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  1. Chapter 11 Class 12 Three Dimensional Geometry
  2. Serial order wise

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 1/π‘Ž2 + 1/𝑏2 + 1/𝑐2 = 1/𝑝2 . Distance of the point (π‘₯_1,𝑦_1,𝑧_1) from the plane Ax + By + Cz = D is |(𝑨𝒙_𝟏 + π‘©π’š_𝟏 + π‘ͺ𝒛_𝟏 βˆ’ 𝑫)/√(𝑨^𝟐 + 𝑩^𝟐 + π‘ͺ^𝟐 )| The equation of a plane having intercepts π‘Ž, b, c on the x βˆ’, y βˆ’ & z βˆ’ axis respectively is 𝒙/𝒂 + π’š/𝒃 + 𝒛/𝒄 = 1 Comparing with Ax + By + Cz = D, A = 1/π‘Ž , B = 1/𝑏 , C = 1/𝑐 , D = 1 Given, the plane is at a distance of β€˜π‘β€™ units from the origin. So, The point is O(0, 0, 0) So, π‘₯_1 = 0, 𝑦_1= 0, 𝑧_1= 0 Now, Distance = |(𝐴π‘₯_1 + 𝐡𝑦_1 + 𝐢𝑧_1 βˆ’ 𝐷)/√(𝐴^2 + 𝐡^2 + 𝐢^2 )| Putting values 𝑝 = |(1/π‘Ž Γ— 0 + 1/𝑏 Γ— 0 + 1/𝑐 Γ— 0 βˆ’ 1)/√((1/π‘Ž)^2+ (1/𝑏)^2+ (1/𝑐)^2 )| 𝑝 = |(0 + 0 + 0 βˆ’ 1)/(√(1/π‘Ž^2 + 1/𝑏^2 + 1/𝑐^2 ) )| 𝑝 = |(βˆ’1)/(√(1/π‘Ž^2 + 1/𝑏^2 + 1/𝑐^2 ) )| 𝑝 = 1/(√(1/π‘Ž^2 + 1/𝑏^2 + 1/𝑐^2 ) ) 1/𝑝 = √(1/π‘Ž^2 + 1/𝑏^2 + 1/𝑐^2 ) Squaring both sides 1/𝑝^2 = 1/π‘Ž^2 + 1/𝑏^2 + 1/𝑐^2 Hence proved.

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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.