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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

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Ex 11.3, 12 Find the angle between the planes whose vector equations are π‘Ÿ βƒ— . (2𝑖 Μ‚ + 2𝑗 Μ‚ – 3π‘˜ Μ‚) = 5 and π‘Ÿ βƒ— . (3𝑖 Μ‚ – 3𝑗 Μ‚ + 5π‘˜ Μ‚) = 3 .Angle between two planes π‘Ÿ βƒ— . (𝑛_1 ) βƒ— = d1 and π‘Ÿ βƒ—.(𝑛2) βƒ— = d2 is given by cos πœƒ = |((π’πŸ) βƒ—. (π’πŸ) βƒ—)/|(π’πŸ) βƒ— ||(π’πŸ) βƒ— | | 𝒓 βƒ—.(2π’Š Μ‚ + 2𝒋 Μ‚ βˆ’ 3π’Œ Μ‚) = 5 Comparing with π‘Ÿ βƒ—.(𝑛1) βƒ— = (𝑑1) βƒ—, (𝑛1) βƒ— = 2𝑖 Μ‚ + 2𝑗 Μ‚βˆ’3π‘˜ Μ‚ Magnitude of (𝑛1) βƒ— = √(2^2+2^2+γ€–(βˆ’3)γ€—^2 ) |(𝑛1) βƒ— | = √(4+4+9) = √17 So, cos ΞΈ = |((2𝑖 Μ‚ + 2𝑗 Μ‚ βˆ’ 3π‘˜ Μ‚ ) . (3𝑖 Μ‚ βˆ’ 3𝑗 Μ‚ + 5π‘˜ Μ‚ ))/(√17 Γ— √43)| = |((2 Γ— 3) + (2 Γ— βˆ’3) + (βˆ’3 Γ— 5))/√(17 Γ— 43)| = |(6 βˆ’ 6 βˆ’ 15)/√731| = |(βˆ’15)/√731| = 15/√731 So, cos ΞΈ = 15/√731 ∴ ΞΈ = cosβˆ’1(πŸπŸ“/βˆšπŸ•πŸ‘πŸ) Therefore, the angle between the planes is cosβˆ’1(15/√731).

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Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.