Ex 11.2, 9 (i) - Chapter 11 Class 12 Three Dimensional Geometry

Transcript

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 ))| 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 − 𝟐)/𝟐 = (𝒚 − 𝟏)/𝟓 = (𝒛 + 𝟑)/( − 𝟑) (𝑥 − 2)/2 = (𝑦 − 1)/5 = (𝑧 − (−3))/( − 3) Comparing with (𝑥 − 𝑥1)/𝑎1 = (𝑦 − 𝑦1)/𝑏1 = (𝑧 − 𝑧1)/𝑐1 x1 = 2, y1 = 1, z1 = –3 & 𝑎1 = 2, b1 = 5, c1 = –3 (𝒙 + 𝟐)/( − 𝟏) = (𝒚 − 𝟒)/𝟖 = (𝒛 − 𝟓)/𝟒 (𝑥 − (− 2))/( − 1) = (𝑦 − 4)/8 = (𝑧 − 5)/4 Comparing with (𝑥 − 𝑥2)/𝑎2 = (𝑦 − 𝑦2)/𝑏2 = (𝑧 − 𝑧2)/𝑐2 𝑥2 = − 2, y2 = 4, z2 = 5 & 𝑎2 = –1, 𝑏2 = 8, 𝑐2 = 4 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 ))| = |((2 × −1) + (5 × 8) + ( − 3 × 4) )/(√(2^2 + 5^2 + 〖(−3)〗^2 ) √(〖(−1)〗^2 + 8^2 + 4^2 ))| = |( −2 + 40 + (−12) )/(√(4 + 25 + 9) √(1 + 64 + 16))| = |26/(√38 √81)| = |26/(√38 × 9)| = 26/(9√38 ) So, cos θ = 26/(9√38 ) ∴ θ = cos−1 (𝟐𝟔/(𝟗√𝟑𝟖 )) Therefore, the angle between the given lines is cos-1 (26/(9√38 )). = |( −2 + 40 + (−12) )/(√(4 + 25 + 9) √(1 + 64 + 16))| = |26/(√38 √81)| = |26/(√38 × 9)| = 26/(9√38 ) So, cos θ = 26/(9√38 ) ∴ θ = cos−1 (𝟐𝟔/(𝟗√𝟑𝟖 )) Therefore, the angle between the given lines is cos-1 (26/(9√38 )).