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Question 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‘Ž/โˆš(๐‘Ž^2 + ๐‘^2 +ใ€– ๐‘ใ€—^2 ) , m = ๐‘/โˆš(๐‘Ž^2 + ๐‘^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 the plane is 5y + 8 = 0 5y = โˆ’8 โˆ’5y = 8 0x โˆ’ 5y + 0z = 8 0x โˆ’ 5y + 0z = 8 Comparing with ax + by + cz = d a = 0, b = โ€“5, c = 0 & d = 8 & โˆš(๐‘Ž^2+๐‘^2+๐‘^2 ) = โˆš(0^2 + ใ€–(โˆ’5)ใ€—^2 + 0^2 ) = โˆš25 = 5 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 = 0/5, m = (โˆ’5)/5, n = ( 0)/5 โˆด Direction cosines of the normal to the plane are = (0, โ€“1, 0) And, Distance form the origin = ๐‘‘/โˆš(๐‘Ž^2 + ๐‘^2 + ๐‘^2 ) = ๐Ÿ–/๐Ÿ“

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

About the Author

Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo