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Ex 11.2
Ex 11.2, 2
Ex 11.2, 3 Important
Ex 11.2, 4
Ex 11.2, 5 Important
Ex 11.2, 6
Ex 11.2, 7 Important
Ex 11.2, 8
Ex 11.2, 9 Important
Ex 11.2, 10 (i) Important Deleted for CBSE Board 2022 Exams You are here
Ex 11.2, 10 (ii)
Ex 11.2, 11 (i) Important Deleted for CBSE Board 2022 Exams
Ex 11.2, 11 (ii)
Ex 11.2, 12 Important
Ex 11.2, 13
Ex 11.2, 14 Important
Ex 11.2, 15 Important
Ex 11.2, 16
Ex 11.2, 17 Important
Last updated at Aug. 24, 2021 by Teachoo
Ex 11.2, 10 Find the angle between the following pairs of lines: (i) 𝑟 ⃗ = 2𝑖 ̂− 5𝑗 ̂ + 𝑘 ̂ + 𝜆 (3𝑖 ̂ + 2𝑗 ̂ + 6𝑘 ̂) and 𝑟 ⃗ = 7𝑖 ̂ – 6𝑘 ̂ + 𝜇(𝑖 ̂ + 2𝑗 ̂ + 2𝑘 ̂) Ex 11.2, 10 Find the angle between the following pairs of lines: (i) 𝑟 ⃗ = 2𝑖 ̂− 5𝑗 ̂ + 𝑘 ̂ + 𝜆 (3𝑖 ̂ + 2𝑗 ̂ + 6𝑘 ̂) and 𝑟 ⃗ = 7𝑖 ̂ – 6𝑘 ̂ + 𝜇(𝑖 ̂ + 2𝑗 ̂ + 2𝑘 ̂) 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 ⃗ = (2𝒊 ̂ − 5𝒋 ̂ + 𝒌 ̂) + 𝜆 (3𝒊 ̂ + 2𝒋 ̂ + 6𝒌 ̂) So, (𝑎1) ⃗ = 2𝑖 ̂ − 5𝑗 ̂ + 1𝑘 ̂ (𝑏1) ⃗ = 3𝑖 ̂ + 2𝑗 ̂ + 6𝑘 ̂ 𝒓 ⃗ = (7𝒊 ̂ − 6𝒌 ̂) + 𝝁 (𝒊 ̂ + 2𝒋 ̂ + 2𝒌 ̂) So, (𝑎2) ⃗ = 7𝑖 ̂ + 0𝑗 ̂ − 6𝑘 ̂ (𝑏2) ⃗ = 1𝑖 ̂ + 2𝑗 ̂ + 2𝑘 ̂ Now, (𝑏1) ⃗.(𝑏2) ⃗ = (3𝑖 ̂ + 2𝑗 ̂ + 6𝑘 ̂) . (1𝑖 ̂ + 2𝑗 ̂ + 2𝑘 ̂) = (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 = |19/(7 × 3 )| cos θ = 19/(21 ) ∴ θ = cos−1 (𝟏𝟗/(𝟐𝟏 )) Therefore, the angle between the given vectors is cos −1(19/(21 ))