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Last updated at Feb. 4, 2020 by Teachoo

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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 = ๐/โ(๐^(2 )+ ๐^2 + ๐^2 ) , m = ๐/โ(๐^2 +ใ ๐ใ^2 + ๐^2 ) , n = ๐/โ(๐^2 + ๐^2 + ๐^2 ) Distance from origin = ๐/โ(๐^2 + ๐^(2 )+ ๐^2 ) 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 & โ(๐^2+๐^2+๐^2 ) = โ(0^2+0^2+1^2 ) = 1 Direction cosines of the normal to the plane are l = ๐/โ(๐^2 + ๐^2 + ๐^2 ) , m = ๐/โ(๐^2 + ๐^2 + ๐^2 ) , n = ๐/โ(๐^2 + ๐^2 + ๐^2 ) l = 0/1 , m = 0/1 , n = 1/1 l = 0, m = 0, n = 1 โด Direction cosines of the normal to the plane are = (0, 0, 1) And, Distance form the origin = ๐/โ(๐^2 + ๐^2 + ๐^2 ) = 2/1 = 2 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. (b) x + y + z = 1 For plane ax + by + cz = d Direction ratios of normal = a, b, c Direction cosines : l = ๐/โ(๐^2 + ๐^2 +ใ ๐ใ^2 ) , m = ๐/โ(๐^2 + ๐^2 + ๐^2 ) , n = ๐/โ(๐^2 + ๐^2 + ๐^2 ) Distance from origin = ๐/โ(๐^2 + ๐^2 + ๐^2 ) Given equation of plane is x + y + z = 1 1x + 1y + 1z = 1 Comparing with ax + by + cz = d a = 1, b = 1, c = 1 & d = 1 & โ(๐^2+๐^2+๐^2 ) = โ(1^2+1^2+1^2 ) = โ3 Direction cosines of the normal to the plane are l = ๐/โ(๐^2 + ๐^2 + ๐^2 ) , m = ๐/โ(๐^2 + ๐^2 + ๐^2 ) , n = ๐/โ(๐^2 + ๐^2 + ๐^2 ) l = 1/โ3 , m = 1/โ3, n = 1/โ3 โด Direction cosines of the normal to the plane are = (๐/โ๐, ๐/โ๐, ๐/โ๐) And, Distance form the origin = ๐/โ(๐^2 + ๐^2 + ๐^2 ) = ๐/โ๐ 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 โ z = 5 For plane ax + by + cz = d Direction ratios of normal = a, b, c Direction cosines : l = ๐/โ(๐^2 + ๐^2 +ใ ๐ใ^2 ) , m = ๐/โ(๐^2 + ๐^2 + ๐^2 ) , n = ๐/โ(๐^2 + ๐^2 + ๐^2 ) Distance from origin = ๐/โ(๐^2 + ๐^2 + ๐^2 ) Given equation of plane is 2x + 3y โ z = 5 2x + 3y โ 1z = 5 Comparing with ax + by + cz = d a = 2, b = 3, c = โ1 & d = 5 & โ(๐^2+๐^2+๐^2 ) = โ(2^2 + 3^2 + ใ(โ1)ใ^2 ) = โ(4+9+1) = โ14 Direction cosines of the normal to the plane are l = ๐/โ(๐^2 + ๐^2 + ๐^2 ) , m = ๐/โ(๐^2 + ๐^2 + ๐^2 ) , n = ๐/โ(๐^2 + ๐^2 + ๐^2 ) l = 2/โ14, m = 3/โ14, n = ( โ 1)/โ14 โด Direction cosines of the normal to the plane are = (๐/โ๐๐, ๐/โ๐๐, ( โ๐)/โ๐๐) And, Distance form the origin = ๐/โ(๐^2 + ๐^2 +ใ ๐ใ^2 ) = ๐/โ๐๐ 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 = ๐/โ(๐^2 + ๐^2 +ใ ๐ใ^2 ) , m = ๐/โ(๐^2 + ๐^2 + ๐^2 ) , n = ๐/โ(๐^2 + ๐^2 + ๐^2 ) Distance from origin = ๐/โ(๐^2 + ๐^2 + ๐^2 ) Given, equation of the plane is 5y + 8 = 0 5y = โ8 โ5y = 8 0x โ 5y + 0z = 8 0x โ 5y + 0z = 8 Comparing with ax + by + cz = d a = 0, b = โ5, c = 0 & d = 8 & โ(๐^2+๐^2+๐^2 ) = โ(0^2 + ใ(โ5)ใ^2 + 0^2 ) = โ25 = 5 Direction cosines of the normal to the plane are l = ๐/โ(๐^2 + ๐^2 + ๐^2 ) , m = ๐/โ(๐^2 + ๐^2 + ๐^2 ) , n = ๐/โ(๐^2 + ๐^2 + ๐^2 ) l = 0/5, m = (โ5)/5, n = ( 0)/5 โด Direction cosines of the normal to the plane are = (0, โ1, 0) And, Distance form the origin = ๐/โ(๐^2 + ๐^2 + ๐^2 ) = ๐/๐

Chapter 11 Class 12 Three Dimensional Geometry

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.