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Ex 11.2, 10 - Class 12th - 3D Geometry - Find angle between

Ex 11.2, 10 - Chapter 11 Class 12 Three Dimensional Geometry - Part 2
Ex 11.2, 10 - Chapter 11 Class 12 Three Dimensional Geometry - Part 3

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Ex 11.2, 10 Find the angle between the following pairs of lines: (i) π‘Ÿ βƒ— = 2𝑖 Μ‚βˆ’ 5𝑗 Μ‚ + π‘˜ Μ‚ + πœ† (3𝑖 Μ‚ + 2𝑗 Μ‚ + 6π‘˜ Μ‚) and π‘Ÿ βƒ— = 7𝑖 Μ‚ – 6π‘˜ Μ‚ + πœ‡(𝑖 Μ‚ + 2𝑗 Μ‚ + 2π‘˜ Μ‚) Ex 11.2, 10 Find the angle between the following pairs of lines: (i) π‘Ÿ βƒ— = 2𝑖 Μ‚βˆ’ 5𝑗 Μ‚ + π‘˜ Μ‚ + πœ† (3𝑖 Μ‚ + 2𝑗 Μ‚ + 6π‘˜ Μ‚) and π‘Ÿ βƒ— = 7𝑖 Μ‚ – 6π‘˜ Μ‚ + πœ‡(𝑖 Μ‚ + 2𝑗 Μ‚ + 2π‘˜ Μ‚) Angle between two vectors π‘Ÿ βƒ— = (π‘Ž1) βƒ— + πœ† (𝑏1) βƒ— & π‘Ÿ βƒ— = (π‘Ž2) βƒ— + πœ‡ (𝑏2) βƒ— is given by cos ΞΈ = |((π’ƒπŸ) βƒ— . (π’ƒπŸ) βƒ—)/|(π’ƒπŸ) βƒ— ||(π’ƒπŸ) βƒ— | | Given, the pair of lines is 𝒓 βƒ— = (2π’Š Μ‚ βˆ’ 5𝒋 Μ‚ + π’Œ Μ‚) + πœ† (3π’Š Μ‚ + 2𝒋 Μ‚ + 6π’Œ Μ‚) So, (π‘Ž1) βƒ— = 2𝑖 Μ‚ βˆ’ 5𝑗 Μ‚ + 1π‘˜ Μ‚ (𝑏1) βƒ— = 3𝑖 Μ‚ + 2𝑗 Μ‚ + 6π‘˜ Μ‚ 𝒓 βƒ— = (7π’Š Μ‚ βˆ’ 6π’Œ Μ‚) + 𝝁 (π’Š Μ‚ + 2𝒋 Μ‚ + 2π’Œ Μ‚) So, (π‘Ž2) βƒ— = 7𝑖 Μ‚ + 0𝑗 Μ‚ βˆ’ 6π‘˜ Μ‚ (𝑏2) βƒ— = 1𝑖 Μ‚ + 2𝑗 Μ‚ + 2π‘˜ Μ‚ Now, (𝑏1) βƒ—.(𝑏2) βƒ— = (3𝑖 Μ‚ + 2𝑗 Μ‚ + 6π‘˜ Μ‚) . (1𝑖 Μ‚ + 2𝑗 Μ‚ + 2π‘˜ Μ‚) = (3 Γ— 1) + (2 Γ— 2) + (6 Γ— 2) = 3 + 4 + 12 = 19 Magnitude of (𝑏1) βƒ— = √(32 + 22 + 62) |(𝑏1) βƒ— | = √(9 + 4 + 36) = √49 = 7 Magnitude of (𝑏2) βƒ— = √(12+22+22) |(𝑏2) βƒ— | = √(1+4+4) = √9 = 3 Now, cos ΞΈ = |((𝑏1) βƒ—.(𝑏2) βƒ—)/|(𝑏1) βƒ— ||(𝑏2) βƒ— | | cos ΞΈ = |19/(7 Γ— 3 )| cos ΞΈ = 19/(21 ) ∴ ΞΈ = cosβˆ’1 (πŸπŸ—/(𝟐𝟏 )) Therefore, the angle between the given vectors is cos βˆ’1(19/(21 ))

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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.