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# Ex 11.2, 10

Last updated at Dec. 8, 2016 by Teachoo

Last updated at Dec. 8, 2016 by Teachoo

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Transcript

Ex 11.2, 10 Find the angle between the following pairs of lines: (i) 𝑟 = 2 𝑖− 5 𝑗 + 𝑘 + 𝜆(3 𝑖 + 2 𝑗 + 6 𝑘) and 𝑟 = 7 𝑖 – 6 𝑘 + 𝜇( 𝑖 + 2 𝑗 + 2 𝑘) Angle between two vectors 𝑟 = 𝑎1 + 𝜆 𝑏1 & 𝑟 = 𝑎2 + 𝜇 𝑏2 is given by cos θ = 𝒃𝟏 . 𝒃𝟐 𝒃𝟏 𝒃𝟐 Given, the pair of lines is Now, 𝑏1. 𝑏2 = (3 𝑖 + 2 𝑗 + 6 𝑘) . (1 𝑖 + 2 𝑗 + 2 𝑘) = (3 × 1) + (2 × 2) + (6 × 2) = 3 + 4 + 12 = 19 Magnitude of 𝑏1 = 32 + 22 + 62 𝑏1 = 9 + 4 + 36 = 49 = 7 Magnitude of 𝑏2 = 12+22+22 𝑏2 = 1+4+4 = 9 = 3 Now, cos θ = 𝑏1. 𝑏2 𝑏1 𝑏2 cos θ = 197 × 3 cos θ = 1921 ∴ θ = cos-1 𝟏𝟗𝟐𝟏 Therefore, the angle between the given vectors is cos − 1 1921 Ex 11.2, 10 Find the angle between the following pairs of lines: (ii) 𝑟 = (3 𝑖 + 𝑗 − 2 𝑘) + 𝜆 ( 𝑖 − 𝑗 − 2 𝑘) and 𝑟 = (2 𝑖 − 𝑗 − 56 𝑘) + 𝜇 (3 𝑖 – 5 𝑗 − 4 𝑘) Angle between two vectors 𝑟 = 𝑎1 + 𝜆 𝑏1 & 𝑟 = 𝑎2 + 𝜇 𝑏2 is given by cos θ = 𝒃𝟏 . 𝒃𝟐 𝒃𝟏 𝒃𝟐 Given, the pair of lines is Now, 𝑏1. 𝑏2 = (1 𝑖 − 1 𝑗 − 2 𝑘).(3 𝑖 − 5 𝑗 − 4 𝑘) = (1 × 3) + ( − 1 × − 5) + ( − 2 × –4) = 3 + 5 + 8 = 16 Magnitude of 𝑏1 = 12+ −12+ −22 𝑏1 = 1+1+4 = 6 Magnitude of 𝑏2 = 32+ −52+ − 42 𝑏2 = 9+25+16 = 50 = 25×2 = 5 2 Now, cos θ = 𝑏1. 𝑏2 𝑏1 𝑏2 = 16 6 × 5 2 = 16 3 × 2 × 5 × 2 = 16 3 × 2 × 5 = 85 3 ∴ θ = cos-1 85 3 Therefore, the angle between the given vectors is cos − 1 85 3

Chapter 11 Class 12 Three Dimensional Geometry

Serial order wise

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CA Maninder Singh

CA Maninder Singh is a Chartered Accountant for the past 8 years. He provides courses for Practical Accounts, Taxation and Efiling at teachoo.com .