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Ex 11.3, 1 - Chapter 11 Class 12 Three Dimensional Geometry - Part 6

Ex 11.3, 1 - Chapter 11 Class 12 Three Dimensional Geometry - Part 7


Transcript

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. (c) 2x + 3y – z = 5 For plane ax + by + cz = d Direction ratios of normal = a, b, c Direction cosines : l = π‘Ž/√(π‘Ž^2 + 𝑏^2 +γ€– 𝑐〗^2 ) , m = 𝑏/√(π‘Ž^2 + 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 2x + 3y βˆ’ z = 5 2x + 3y βˆ’ 1z = 5 Comparing with ax + by + cz = d a = 2, b = 3, c = –1 & d = 5 & √(π‘Ž^2+𝑏^2+𝑐^2 ) = √(2^2 + 3^2 + γ€–(βˆ’1)γ€—^2 ) = √(4+9+1) = √14 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 = 2/√14, m = 3/√14, n = ( βˆ’ 1)/√14 ∴ Direction cosines of the normal to the plane are = (𝟐/βˆšπŸπŸ’, πŸ‘/βˆšπŸπŸ’, ( βˆ’πŸ)/βˆšπŸπŸ’) And, Distance form the origin = 𝑑/√(π‘Ž^2 + 𝑏^2 +γ€– 𝑐〗^2 ) = πŸ“/βˆšπŸπŸ’

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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.