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Three Dimensional Geometry — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsThree Dimensional Geometry3 Marks
Question
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 $\frac{1}{\text{a}^2}+\frac{1}{\text{b}^2}+\frac{1}{\text{c}^2}=\frac{1}{\text{a}'^2}+\frac{1}{\text{b}'^2}+\frac{1}{\text{c}'^2}.$
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Answer

Consider OX, OY, OZ and ox, oy, oz are two system of rectangular axes. Let their corresponding equation of plane be
$\frac{\text{x}}{\text{a}}+\frac{\text{y}}{\text{b}}+\frac{\text{z}}{\text{c}}=1\ ....(\text{i})$
And $\frac{\text{x}}{\text{a}'}+\frac{\text{y}}{\text{b}'}+\frac{\text{z}}{\text{c}'}=1\ ....(\text{ii})$
Also the length of perpendicular from origin to equations (i) and (ii) must be same.
$\therefore\frac{\frac{0}{\text{a}}+\frac{0}{\text{b}}+\frac{0}{\text{c}}-1}{\sqrt{\frac{1}{\text{a}^2}+\frac{1}{\text{b}^2}+\frac{1}{\text{c}^2}}}=\frac{\frac{0}{\text{a}'}+\frac{0}{\text{b}'}+\frac{0}{\text{c}'}-1}{\sqrt{\frac{1}{\text{a}'^2}+\frac{1}{\text{b}'^2}+\frac{1}{\text{c}'^2}}}$
$\frac{1}{\text{a}^2}+\frac{1}{\text{b}^2}+\frac{1}{\text{c}^2}=\frac{1}{\text{a}'^2}+\frac{1}{\text{b}'^2}+\frac{1}{\text{c}'^2}$

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