Last updated at May 29, 2018 by Teachoo

Transcript

Misc 21 Prove that if a plane has the intercepts a, b, c and is at a distance of p units from the origin, then 1𝑎2 + 1𝑏2 + 1𝑐2 = 1𝑝2 . Distance of the point ( 𝑥1, 𝑦1, 𝑧1) from the plane Ax + By + Cz = D is 𝑨 𝒙𝟏 + 𝑩 𝒚𝟏 + 𝑪 𝒛𝟏 − 𝑫 𝑨𝟐 + 𝑩𝟐 + 𝑪𝟐 The equation of a plane having intercepts 𝑎, b, c on the x −, y− and z − axis respectively is 𝒙𝒂 + 𝒚𝒃 + 𝒛𝒄 = 1 Comparing with Ax + By + Cz = D, A = 1𝑎 , B = 1𝑏 , C = 1𝑐 , D = 1 Given, the plane is at a distance of ‘𝑝’ units from the origin. So, The point is O(0, 0, 0) So, 𝑥1 = 0, 𝑦1= 0, 𝑧1= 0 Now, Distance = 𝐴 𝑥1 + 𝐵 𝑦1 + 𝐶 𝑧1 − 𝐷 𝐴2 + 𝐵2 + 𝐶2 Putting values 𝑝 = 1𝑎 × 0 + 1𝑏×0 + 1𝑐 ×0−1 1𝑎2+ 1𝑏2+ 1𝑐2 𝑝 = 0 + 0 + 0 − 1 1 𝑎2 + 1 𝑏2 + 1 𝑐2 𝑝 = −1 1 𝑎2 + 1 𝑏2 + 1 𝑐2 𝑝 = 1 1 𝑎2 + 1 𝑏2 + 1 𝑐2 1𝑝 = 1 𝑎2 + 1 𝑏2 + 1 𝑐2 Squaring both sides 1 𝑝2 = 1 𝑎2 + 1 𝑏2 + 1 𝑐2 Hence proved.

Chapter 11 Class 12 Three Dimensional Geometry

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.