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Question 35 (A) - Find the shortest distance between lines l1 and l2 - CBSE Class 12 Sample Paper for 2025 Boards

part 2 - Question 35 (A) - CBSE Class 12 Sample Paper for 2025 Boards - Solutions of Sample Papers and Past Year Papers - for Class 12 Boards - Class 12
part 3 - Question 35 (A) - CBSE Class 12 Sample Paper for 2025 Boards - Solutions of Sample Papers and Past Year Papers - for Class 12 Boards - Class 12

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Question 35 (A) Find the shortest distance between the lines 𝑙_1 and 𝑙_2 whose vector equations are π‘Ÿ βƒ—=(βˆ’Δ± Λ†βˆ’Θ· Λ†βˆ’π‘˜ Λ†)+πœ†(7Δ± Λ†βˆ’6Θ· Λ†+π‘˜ Λ†)" and " π‘Ÿ βƒ—=(3Δ± Λ†+5Θ· Λ†+7π‘˜ Λ†)+πœ‡(Δ± Λ†βˆ’2Θ· Λ†+π‘˜ Λ†) where πœ† and πœ‡ are parameters.Shortest distance between the lines with vector equations π‘Ÿ βƒ— = (π‘Ž1) βƒ— + πœ† (𝑏1) βƒ—and π‘Ÿ βƒ— = (π‘Ž2) βƒ— + πœ‡(𝑏2) βƒ— is |(((π’ƒπŸ) βƒ— Γ— (π’ƒπŸ) βƒ— ).((π’‚πŸ) βƒ— βˆ’ (π’‚πŸ) βƒ— ))/|(π’ƒπŸ) βƒ— Γ— (π’ƒπŸ) βƒ— | | 𝒓 βƒ— = (β€“π’Š Μ‚ – 𝒋 Μ‚ – π’Œ Μ‚) + πœ†(7π’Š Μ‚ βˆ’ 6𝒋 Μ‚ + π’Œ Μ‚) Comparing with π‘Ÿ βƒ— = (π‘Ž1) βƒ— + πœ† (𝑏1) βƒ—, (π‘Ž1) βƒ— = 𝑖 Μ‚ – 𝑗 Μ‚ – π‘˜ Μ‚ & (𝑏1) βƒ— = 7𝑖 Μ‚ – 6𝑗 Μ‚ + π‘˜ Μ‚ 𝒓 βƒ— = (3π’Š Μ‚ + 5𝒋 Μ‚ + 7π’Œ Μ‚) + 𝝁 (π’Š Μ‚ – 2𝒋 Μ‚ + π’Œ Μ‚) Comparing with π‘Ÿ βƒ— = (π‘Ž2) βƒ— + πœ‡(𝑏2) βƒ— , (π‘Ž2) βƒ— = 3𝑖 Μ‚ + 5𝑗 Μ‚ + 7π‘˜ Μ‚ & (𝑏2) βƒ— = 𝑖 Μ‚ – 2𝑗 Μ‚ + π‘˜ Μ‚ Now, (π’‚πŸ) βƒ— βˆ’ (π’‚πŸ) βƒ— = (3𝑖 Μ‚ + 5𝑗 Μ‚ + 7π‘˜ Μ‚) βˆ’ (–𝑖 Μ‚ – 𝑗 Μ‚ – π‘˜ Μ‚) = (3 + 1) 𝑖 Μ‚ + (5 + 1)𝑗 Μ‚ + (7 + 1) π‘˜ Μ‚ = 4π’Š Μ‚ + 6𝒋 Μ‚ + 8π’Œ Μ‚ (π’ƒπŸ) βƒ— Γ— (π’ƒπŸ) βƒ— = |β– 8(𝑖 Μ‚&𝑗 Μ‚&π‘˜ Μ‚@7& βˆ’6&1@1&βˆ’2&1)| = 𝑖 Μ‚ [(βˆ’6 Γ— 1)βˆ’(βˆ’2Γ—1)] βˆ’ 𝑗 Μ‚ [(7Γ—1)βˆ’(1Γ—1)] + π‘˜ Μ‚ [(7Γ—βˆ’2)βˆ’(1Γ—βˆ’6)] = 𝑖 Μ‚ [βˆ’6+2] βˆ’ 𝑗 Μ‚ [7βˆ’1] + π‘˜ Μ‚ [βˆ’14+6] = βˆ’4π’Š Μ‚ βˆ’ 6𝒋 Μ‚ – 8π’Œ Μ‚ Magnitude of ((𝑏1) βƒ— Γ— (𝑏2) βƒ—) = √((βˆ’4)2+(βˆ’6)2+(βˆ’8)2) |(π’ƒπŸ) βƒ— Γ— (π’ƒπŸ) βƒ— | = √(16+36+64) = βˆšπŸπŸπŸ” Also, ((π’ƒπŸ) βƒ— Γ— (π’ƒπŸ) βƒ—) . ((π’‚πŸ) βƒ— – (π’‚πŸ) βƒ—) = ("βˆ’4" 𝑖 Μ‚" βˆ’ 6" 𝑗 Μ‚" – 8" π‘˜ Μ‚).(4𝑖 Μ‚ + 6𝑗 Μ‚ + 8π‘˜ Μ‚) = –4 Γ— 4 + (–6) Γ— 6 + (–8) Γ— 8 = –16 – 36 – 64 = – 116 So, Shortest distance = |(((𝑏_1 ) βƒ— Γ— (𝑏_2 ) βƒ— ).((π‘Ž_2 ) βƒ— βˆ’ (π‘Ž_1 ) βƒ— ))/|(𝑏_1 ) βƒ— Γ— (𝑏_2 ) βƒ— | | = |( βˆ’116)/√116| = βˆšπŸπŸπŸ” Therefore, shortest distance between the given two lines is (3√2)/2.

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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 15 years. He provides courses for Maths, Science and Computer Science at Teachoo