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Chapter 11 Class 12 Three Dimensional Geometry
Serial order wise

Ex 11.3, 12 - Find angle between planes r.(2i + 2j - 3k) = 5

Ex 11.3, 12 - Chapter 11 Class 12 Three Dimensional Geometry - Part 2

Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class


Transcript

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).

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.