Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class
Angle between two planes
Question 12 Important Deleted for CBSE Board 2024 Exams You are here
Question 18 (MCQ) Important Deleted for CBSE Board 2024 Exams
Question 13 Important Deleted for CBSE Board 2024 Exams
Question 13 (a) Important Deleted for CBSE Board 2024 Exams
Question 17 Important Deleted for CBSE Board 2024 Exams
Question 9 Important Deleted for CBSE Board 2024 Exams
Question 13 Important Deleted for CBSE Board 2024 Exams
Angle between two planes
Last updated at May 29, 2023 by Teachoo
Question 12 Find the angle between the planes whose vector equations are 𝑟 ⃗ . (2𝑖 ̂ + 2𝑗 ̂ – 3𝑘 ̂) = 5 and 𝑟 ⃗ . (3𝑖 ̂ – 3𝑗 ̂ + 5𝑘 ̂) = 3 .Angle between two planes 𝑟 ⃗ . (𝑛_1 ) ⃗ = d1 and 𝑟 ⃗.(𝑛2) ⃗ = d2 is given by cos 𝜃 = |((𝒏𝟏) ⃗. (𝒏𝟐) ⃗)/|(𝒏𝟏) ⃗ ||(𝒏𝟐) ⃗ | | 𝒓 ⃗.(2𝒊 ̂ + 2𝒋 ̂ − 3𝒌 ̂) = 5 Comparing with 𝑟 ⃗.(𝑛1) ⃗ = (𝑑1) ⃗, (𝑛1) ⃗ = 2𝑖 ̂ + 2𝑗 ̂−3𝑘 ̂ Magnitude of (𝑛1) ⃗ = √(2^2+2^2+〖(−3)〗^2 ) |(𝑛1) ⃗ | = √(4+4+9) = √17 So, cos θ = |((2𝑖 ̂ + 2𝑗 ̂ − 3𝑘 ̂ ) . (3𝑖 ̂ − 3𝑗 ̂ + 5𝑘 ̂ ))/(√17 × √43)| = |((2 × 3) + (2 × −3) + (−3 × 5))/√(17 × 43)| = |(6 − 6 − 15)/√731| = |(−15)/√731| = 15/√731 So, cos θ = 15/√731 ∴ θ = cos−1(𝟏𝟓/√𝟕𝟑𝟏) Therefore, the angle between the planes is cos−1(15/√731).