Maths Crash Course - Live lectures + all videos + Real time Doubt solving!

Ex 11.2

Ex 11.2, 1

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

Ex 11.2, 10 (ii) You are here

Ex 11.2, 11 (i) Important

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

Chapter 11 Class 12 Three Dimensional Geometry

Serial order wise

Last updated at Aug. 24, 2021 by Teachoo

Maths Crash Course - Live lectures + all videos + Real time Doubt solving!

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 ))