

Ex 11.3
Ex 11.3, 1 (b)
Ex 11.3, 1 (c) Important
Ex 11.3, 1 (d) Important
Ex 11.3, 2
Ex 11.3, 3 (a)
Ex 11.3, 3 (b)
Ex 11.3, 3 (c) Important
Ex 11.3, 4 (a) Important
Ex 11.3, 4 (b)
Ex 11.3, 4 (c)
Ex 11.3, 4 (d) Important
Ex 11.3, 5 (a) Important
Ex 11.3, 5 (b)
Ex 11.3, 6 (a) Important
Ex 11.3, 6 (b)
Ex 11.3, 7
Ex 11.3, 8
Ex 11.3, 9
Ex 11.3, 10 Important
Ex 11.3, 11 Important
Ex 11.3, 12 Important Deleted for CBSE Board 2022 Exams
Ex 11.3, 13 (a) Important Deleted for CBSE Board 2022 Exams
Ex 11.3, 13 (b) Important
Ex 11.3, 13 (c)
Ex 11.3, 13 (d)
Ex 11.3, 13 (e) Deleted for CBSE Board 2022 Exams
Ex 11.3, 14 (a) Important
Ex 11.3, 14 (b)
Ex 11.3, 14 (c)
Ex 11.3, 14 (d) Important
Last updated at Dec. 11, 2021 by Teachoo
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 = π/β(π^(2 )+ π^2 + π^2 ) , m = π/β(π^2 +γ πγ^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 z = 2 0x + 0y + 1z = 2 Comparing with ax + by + cz = d a = 0, b = 0, c = 1 & d = 2 And, β(π^π+π^π+π^π ) = β(0^2+0^2+1^2 ) = 1 Direction cosines 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 = 0/1 , m = 0/1 , n = 1/1 l = 0, m = 0, n = 1 β΄ Direction cosines of the normal to the plane are = (0, 0, 1) Distance from origin Distance form the origin = π/β(π^2 + π^2 + π^2 ) = 2/1 = 2