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Last updated at March 22, 2023 by Teachoo
Misc 9 Find the shortest distance between lines π β = 6π Μ + 2π Μ + 2π Μ + π (π Μ β 2π Μ + 2π Μ) and π β = β4π Μ β π Μ + π (3π Μ β 2π Μ β 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 |(((ππ) β Γ (ππ) β ).((ππ) β β (ππ) β ))/|(ππ) β Γ (ππ) β | | π β = (6π Μ + 2π Μ + 2π Μ) + π (π Μ β 2π Μ + 2π Μ) Comparing with π β = (π1) β + π(π1) β , (π1) β = 6π Μ + 2π Μ + 2π Μ & (π1) β = 1π Μ β 2π Μ + 2π Μ π β = (β4π Μ β π Μ) + π (3π Μ β 2π Μ β 2π Μ) Comparing with π β = (π2) β + π(π2) β , (π2) β = β 4π Μ + 0π Μ β 1π Μ & (π2) β = 3π Μ β 2π Μ β 2π Μ Now, ((ππ) β β (ππ) β) = (β4π Μ + 0π Μ β 1π Μ) β (6π Μ + 2π Μ + 2π Μ) = (β4 β 6) π Μ + (0 β 2)π Μ + (β1 β 2) π Μ = β 10π Μ β 2π Μ β 3π Μ ((ππ) β Γ (ππ) β) = |β 8(π Μ&π Μ&π Μ@1& β2&[email protected]&β2&β2)| = π Μ [(β2Γβ2)β(β2Γ2)] β π Μ [(1Γβ2)β(3Γ2)] + π Μ [(1Γβ2)β(3Γβ2)] = π Μ [ 4+4] β π Μ [β2β6] + π Μ [β2+6] = π Μ (8) β π Μ (β8) + π Μ(4) = 8π Μ + 8π Μ + 4π Μ Magnitude of (π1) β Γ (π2) β = β(8^2+8^2+4^2 ) |(ππ) β Γ (ππ) β | = β(64+64+16) = β144 = ππ Also, ((ππ) βΓ(ππ) β ) . ((ππ) β β (ππ) β ) = (8π Μ + 8π Μ + 4π Μ).(β 10π Μ β 2π Μ β 3π Μ) = (8 Γ β 10) + (8 Γ β 2) + (4 Γ β 3) = β 80 + (β16) + (-12) = β 108 Shortest distance = |(((π1) β Γ (π2) β ) . ((π2) β β (π1) β ))/|(π1) β Γ (π2) β | | = |( β108)/12| = |β9| = 9 Therefore, the shortest distance between the given two lines is 9.