Ex 11.2

Chapter 11 Class 12 Three Dimensional Geometry
Serial order wise

Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class

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Ex 11.2, 14 Find the shortest distance between the lines whose vector equations are π β = (π Μ + 2π Μ + 3π Μ) + π (π Μ β 3π Μ + 2π Μ) and π β = (4π Μ + 5π Μ + 6π Μ) + π (2π Μ + 3π Μ + π Μ)Shortest distance between the lines with vector equations π β = (π_1 ) β + π (π"1" ) β and π β = (π"2" ) β + π(π"2" ) β is |(((ππ) β Γ (ππ) β ).((ππ) β β (ππ) β ))/|(ππ) β Γ (ππ) β | | Given, π β = (π Μ + 2π Μ + 3π Μ) + π (π Μ β 3π Μ + 2π Μ) Comparing with π β = (π1) β + π (π1) β, (π1) β = 1π Μ + 2π Μ + 3π Μ & (π1) β = 1π Μ β 3π Μ + 2π Μ π β = (4π Μ + 5π Μ + 6π Μ) + π(2π Μ + 3π Μ + π Μ) Comparing with π β = (π2) β + π(π2) β, (π2) β = 4π Μ + 5π Μ + 6π Μ & (π2) β = 2π Μ + 3π Μ + 1π Μ Now, ((ππ) β β (ππ) β) = (4π Μ + 5π Μ + 6π Μ) β (1π Μ + 2π Μ + 3π Μ) = (4 β 1)π Μ + (5 β 2) π Μ + (6 β 3) π = 3π Μ + 3π Μ + 3π Μ ((ππ) β Γ (ππ) β) = |β 8(π Μ&π Μ&π Μ@1& β3&[email protected]&3&1)| = π Μ [(β3Γ1)β(3Γ2)] β π Μ [(1Γ1)β(2Γ2)] + π Μ [(1Γ3)β(2Γβ3)] = π Μ [β3β6] β π Μ [1β4] + π Μ [3+6] = π Μ (β9) β π Μ (β3) + π Μ(9) = β 9π Μ + 3π Μ + 9π Μ Magnitude of ((π1) β Γ (π2) β) = β((β 9)2+32+92) |(ππ) β" Γ " (ππ) β | = β(81+9+81) = β171 = β(9Γ19 ) = 3βππ Also, ((ππ) β Γ (ππ) β) . ((ππ) β β (ππ) β) = (β9π Μ + 3π Μ + 9π Μ).(3π Μ + 3π Μ + 3π Μ) = (β9 Γ 3) + (3 Γ 3) + (9 Γ 3) = β27 + 9 + 27 = 9 So, shortest distance = |("(" (π1) βΓ" " (π2) β")" ."(" (π2) β β" " (π1) β")" )/|(π1) βΓ (π2) β | | = |9/(3β19)| = π/βππ Therefore, shortest distance between the given two lines is 3/β19.