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Last updated at Aug. 24, 2021 by Teachoo

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 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

Ex 11.3, 1 (a)

Ex 11.3, 1 (b)

Ex 11.3, 1 (c) Important You are here

Ex 11.3, 1 (d) Important

Ex 11.3, 2

Ex 11.3, 3 (a)

Ex 11.3, 3 (b)

Ex 11.3, 3 (c) Important

Ex 11.3, 4 (a) Important

Ex 11.3, 4 (b)

Ex 11.3, 4 (c)

Ex 11.3, 4 (d) Important

Ex 11.3, 5 (a) Important

Ex 11.3, 5 (b)

Ex 11.3, 6 (a) Important

Ex 11.3, 6 (b)

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 2022 Exams

Ex 11.3, 13 (a) Important Deleted for CBSE Board 2022 Exams

Ex 11.3, 13 (b) Important

Ex 11.3, 13 (c)

Ex 11.3, 13 (d)

Ex 11.3, 13 (e) Deleted for CBSE Board 2022 Exams

Ex 11.3, 14 (a) Important

Ex 11.3, 14 (b)

Ex 11.3, 14 (c)

Ex 11.3, 14 (d) Important

Chapter 11 Class 12 Three Dimensional Geometry (Term 2)

Serial order wise

About the Author

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.