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Misc 21 - Prove that if a plane has intercepts a, b, c, p from

Misc 21 - Chapter 11 Class 12 Three Dimensional Geometry - Part 2
Misc 21 - Chapter 11 Class 12 Three Dimensional Geometry - Part 3

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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 has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.