Ex 11.3, 1 - Chapter 11 Class 12 Three Dimensional Geometry - Part 8

Ex 11.3, 1 - Chapter 11 Class 12 Three Dimensional Geometry - Part 9
Ex 11.3, 1 - Chapter 11 Class 12 Three Dimensional Geometry - Part 10

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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 = š‘Ž/√(š‘Ž^2 + š‘^2 +怖 š‘ć€—^2 ) , m = š‘/√(š‘Ž^2 + š‘^2 + š‘^2 ) , n = š‘/√(š‘Ž^2 + š‘^2 + š‘^2 ) Distance from origin = š‘‘/√(š‘Ž^2 + š‘^2 + š‘^2 ) Given, equation of the plane is 5y + 8 = 0 5y = āˆ’8 āˆ’5y = 8 0x āˆ’ 5y + 0z = 8 0x āˆ’ 5y + 0z = 8 Comparing with ax + by + cz = d a = 0, b = –5, c = 0 & d = 8 & √(š‘Ž^2+š‘^2+š‘^2 ) = √(0^2 + 怖(āˆ’5)怗^2 + 0^2 ) = √25 = 5 Direction cosines of the normal to the plane are l = š‘Ž/√(š‘Ž^2 + š‘^2 + š‘^2 ) , m = š‘/√(š‘Ž^2 + š‘^2 + š‘^2 ) , n = š‘/√(š‘Ž^2 + š‘^2 + š‘^2 ) l = 0/5, m = (āˆ’5)/5, n = ( 0)/5 ∓ Direction cosines of the normal to the plane are = (0, –1, 0) And, Distance form the origin = š‘‘/√(š‘Ž^2 + š‘^2 + š‘^2 ) = šŸ–/šŸ“

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