This question is similar to Chapter 11 Class 12 Three Dimensional Geometry - Ex 11.2
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CBSE Class 12 Sample Paper for 2025 Boards
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Question 35 (A) You are here
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CBSE Class 12 Sample Paper for 2025 Boards
Last updated at Oct. 3, 2024 by Teachoo
This question is similar to Chapter 11 Class 12 Three Dimensional Geometry - Ex 11.2
Please check the question here
Question 35 (A) Find the shortest distance between the lines π_1 and π_2 whose vector equations are π β=(βΔ± ΛβΘ· Λβπ Λ)+π(7Δ± Λβ6Θ· Λ+π Λ)" and " π β=(3Δ± Λ+5Θ· Λ+7π Λ)+π(Δ± Λβ2Θ· Λ+π Λ) where π and π are parameters.Shortest distance between the lines with vector equations π β = (π1) β + π (π1) βand π β = (π2) β + π(π2) β is |(((ππ) β Γ (ππ) β ).((ππ) β β (ππ) β ))/|(ππ) β Γ (ππ) β | | π β = (βπ Μ β π Μ β π Μ) + π(7π Μ β 6π Μ + π Μ) Comparing with π β = (π1) β + π (π1) β, (π1) β = π Μ β π Μ β π Μ & (π1) β = 7π Μ β 6π Μ + π Μ π β = (3π Μ + 5π Μ + 7π Μ) + π (π Μ β 2π Μ + π Μ) Comparing with π β = (π2) β + π(π2) β , (π2) β = 3π Μ + 5π Μ + 7π Μ & (π2) β = π Μ β 2π Μ + π Μ Now, (ππ) β β (ππ) β = (3π Μ + 5π Μ + 7π Μ) β (βπ Μ β π Μ β π Μ) = (3 + 1) π Μ + (5 + 1)π Μ + (7 + 1) π Μ = 4π Μ + 6π Μ + 8π Μ (ππ) β Γ (ππ) β = |β 8(π Μ&π Μ&π Μ@7& β6&1@1&β2&1)| = π Μ [(β6 Γ 1)β(β2Γ1)] β π Μ [(7Γ1)β(1Γ1)] + π Μ [(7Γβ2)β(1Γβ6)] = π Μ [β6+2] β π Μ [7β1] + π Μ [β14+6] = β4π Μ β 6π Μ β 8π Μ Magnitude of ((π1) β Γ (π2) β) = β((β4)2+(β6)2+(β8)2) |(ππ) β Γ (ππ) β | = β(16+36+64) = βπππ Also, ((ππ) β Γ (ππ) β) . ((ππ) β β (ππ) β) = ("β4" π Μ" β 6" π Μ" β 8" π Μ).(4π Μ + 6π Μ + 8π Μ) = β4 Γ 4 + (β6) Γ 6 + (β8) Γ 8 = β16 β 36 β 64 = β 116 So, Shortest distance = |(((π_1 ) β Γ (π_2 ) β ).((π_2 ) β β (π_1 ) β ))/|(π_1 ) β Γ (π_2 ) β | | = |( β116)/β116| = βπππ Therefore, shortest distance between the given two lines is (3β2)/2.