Ex 11.2

Chapter 11 Class 12 Three Dimensional Geometry
Serial order wise

### Transcript

Ex 11.2, 12 Find the shortest distance between the lines π β = (π Μ + 2π Μ + π Μ) + π (π Μ β π Μ + π Μ) and π β = (2π Μ β π Μ β π Μ) + π (2π Μ + π Μ + 2π 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 + π Μ) + π(π Μ β π Μ + π Μ) Comparing with π β = (π1) β + π (π1) β, (π1) β = 1π Μ + 2π Μ + 1π Μ & (π1) β = 1π Μ β 1π Μ + 1π Μ π β = (2π Μ β π Μ β π Μ) + π (2π Μ + π Μ + 2π Μ) Comparing with π β = (π2) β + π(π2) β , (π2) β = 2π Μ β 1π Μ β 1π Μ & (π2) β = 2π Μ + 1π Μ + 2π Μ Now, (ππ) β β (ππ) β = (2π Μ β 1π Μ β 1π Μ) β (1π Μ + 2π Μ + 1π Μ) = (2 β 1) π Μ + (β1β 2)π Μ + (β1 β 1) π Μ = 1π Μ β 3π Μ β 2π Μ (ππ) β Γ (ππ) β = |β 8(π Μ&π Μ&π Μ@1& β1&1@2&1&2)| = π Μ [(β1Γ 2)β(1Γ1)] β π Μ [(1Γ2)β(2Γ1)] + π Μ [(1Γ1)β(2Γβ1)] = π Μ [β2β1] β π Μ [2β2] + π Μ [1+2] = β3π Μ β 0π Μ + 3π Μ Magnitude of ((π1) β Γ (π2) β) = β((β3)2+(0)2+32) |(ππ) β Γ (ππ) β | = β(9+0+9) = β18 = β(9 Γ 2) = 3βπ Also, ((ππ) β Γ (ππ) β) . ((ππ) β β (ππ) β) = (β 3π Μβ0π Μ+3π Μ).(1π Μ β 3π Μ β 2π Μ) = (β3Γ1)".(" 0Γβ"3)" + (3 Γ β2) = β3 β 0 β 6 = β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 | | = |( βπ)/(πβπ)| = 3/β2 = 3/β2 Γ β2/β2 = (πβπ)/π Therefore, shortest distance between the given two lines is (3β2)/2.