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

Ex 11.3

Ex 11.3, 1
Important
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Ex 11.3, 2

Ex 11.3, 3

Ex 11.3, 4 Important

Ex 11.3, 5 Important

Ex 11.3, 6 Important

Ex 11.3, 7

Ex 11.3, 8

Ex 11.3, 9

Ex 11.3, 10 Important

Ex 11.3, 11 Important

Ex 11.3, 12 Important Deleted for CBSE Board 2021 Exams only

Ex 11.3, 13 (a) Important

Ex 11.3, 13 (b)

Ex 11.3, 13 (c)

Ex 11.3, 13 (d)

Ex 11.3, 13 (e)

Ex 11.3, 14 (a) Important

Ex 11.3, 14 (b)

Ex 11.3, 14 (c)

Ex 11.3, 14 (d)

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 10 years. He provides courses for Maths and Science at Teachoo.