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Question 1 (a) - Equation of plane - In Normal Form - Chapter 11 Class 12 Three Dimensional Geometry
Last updated at August 13, 2026 by Teachoo
Chapter 11 Class 12 Three Dimensional Geometry
Concept wise
- Direction cosines and ratios
- Equation of line - given point and //vector
- Equation of line - given 2 points
- Angle between two lines - Vector
- Angle between two lines - Cartisian
- Angle between two lines - Direction ratios or cosines
- Shortest distance between two skew lines
- Shortest distance between two parallel lines
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Equation of plane - In Normal Form
- Equation of plane - Prependicular to Vector & Passing Through Point
- Equation of plane - Passing Through 3 Non Collinear Points
- Equation of plane - Intercept Form
- Equation of plane - Passing Through Intersection Of Planes
- Coplanarity of 2 lines
- Angle between two planes
- Distance of point from plane
- Angle between Line and Plane
- Equation of line under planes condition
- Point with Lines and Planes
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Transcript
Question 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 And, โ(๐^๐+๐^๐+๐^๐ ) = โ(0^2+0^2+1^2 ) = 1 Direction cosines 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) Distance from origin Distance form the origin = ๐/โ(๐^2 + ๐^2 + ๐^2 ) = 2/1 = 2
