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Ex 11.3
Ex 11.3, 1 (b)
Ex 11.3, 1 (c) Important
Ex 11.3, 1 (d) Important
Ex 11.3, 2
Ex 11.3, 3 (a)
Ex 11.3, 3 (b)
Ex 11.3, 3 (c) Important
Ex 11.3, 4 (a) Important
Ex 11.3, 4 (b)
Ex 11.3, 4 (c)
Ex 11.3, 4 (d) Important
Ex 11.3, 5 (a) Important
Ex 11.3, 5 (b)
Ex 11.3, 6 (a) Important
Ex 11.3, 6 (b)
Ex 11.3, 7
Ex 11.3, 8
Ex 11.3, 9
Ex 11.3, 10 Important
Ex 11.3, 11 Important
Ex 11.3, 12 Important Deleted for CBSE Board 2022 Exams You are here
Ex 11.3, 13 (a) Important Deleted for CBSE Board 2022 Exams
Ex 11.3, 13 (b) Important
Ex 11.3, 13 (c)
Ex 11.3, 13 (d)
Ex 11.3, 13 (e) Deleted for CBSE Board 2022 Exams
Ex 11.3, 14 (a) Important
Ex 11.3, 14 (b)
Ex 11.3, 14 (c)
Ex 11.3, 14 (d) Important
Last updated at Feb. 1, 2020 by Teachoo
Ex 11.3, 12 Find the angle between the planes whose vector equations are π β . (2π Μ + 2π Μ β 3π Μ) = 5 and π β . (3π Μ β 3π Μ + 5π Μ) = 3 .Angle between two planes π β . (π_1 ) β = d1 and π β.(π2) β = d2 is given by cos π = |((ππ) β. (ππ) β)/|(ππ) β ||(ππ) β | | π β.(2π Μ + 2π Μ β 3π Μ) = 5 Comparing with π β.(π1) β = (π1) β, (π1) β = 2π Μ + 2π Μβ3π Μ Magnitude of (π1) β = β(2^2+2^2+γ(β3)γ^2 ) |(π1) β | = β(4+4+9) = β17 So, cos ΞΈ = |((2π Μ + 2π Μ β 3π Μ ) . (3π Μ β 3π Μ + 5π Μ ))/(β17 Γ β43)| = |((2 Γ 3) + (2 Γ β3) + (β3 Γ 5))/β(17 Γ 43)| = |(6 β 6 β 15)/β731| = |(β15)/β731| = 15/β731 So, cos ΞΈ = 15/β731 β΄ ΞΈ = cosβ1(ππ/βπππ) Therefore, the angle between the planes is cosβ1(15/β731).