Last updated at April 21, 2021 by Teachoo

Transcript

Ex 10.3, 2 Find the angle between the vectors π Μ β 2π Μ + 3π Μ and 3π Μ - 2π Μ + π ΜLet π β = π Μ β 2π Μ + 3π Μ = 1π Μ β 2π Μ + 3π Μ and π β = 3π Μ β 2π Μ + π Μ = 3π Μ β 2π Μ + 1π Μ We know that π β . π β = |π β ||π β | cos ΞΈ ; ΞΈ is the angle between π β & π β Now, π β. π β = (1π Μ β 2π Μ + 3π Μ). (3π Μ β 2π Μ + 1π Μ) = 1.3 + (β2).(β2) + 3.1 = 3 + 4 + 3 = 10 Magnitude of π β = β(12+(β2)2+32) |π β |= β(1+4+9) = β14 Magnitude of π β = β(32+(β2)2+12) |π β |= β(9+4+1) = β14 Now, π β . π β = |π β ||π β | cos ΞΈ 10 = β14 Γ β14 x cos ΞΈ 10 = 14 Γ cos ΞΈ cos ΞΈ = 10/14 ΞΈ = cos-1(π/π) Thus, the angle between π β and π β is cos-1(5/7)

Ex 10.3

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Ex 10.3, 2 You are here

Ex 10.3, 3 Important

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Ex 10.3, 18 (MCQ) Important

About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.