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  1. Chapter 11 Class 12 Three Dimensional Geometry
  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 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‘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 plane is x + y + z = 1 1x + 1y + 1z = 1 Comparing with ax + by + cz = d a = 1, b = 1, c = 1 & d = 1 & โˆš(๐‘Ž^2+๐‘^2+๐‘^2 ) = โˆš(1^2+1^2+1^2 ) = โˆš3 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 = 1/โˆš3 , m = 1/โˆš3, n = 1/โˆš3 โˆด Direction cosines of the normal to the plane are = (๐Ÿ/โˆš๐Ÿ‘, ๐Ÿ/โˆš๐Ÿ‘, ๐Ÿ/โˆš๐Ÿ‘) And, Distance form the origin = ๐‘‘/โˆš(๐‘Ž^2 + ๐‘^2 + ๐‘^2 ) = ๐Ÿ/โˆš๐Ÿ‘ 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‘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 plane is 2x + 3y โˆ’ z = 5 2x + 3y โˆ’ 1z = 5 Comparing with ax + by + cz = d a = 2, b = 3, c = โ€“1 & d = 5 & โˆš(๐‘Ž^2+๐‘^2+๐‘^2 ) = โˆš(2^2 + 3^2 + ใ€–(โˆ’1)ใ€—^2 ) = โˆš(4+9+1) = โˆš14 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 = 2/โˆš14, m = 3/โˆš14, n = ( โˆ’ 1)/โˆš14 โˆด Direction cosines of the normal to the plane are = (๐Ÿ/โˆš๐Ÿ๐Ÿ’, ๐Ÿ‘/โˆš๐Ÿ๐Ÿ’, ( โˆ’๐Ÿ)/โˆš๐Ÿ๐Ÿ’) And, Distance form the origin = ๐‘‘/โˆš(๐‘Ž^2 + ๐‘^2 +ใ€– ๐‘ใ€—^2 ) = ๐Ÿ“/โˆš๐Ÿ๐Ÿ’ 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‘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“

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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.