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Chapter 11 Class 12 Three Dimensional Geometry

Serial order wise

Last updated at Feb. 1, 2020 by Teachoo

Example 11 Find the shortest distance between the lines l1 and l2 whose vector equations are π β = π Μ + π Μ + π(2π Μ β π Μ + π Μ ) and π β = 2π Μ + π Μ β π Μ + π (3π Μ β 5π Μ + 2π Μ )Shortest distance between lines with vector equations π β = (π1) β + π (π1) β and π β = (π2) β + π(π2) β is |(((π1) β Γ (π2) β ).((π2) β β (π1) β ))/|(π1) β Γ (π2) β | | π β = (π Μ + π Μ) + π (2π Μ β π Μ + π Μ) Comparing with π β = (π1) β + π (π1) β (π1) β = 1π Μ + 1π Μ + 0π Μ & (π1) β = 2π Μ β 1π Μ + 1π Μ π β = (2π Μ + π Μ β π Μ) + π (3π Μ β 5π Μ + 2π Μ) Comparing with π β = (π2) β + π(π2) β (π2) β = 2π Μ + 1π Μ β 1π Μ & (π2) β = 3π Μ β 5π Μ + 2π Μ Now (ππ) β β (ππ) β = (2π Μ + 1π Μ β 1π Μ) β (1π Μ + 1π Μ + 0π Μ) = (2 β 1) π Μ + (1 β 1)π Μ + (β1 β 0) π Μ = 1π Μ + 0π Μ β 1π Μ (ππ) β Γ (ππ) β = |β 8(π Μ&π Μ&π Μ@2& β1&1@3& β5&2)| = π Μ [(β1Γ2)β(β5Γ1)] β π Μ [(2Γ2)β(3Γ1)] + π Μ[(2Γβ5)β(3Γβ1)] = π Μ [β2+5] β π Μ [4β3] + π Μ [β10+3] = π Μ (3) β π Μ (1) + π Μ(β7) = 3π Μ β π Μ β 7π Μ Magnitude of ((π1) β Γ (π2) β) = β(32+(β1)2+(β7)^2 ) |(ππ) βΓ (ππ) β | = β(9+1+49) = βππ Also, ((ππ) β Γ (ππ) β) .((ππ) β β (ππ) β) = (3π Μ β π Μ β 7π Μ) . (1π Μ + 0π Μ β 1π Μ) = (3 Γ 1) + (β1 Γ 0) + (β7 Γ β1) = 3 + 0 + 7 = 10 β΄ Shortest distance = |(((π1) β Γ (π2) β ).((π2) β β (π1) β ))/|(π1) β Γ (π2) β | | = |10/β59| = ππ/βππ .