Ex 11.2

Chapter 11 Class 12 Three Dimensional Geometry
Serial order wise

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Ex 11.2, 10 Find the angle between the following pairs of lines: (ii) π β = (3π Μ + π Μ β 2π Μ) + π (π Μ β π Μ β 2π Μ) and π β = (2π Μ β π Μ β 56π Μ) + π (3π Μ β 5π Μ β 4π Μ) Angle between two vectors π β = (π1) β + π (π1) β & π β = (π2) β + π (π2) β is given by cos ΞΈ = |((ππ) β . (ππ) β)/|(ππ) β ||(ππ) β | | Given, the pair of lines is π β = (3π Μ + π Μ β 2π Μ) + π (π Μ β π Μ β 2π Μ) So , (π1) β= 3π Μ + 1π Μ β 2π Μ (π1) β = 1π Μ β 1π Μ β 2π Μ π β = (2π Μ β π Μ β 56π Μ) + π (ππ Μ β 5π Μ β 4π Μ) So, (π2) β = 2π Μ β 1π Μ β 56π Μ (π2) β = 3π Μ β 5π Μ β 4π Μ Now, (π1) β. (π2) β = (1π Μ β 1π Μ β 2π Μ).(3π Μ β 5π Μ β 4π Μ) = (1 Γ 3) + ( β1 Γ β5) + ( β2 Γ β4) = 3 + 5 + 8 = 16 Magnitude of (π1) β = β(1^2+(β1)^2+(β2)^2 ) |(π1) β | = β(1+1+4) = β6 Magnitude of (π2) β = β(3^2+(β5)^2+( β4)2) |(π2) β | = β(9+25+16) = β50 = β(25Γ2) = 5β2 Now, cos ΞΈ = |((π1) β.(π2) β)/|(π1) β ||(π2) β | | = |16/(β6 Γ 5β2)| = |16/(β3 Γ β2 Γ 5 Γ β2)| = |16/(β3 Γ 2 Γ 5 )| = 8/(5β3 ) β΄ ΞΈ = cos-1 (8/(5β3 )) Therefore, the angle between the given vectors is cos β 1(8/(5β3 ))