Ex 11.3, 1 - Determine direction cosines of normal to plane and dist

Ex 11.3, 1 - Chapter 11 Class 12 Three Dimensional Geometry - Part 2
Ex 11.3, 1 - Chapter 11 Class 12 Three Dimensional Geometry - Part 3

  1. Chapter 11 Class 12 Three Dimensional Geometry (Term 2)
  2. Serial order wise

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. (a) z = 2 For plane ax + by + cz = d Direction ratios of normal = a, b, c Direction cosines : l = ๐‘Ž/โˆš(๐‘Ž^(2 )+ ๐‘^2 + ๐‘^2 ) , m = ๐‘/โˆš(๐‘Ž^2 +ใ€– ๐‘ใ€—^2 + ๐‘^2 ) , n = ๐‘/โˆš(๐‘Ž^2 + ๐‘^2 + ๐‘^2 ) Distance from origin = ๐‘‘/โˆš(๐‘Ž^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 & โˆš(๐‘Ž^2+๐‘^2+๐‘^2 ) = โˆš(0^2+0^2+1^2 ) = 1 Direction cosines of the normal to the plane are l = ๐‘Ž/โˆš(๐‘Ž^2 + ๐‘^2 + ๐‘^2 ) , m = ๐‘/โˆš(๐‘Ž^2 + ๐‘^2 + ๐‘^2 ) , n = ๐‘/โˆš(๐‘Ž^2 + ๐‘^2 + ๐‘^2 ) l = 0/1 , m = 0/1 , n = 1/1 l = 0, m = 0, n = 1 โˆด Direction cosines of the normal to the plane are = (0, 0, 1) And, Distance form the origin = ๐‘‘/โˆš(๐‘Ž^2 + ๐‘^2 + ๐‘^2 ) = 2/1 = 2

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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.