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Ex 11.2

Ex 11.2, 1

Ex 11.2, 2

Ex 11.2, 3 Important

Ex 11.2, 4

Ex 11.2, 5 Important

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Ex 11.2, 7 Important

Ex 11.2, 8 (i) Important

Ex 11.2, 8 (ii) You are here

Ex 11.2, 9 (i) Important

Ex 11.2, 9 (ii)

Ex 11.2, 10 Important

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Ex 11.2, 12 Important

Ex 11.2, 13 Important

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Question 1 Important Deleted for CBSE Board 2024 Exams

Question 2 Deleted for CBSE Board 2024 Exams

Chapter 11 Class 12 Three Dimensional Geometry

Serial order wise

Last updated at May 29, 2023 by Teachoo

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 = (π1) β + π (π1) β & π β = (π2) β + π (π2) β is given by cos ΞΈ = |((ππ) β . (ππ) β)/|(ππ) β ||(ππ) 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 , (π1) β= 3π Μ + 1π Μ β 2π Μ (π1) β = 1π Μ β 1π Μ β 2π Μ π β = (2π Μ β π Μ β 56π Μ) + π (ππ Μ β 5π Μ β 4π Μ) So, (π2) β = 2π Μ β 1π Μ β 56π Μ (π2) β = 3π Μ β 5π Μ β 4π Μ Now, (π1) β. (π2) β = (1π Μ β 1π Μ β 2π Μ).(3π Μ β 5π Μ β 4π Μ) = (1 Γ 3) + ( β1 Γ β5) + ( β2 Γ β4) = 3 + 5 + 8 = 16 Magnitude of (π1) β = β(1^2+(β1)^2+(β2)^2 ) |(π1) β | = β(1+1+4) = β6 Magnitude of (π2) β = β(3^2+(β5)^2+( β4)2) |(π2) β | = β(9+25+16) = β50 = β(25Γ2) = 5β2 Now, cos ΞΈ = |((π1) β.(π2) β)/|(π1) β ||(π2) β | | = |16/(β6 Γ 5β2)| = |16/(β3 Γ β2 Γ 5 Γ β2)| = |16/(β3 Γ 2 Γ 5 )| = 8/(5β3 ) β΄ ΞΈ = cos-1 (8/(5β3 )) Therefore, the angle between the given vectors is cos β 1(8/(5β3 ))