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.

Miscellaneous

Misc 1
Important

Misc 2

Misc 3

Misc 4 Important

Misc 5 Important

Misc 6 Important

Misc 7

Misc 8 Important

Misc 9 Important

Misc 10

Misc 11 Important

Misc 12 Important

Misc 13 Important

Misc 14 Important

Misc 15 Important

Misc 16 Important

Misc 17 Important

Misc 18 Important

Misc 19 Important

Misc 20 Important

Misc 21 Important You are here

Misc 22 Important

Misc 23 Important

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.