Last updated at Feb. 1, 2020 by Teachoo
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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/π2 + 1/π2 + 1/π2 = 1/π2 . Distance of the point (π₯_1,π¦_1,π§_1) from the plane Ax + By + Cz = D is |(π¨π_π + π©π_π + πͺπ_π β π«)/β(π¨^π + π©^π + πͺ^π )| The equation of a plane having intercepts π, b, c on the x β, y β & z β axis respectively is π/π + π/π + π/π = 1 Comparing with Ax + By + Cz = D, A = 1/π , B = 1/π , C = 1/π , D = 1 Given, the plane is at a distance of βπβ units from the origin. So, The point is O(0, 0, 0) So, π₯_1 = 0, π¦_1= 0, π§_1= 0 Now, Distance = |(π΄π₯_1 + π΅π¦_1 + πΆπ§_1 β π·)/β(π΄^2 + π΅^2 + πΆ^2 )| Putting values π = |(1/π Γ 0 + 1/π Γ 0 + 1/π Γ 0 β 1)/β((1/π)^2+ (1/π)^2+ (1/π)^2 )| π = |(0 + 0 + 0 β 1)/(β(1/π^2 + 1/π^2 + 1/π^2 ) )| π = |(β1)/(β(1/π^2 + 1/π^2 + 1/π^2 ) )| π = 1/(β(1/π^2 + 1/π^2 + 1/π^2 ) ) 1/π = β(1/π^2 + 1/π^2 + 1/π^2 ) Squaring both sides 1/π^2 = 1/π^2 + 1/π^2 + 1/π^2 Hence proved.
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