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  1. Chapter 11 Class 12 Three Dimensional Geometry
  2. Serial order wise

Transcript

Example 12 Find the distance between the lines 𝑙_1 and 𝑙_2 given by π‘Ÿ βƒ— = 𝑖 Μ‚ + 2𝑗 Μ‚ – 4π‘˜ Μ‚ + πœ† (2π’Š Μ‚ + 3𝒋 Μ‚ + 6π’Œ Μ‚ ) and π‘Ÿ βƒ— = 3𝑖 Μ‚ + 3𝑗 Μ‚ βˆ’ 5π‘˜ Μ‚ + ΞΌ (2π’Š Μ‚ + 3𝒋 Μ‚ + 6π’Œ Μ‚)Since they are same, they are parallel lines Distance between two parallel lines with vector equations π‘Ÿ βƒ— = (π‘Ž_1 ) βƒ— + πœ†π‘ βƒ— and π‘Ÿ βƒ— = (π‘Ž_2 ) βƒ— + μ𝑏 βƒ— is |(𝑏 βƒ— Γ— ((π‘Ž_2 ) βƒ— βˆ’ (π‘Ž_1 ) βƒ—))/|𝑏 βƒ— | | π‘Ÿ βƒ— = (𝑖 Μ‚ + 2𝑗 Μ‚ βˆ’ 4π‘˜ Μ‚) + πœ† (2π’Š Μ‚ + 3𝒋 Μ‚ + 6π’Œ Μ‚) Comparing with π‘Ÿ βƒ— = (π‘Ž1) βƒ— + πœ† 𝑏 βƒ—, (π‘Ž1) βƒ— = 1𝑖 Μ‚ + 2𝑗 Μ‚ – 4π‘˜ Μ‚ & 𝑏 βƒ— = 2𝑖 Μ‚ + 3𝑗 Μ‚ + 6π‘˜ Μ‚ π‘Ÿ βƒ— = (3𝑖 Μ‚ + 3𝑗 Μ‚ βˆ’ 5π‘˜ Μ‚) + πœ‡ (2π’Š Μ‚ + 3𝒋 Μ‚ + 6π’Œ Μ‚) Comparing with π‘Ÿ βƒ— = (π‘Ž2) βƒ— + πœ‡π‘ βƒ—, (π‘Ž2) βƒ— = 3𝑖 Μ‚ + 3𝑗 Μ‚ βˆ’ 5π‘˜ Μ‚ & 𝑏 βƒ— = 2𝑖 Μ‚ + 3𝑗 Μ‚ + 6π‘˜ Μ‚ Now, ((π’‚πŸ) βƒ— βˆ’ (π’‚πŸ) βƒ—) = (3𝑖 Μ‚ + 3𝑗 Μ‚ βˆ’ 5π‘˜ Μ‚) βˆ’ (1𝑖 Μ‚ + 2𝑗 Μ‚ βˆ’ 4π‘˜ Μ‚) = (3 βˆ’ 1) 𝑖 Μ‚ + (3 βˆ’ 2)𝑗 Μ‚ + ( βˆ’ 5 + 4)π‘˜ Μ‚ = 2π’Š Μ‚ + 1𝒋 Μ‚ βˆ’ 1π’Œ Μ‚ Magnitude of 𝑏 βƒ— = √(22 + 32 + 62) |𝒃 βƒ— | = √(4+9+36) = √49 = 7 Also, 𝒃 βƒ— Γ— ((π’‚πŸ) βƒ— βˆ’ (π’‚πŸ) βƒ—) = |β– 8(𝑖 Μ‚&𝑗 Μ‚&π‘˜ Μ‚@2&3&6@2&1&βˆ’1)| = 𝑖 Μ‚ [(3Γ—βˆ’1)βˆ’(1Γ—6)] βˆ’ 𝑗 Μ‚ [(2Γ—βˆ’1)βˆ’(2Γ—6)] + π‘˜ Μ‚ [(2Γ—1)βˆ’(2Γ—3)] = 𝑖 Μ‚ [βˆ’3βˆ’6] βˆ’ 𝑗 Μ‚ [βˆ’2βˆ’12] + π‘˜ Μ‚ [2βˆ’6] = 𝑖 Μ‚ (–9) βˆ’ 𝑗 Μ‚ (–14) + π‘˜ Μ‚(βˆ’4) = βˆ’πŸ—π’Š Μ‚ + 14𝒋 Μ‚ βˆ’ 4π’Œ Μ‚ Now, |𝒃 βƒ—" Γ— (" (π’‚πŸ) βƒ—" βˆ’ " (π’‚πŸ) βƒ—")" | = √((βˆ’9)^2+(14)^2+(βˆ’4)^2 ) = √(81+196+16) = βˆšπŸπŸ—πŸ‘ So, distance = |(𝑏 βƒ— Γ— ((π‘Ž_2 ) βƒ— βˆ’ (π‘Ž_1 ) βƒ—))/|𝑏 βƒ— | | = |√293/7| = βˆšπŸπŸ—πŸ‘/πŸ• Therefore, the distance between the given two parallel lines is √293/7.

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.