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Ex 11.2, 10 - Chapter 11 Class 12 Three Dimensional Geometry - Part 4

Ex 11.2, 10 - Chapter 11 Class 12 Three Dimensional Geometry - Part 5
Ex 11.2, 10 - Chapter 11 Class 12 Three Dimensional Geometry - Part 6

 

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Ex 11.2, 10 Find the angle between the following pairs of lines: (ii) 𝑟 ⃗ = (3𝑖 ̂ + 𝑗 ̂ − 2𝑘 ̂) + 𝜆 (𝑖 ̂ − 𝑗 ̂ − 2𝑘 ̂) and 𝑟 ⃗ = (2𝑖 ̂ − 𝑗 ̂ − 56𝑘 ̂) + 𝜇 (3𝑖 ̂ – 5𝑗 ̂ − 4𝑘 ̂) 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 𝒓 ⃗ = (3𝒊 ̂ + 𝒋 ̂ − 2𝒌 ̂) + 𝜆 (𝒊 ̂ − 𝒋 ̂ − 2𝒌 ̂) So , (𝑎1) ⃗= 3𝑖 ̂ + 1𝑗 ̂ − 2𝑘 ̂ (𝑏1) ⃗ = 1𝑖 ̂ − 1𝑗 ̂ − 2𝑘 ̂ 𝒓 ⃗ = (2𝒊 ̂ − 𝒋 ̂ − 56𝒌 ̂) + 𝝁 (𝟑𝒊 ̂ – 5𝒋 ̂ − 4𝒌 ̂) So, (𝑎2) ⃗ = 2𝑖 ̂ − 1𝑗 ̂ − 56𝑘 ̂ (𝑏2) ⃗ = 3𝑖 ̂ − 5𝑗 ̂ − 4𝑘 ̂ Now, (𝑏1) ⃗. (𝑏2) ⃗ = (1𝑖 ̂ − 1𝑗 ̂ − 2𝑘 ̂).(3𝑖 ̂ − 5𝑗 ̂ − 4𝑘 ̂) = (1 × 3) + ( −1 × −5) + ( −2 × –4) = 3 + 5 + 8 = 16 Magnitude of (𝑏1) ⃗ = √(1^2+(−1)^2+(−2)^2 ) |(𝑏1) ⃗ | = √(1+1+4) = √6 Magnitude of (𝑏2) ⃗ = √(3^2+(−5)^2+( −4)2) |(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 )| = 8/(5√3 ) ∴ θ = cos-1 (8/(5√3 )) Therefore, the angle between the given vectors is cos − 1(8/(5√3 ))

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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.