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Ex 11.3

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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


Transcript

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

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.