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Question 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 = 𝑎/√(𝑎^(2 )+ 𝑏^2 + 𝑐^2 ) , m = 𝑏/√(𝑎^2 +〖 𝑏〗^2 + 𝑐^2 ) , n = 𝑐/√(𝑎^2 + 𝑏^2 + 𝑐^2 ) Distance from origin = 𝑑/√(𝑎^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 = 𝑎/√(𝑎^2 + 𝑏^2 + 𝑐^2 ) , m = 𝑏/√(𝑎^2 + 𝑏^2 + 𝑐^2 ) , n = 𝑐/√(𝑎^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/√(𝑎^2 + 𝑏^2 + 𝑐^2 ) = 2/1 = 2

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo