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Home  >>  CBSE XII  >>  Math  >>  Three Dimensional Geometry
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Two systems of rectangular axis have the same origin.If a plane cuts them at distances a,b,c and a',b',c',respectively,from the origin,prove that\[\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=\frac{1}{{a'}^2}+\frac{1}{{b'}^2}+\frac{1}{{c'}^2}\]

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Toolbox:
  • Perpendicular diatance of the plane $ax+by+cz+d=0$ from origin is $\bigg|\large\frac{d}{\sqrt {a^2+b^2+c^2}}\bigg|$
  • If two system of lines have the same origin then their $ \perp$ distance from origin to the plane in both the system are equal.
Let the equation of the plane in both the systems be
$ \large\frac{x}{a}+\large\frac{y}{b}+\large\frac{z}{c}$$=1$ and $ \large\frac{X}{a'}+\large\frac{Y}{b'}+\large\frac{Z}{c'}$$=1$
We know that $\perp$ diatance of the plane $ax+by+cz+d=0$ from origin is $\bigg|\large\frac{d}{\sqrt {a^2+b^2+c^2}}\bigg|$
It is given that the origin is same for both the system.
Hence $\perp$ distance of the plane from origin are equal for both the system.
$(ie)\Bigg| \Large\frac{-1}{\sqrt{\Large\frac{1}{a^2}+\Large\frac{1}{b^2}+\Large\frac{1}{c^2}}} \Bigg|$$ = \Bigg| \Large\frac{-1}{\sqrt{\Large\frac{1}{a'^2}+\Large\frac{1}{b'^2}+\Large\frac{1}{c'^2}}} \Bigg| $
$\Rightarrow\: \large\frac{1}{a^2}+\large\frac{1}{b^2}+\large\frac{1}{c^2}=\large\frac{1}{a'^2}+\large\frac{1}{b'^2}+\large\frac{1}{c'^2}$
Hence proved.
answered Jun 11, 2013 by meena.p
 

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