Plane

Chapter 11 Class 12 Three Dimensional Geometry
Serial order wise

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Question 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.(π2) β = d2 is given by cos π = |((ππ) β. (ππ) β)/|(ππ) β ||(ππ) β | | π β.(2π Μ + 2π Μ β 3π Μ) = 5 Comparing with π β.(π1) β = (π1) β, (π1) β = 2π Μ + 2π Μβ3π Μ Magnitude of (π1) β = β(2^2+2^2+γ(β3)γ^2 ) |(π1) β | = β(4+4+9) = β17 So, cos ΞΈ = |((2π Μ + 2π Μ β 3π Μ ) . (3π Μ β 3π Μ + 5π Μ ))/(β17 Γ β43)| = |((2 Γ 3) + (2 Γ β3) + (β3 Γ 5))/β(17 Γ 43)| = |(6 β 6 β 15)/β731| = |(β15)/β731| = 15/β731 So, cos ΞΈ = 15/β731 β΄ ΞΈ = cosβ1(ππ/βπππ) Therefore, the angle between the planes is cosβ1(15/β731).