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Last updated at March 22, 2023 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&[email protected]& 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 .