos-1((𝟖√𝟑)/(𝟏𝟓 )) Therefore, the angle between the given pair of line is cos−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 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.