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Chapter 11 Class 12 Three Dimensional Geometry

Serial order wise

Last updated at Feb. 1, 2020 by Teachoo

Example 12 Find the distance between the lines π_1 and π_2 given by π β = π Μ + 2π Μ β 4π Μ + π (2π Μ + 3π Μ + 6π Μ ) and π β = 3π Μ + 3π Μ β 5π Μ + ΞΌ (2π Μ + 3π Μ + 6π Μ)Since they are same, they are parallel lines Distance between two parallel lines with vector equations π β = (π_1 ) β + ππ β and π β = (π_2 ) β + ΞΌπ β is |(π β Γ ((π_2 ) β β (π_1 ) β))/|π β | | π β = (π Μ + 2π Μ β 4π Μ) + π (2π Μ + 3π Μ + 6π Μ) Comparing with π β = (π1) β + π π β, (π1) β = 1π Μ + 2π Μ β 4π Μ & π β = 2π Μ + 3π Μ + 6π Μ π β = (3π Μ + 3π Μ β 5π Μ) + π (2π Μ + 3π Μ + 6π Μ) Comparing with π β = (π2) β + ππ β, (π2) β = 3π Μ + 3π Μ β 5π Μ & π β = 2π Μ + 3π Μ + 6π Μ Now, ((ππ) β β (ππ) β) = (3π Μ + 3π Μ β 5π Μ) β (1π Μ + 2π Μ β 4π Μ) = (3 β 1) π Μ + (3 β 2)π Μ + ( β 5 + 4)π Μ = 2π Μ + 1π Μ β 1π Μ Magnitude of π β = β(22 + 32 + 62) |π β | = β(4+9+36) = β49 = 7 Also, π β Γ ((ππ) β β (ππ) β) = |β 8(π Μ&π Μ&π Μ@2&3&[email protected]&1&β1)| = π Μ [(3Γβ1)β(1Γ6)] β π Μ [(2Γβ1)β(2Γ6)] + π Μ [(2Γ1)β(2Γ3)] = π Μ [β3β6] β π Μ [β2β12] + π Μ [2β6] = π Μ (β9) β π Μ (β14) + π Μ(β4) = βππ Μ + 14π Μ β 4π Μ Now, |π β" Γ (" (ππ) β" β " (ππ) β")" | = β((β9)^2+(14)^2+(β4)^2 ) = β(81+196+16) = βπππ So, distance = |(π β Γ ((π_2 ) β β (π_1 ) β))/|π β | | = |β293/7| = βπππ/π Therefore, the distance between the given two parallel lines is β293/7.