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: (i) π β = 2π Μβ 5π Μ + π Μ + π (3π Μ + 2π Μ + 6π Μ) and π β = 7π Μ β 6π Μ + π(π Μ + 2π Μ + 2π Μ) Ex 11.2, 10 Find the angle between the following pairs of lines: (i) π β = 2π Μβ 5π Μ + π Μ + π (3π Μ + 2π Μ + 6π Μ) and π β = 7π Μ β 6π Μ + π(π Μ + 2π Μ + 2π 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 = (2π Μ β 5π Μ + π Μ) + π (3π Μ + 2π Μ + 6π Μ) So, (π1) β = 2π Μ β 5π Μ + 1π Μ (π1) β = 3π Μ + 2π Μ + 6π Μ π β = (7π Μ β 6π Μ) + π (π Μ + 2π Μ + 2π Μ) So, (π2) β = 7π Μ + 0π Μ β 6π Μ (π2) β = 1π Μ + 2π Μ + 2π Μ Now, (π1) β.(π2) β = (3π Μ + 2π Μ + 6π Μ) . (1π Μ + 2π Μ + 2π Μ) = (3 Γ 1) + (2 Γ 2) + (6 Γ 2) = 3 + 4 + 12 = 19 Magnitude of (π1) β = β(32 + 22 + 62) |(π1) β | = β(9 + 4 + 36) = β49 = 7 Magnitude of (π2) β = β(12+22+22) |(π2) β | = β(1+4+4) = β9 = 3 Now, cos ΞΈ = |((π1) β.(π2) β)/|(π1) β ||(π2) β | | cos ΞΈ = |19/(7 Γ 3 )| cos ΞΈ = 19/(21 ) β΄ ΞΈ = cosβ1 (ππ/(ππ )) Therefore, the angle between the given vectors is cos β1(19/(21 ))