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Ex 11.3, 1 In each of the following cases, determine the direction cosines of the normal to the plane and the distance from the origin. (a) z = 2 For plane ax + by + cz = d Direction ratios of normal = a, b, c Direction cosines : l = π‘Ž/√(π‘Ž^(2 )+ 𝑏^2 + 𝑐^2 ) , m = 𝑏/√(π‘Ž^2 +γ€– 𝑏〗^2 + 𝑐^2 ) , n = 𝑐/√(π‘Ž^2 + 𝑏^2 + 𝑐^2 ) Distance from origin = 𝑑/√(π‘Ž^2 + 𝑏^(2 )+ 𝑐^2 ) Given equation of plane is z = 2 0x + 0y + 1z = 2 Comparing with ax + by + cz = d a = 0, b = 0, c = 1 & d = 2 And, √(𝒂^𝟐+𝒃^𝟐+𝒄^𝟐 ) = √(0^2+0^2+1^2 ) = 1 Direction cosines Direction cosines of the normal to the plane are l = π‘Ž/√(π‘Ž^2 + 𝑏^2 + 𝑐^2 ) , m = 𝑏/√(π‘Ž^2 + 𝑏^2 + 𝑐^2 ) , n = 𝑐/√(π‘Ž^2 + 𝑏^2 + 𝑐^2 ) l = 0/1 , m = 0/1 , n = 1/1 l = 0, m = 0, n = 1 ∴ Direction cosines of the normal to the plane are = (0, 0, 1) Distance from origin Distance form the origin = 𝑑/√(π‘Ž^2 + 𝑏^2 + 𝑐^2 ) = 2/1 = 2

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.