Shortest distance between two skew lines
Shortest distance between two skew lines
Last updated at April 16, 2024 by Teachoo
Ex 11.2, 15 Find the shortest distance between the lines whose vector equations are π β = (1 β t) π Μ + (t β 2) π Μ + (3 β 2t) π Μ and π β = (s + 1) π Μ + (2s β 1) π Μ β (2s + 1) π 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) βΓ (ππ) β")" ."(" (ππ) β β (ππ) β")" )/|(ππ) β Γ (ππ) β | | π β = (π β t) π Μ + (πβπ)π Μ + (3 β 2t) π Μ = 1π Μ β tπ Μ + tπ Μ β 2π Μ + 3π Μ β 2tπ Μ = (1π Μ β 2π Μ + 3π Μ) + t(β1π Μ + 1π Μ β 2π Μ) Comparing with π β = (π1) β + t (π1) β, (π1) β = 1π Μ β 2π Μ + 3π Μ & (π1) β = β 1π Μ + 1π Μ β 2π Μ π β = (π + 1) π Μ + (ππ" β " π)π Μ β (2s + 1) π Μ = sπ Μ + 1π Μ + 2sπ Μ β 1π Μ β 2sπ Μ β 1π Μ = (1π Μ β 1π Μ β 1π Μ) + s(1π Μ + 2π Μ β 2π Μ) Comparing with π β = (π2) β + s (π2) β, (π2) β = 1π Μ β 1π Μ β 1π Μ & (π2) β = 1π Μ + 2π Μ β 2π Μ Now, ((ππ) β β (π_π ) β) = (1π Μ β 1π Μ β 1π Μ) β (1π Μ β 2π + 3π Μ) = (1 β 1) π Μ + ( β 1 + 2)π Μ + ( β 1 β 3)π Μ = 0π Μ + 1π Μ β 4π Μ ( (π_π ) βΓ (π_π ) β ) = |β 8(π Μ&π Μ&π Μ@ β1&1& β2@1&2& β2)| = π Μ[(1Γβ 2)β(2Γβ 2)] β π Μ[(β1Γβ2)β(1Γβ 2)] + π Μ[(β 1Γ2)β(1Γ1)] = π Μ[β2+4] β π Μ[2+2] A + π Μ[β2β1] = 2π Μ β 4π Μ β 3π Μ Magnitude of ((π1) βΓ(π2) β) = β(22+(β 4)2+(β 3)2) |(ππ) βΓ(ππ) β | = β(4+16+9) = βππ Also, ((ππ) β Γ (ππ) β) . ((ππ) β β (ππ) β) = (2π Μ β 4π Μ β 3π Μ) . (0π Μ + 1π Μ β 4π Μ) = (2 Γ 0) + (β4 Γ 1) + (β3 Γ β4) = β0 + (β4) + 12 = 8 So, shortest distance = |(((π_1 ) β Γ (π_2 ) β ) . ((π_2 ) β Γ (π_1 ) β ).)/((π_1 ) β Γ (π_2 ) β )| = |8/β29| = π/βππ Therefore, the shortest distance between the given two lines is 8/β29 .