Example 15  - Find distance of plane from origin - Class 12 - Examples

Example 15  - Chapter 11 Class 12 Three Dimensional Geometry - Part 2

Example 15  - Chapter 11 Class 12 Three Dimensional Geometry - Part 3

Example 15  - Chapter 11 Class 12 Three Dimensional Geometry - Part 4

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Question 5(Method 1) Find the distance of the plane 2x 3y + 4z 6 = 0 from the origin. Given, the equation of plane is 2x 3y + 4z 6 = 0 2x 3y + 4z = 6 Direction ratios of = , , a = 2, b = 3, c = 4 Also, 2 + 2 + 2 = 2 2 + ( 3) 2 + 4 2 = 4+9+16 = 29 Direction cosines are l = 2 + 2 + 2 , m = 2 + 2 + 2 , n = 2 + 2 + 2 l = 2 29 , m = 3 29 ,n = 4 29 Equation of plane is lx + my + nz = d 2 29 x 3 29 y + 4 29 z = d 2x 3y + 4z = d 29 Comparing with (1) i.e. 2x 3y + 4z = 6, d 29 = 6 d = Question 5(Method 2) Find the distance of the plane 2x 3y + 4z 6 = 0 from the origin. Distance of point P(x1, y1, z1) from plane Ax + By + Cz = D is d = 1 + 1 + 1 2 + 2 + 2 Since we have to find distance from Origin P(x1, y1, z1) = O(0, 0, 0) x1 = 0, y1 = 0, z1 = 0 & plane is 2x 3y + 4z 6 = 0 2x 3y + 4z = 6 Comparing with Ax + By + Cz = D A = 2, B = 3, C = 4 & D = 6 Putting values in formula d = 1 + 1 + 1 2 + 2 + 2 d = 2 0 3 0 + 4 0 6 2 2 + ( 3) 2 + 4 2 d = 6 4 + 9 + 16 d = 6 29 d =

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo