This question is similar to Chapter 11 Class 12 Three Dimensional Geometry - Ex 11.2

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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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