This question is similar to Chapter 11 Class 12 Three Dimensional Geometry - Ex 11.2
Please check the question here


CBSE Class 12 Sample Paper for 2025 Boards
CBSE Class 12 Sample Paper for 2025 Boards
Last updated at Dec. 13, 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.