Ex 11.2, 10 - Chapter 11 Class 12 Three Dimensional Geometry (Term 2)

Transcript

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 )) 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 ))