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(D) Let the equation of the plane in the first system of rectangular axes be $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$.The perpendicular distance

Vedclass · Accepted Answer

$\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2} - \frac{1}{a'^2} - \frac{1}{b'^2} - \frac{1}{c'^2} = 0$

Vedclass · Answer

(D) Let the equation of the plane in the first system of rectangular axes be $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$.The perpendicular distance $p$ from the origin $(0, 0, 0)$ to this plane is given by $p = \frac{|-1|}{\sqrt{\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2}}}$,which implies $\frac{1}{p^2} = \frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2}$.Similarly,for the second system of rectangular axes,the equation of the same plane is $\frac{x}{a'} + \frac{y}{b'} + \frac{z}{c'} = 1$.The perpendicular distance $p$ from the origin to this plane is the same,so $p = \frac{|-1|}{\sqrt{\frac{1}{a'^2} + \frac{1}{b'^2} + \frac{1}{c'^2}}}$,which implies $\frac{1}{p^2} = \frac{1}{a'^2} + \frac{1}{b'^2} + \frac{1}{c'^2}$.Equating the two expressions for $\frac{1}{p^2}$,we get $\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2} = \frac{1}{a'^2} + \frac{1}{b'^2} + \frac{1}{c'^2}$.Therefore,$\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2} - \frac{1}{a'^2} - \frac{1}{b'^2} - \frac{1}{c'^2} = 0$.

Two systems of rectangular axes have the same origin. If a plane cuts them at distances $a, b, c$ and $a', b', c'$ from the origin along the axes,then:

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