Plane

Chapter 11 Class 12 Three Dimensional Geometry
Serial order wise

Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class

### 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. (c) 2x + 3y β z = 5 For plane ax + by + cz = d Direction ratios of normal = a, b, c Direction cosines : l = π/β(π^2 + π^2 +γ πγ^2 ) , m = π/β(π^2 + π^2 + π^2 ) , n = π/β(π^2 + π^2 + π^2 ) Distance from origin = π/β(π^2 + π^2 + π^2 ) Given equation of plane is 2x + 3y β z = 5 2x + 3y β 1z = 5 Comparing with ax + by + cz = d a = 2, b = 3, c = β1 & d = 5 & β(π^2+π^2+π^2 ) = β(2^2 + 3^2 + γ(β1)γ^2 ) = β(4+9+1) = β14 Direction cosines of the normal to the plane are l = π/β(π^2 + π^2 + π^2 ) , m = π/β(π^2 + π^2 + π^2 ) , n = π/β(π^2 + π^2 + π^2 ) l = 2/β14, m = 3/β14, n = ( β 1)/β14 β΄ Direction cosines of the normal to the plane are = (π/βππ, π/βππ, ( βπ)/βππ) And, Distance form the origin = π/β(π^2 + π^2 +γ πγ^2 ) = π/βππ