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Ex 11.3, 1 - Chapter 11 Class 12 Three Dimensional Geometry - Part 8

Ex 11.3, 1 - Chapter 11 Class 12 Three Dimensional Geometry - Part 9
Ex 11.3, 1 - Chapter 11 Class 12 Three Dimensional Geometry - Part 10

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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. (d) 5y + 8 = 0 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 the plane is 5y + 8 = 0 5y = βˆ’8 βˆ’5y = 8 0x βˆ’ 5y + 0z = 8 0x βˆ’ 5y + 0z = 8 Comparing with ax + by + cz = d a = 0, b = –5, c = 0 & d = 8 & √(π‘Ž^2+𝑏^2+𝑐^2 ) = √(0^2 + γ€–(βˆ’5)γ€—^2 + 0^2 ) = √25 = 5 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 = 0/5, m = (βˆ’5)/5, n = ( 0)/5 ∴ Direction cosines of the normal to the plane are = (0, –1, 0) And, Distance form the origin = 𝑑/√(π‘Ž^2 + 𝑏^2 + 𝑐^2 ) = π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.