Q. Two systems of rectangular axes have the same origin. If a plane cuts them at distance $a,\ b,\ c$ and $a^{'},\ b^{'},\ c^{'}$ from the origin, then:
A) $\frac{1}{a^{2}} + \frac{1}{b^{2}} + \frac{1}{c^{2}} + \frac{1}{a^{'^{2}}} + \frac{1}{b^{'^{2}}} + \frac{1}{c^{'^{2}}} = 0$
B) $\frac{1}{a^{2}} - \frac{1}{b^{2}} - \frac{1}{c^{2}} + \frac{1}{a^{'^{2}}} - \frac{1}{b^{'^{2}}} - \frac{1}{c^{'^{2}}} = 0$
C) $\frac{1}{a^{2}} + \frac{1}{b^{2}} + \frac{1}{c^{2}} - \frac{1}{a^{'^{2}}} - \frac{1}{b^{'^{2}}} - \frac{1}{c^{'^{2}}} = 0$
D) $\frac{1}{a^{2}} + \frac{1}{b^{2}} + \frac{1}{c^{2}} + \frac{1}{a^{'^{2}}} + \frac{1}{b^{'^{2}}} + \frac{1}{c^{'^{2}}} = 0$
Explanation
(c)
The planes are $\frac{x}{a} + \frac{y}{b} + \frac{\mathcal{z}}{c} = 1$ and $\frac{x}{a^{'}} + \frac{y}{b^{'}} + \frac{\mathcal{z}}{c^{'}} = 1$
Since the perpendicular distance of the origin on the planes is same, therefore
$\left| \frac{- 1}{\sqrt{\frac{1}{a^{2}} + \frac{1}{b^{2}} + \frac{1}{c^{2}}}} \right| = \left| \frac{- 1}{\sqrt{\frac{1}{a^{'2}} + \frac{1}{b^{'2}} + \frac{1}{c^{'2}}}} \right|$
$\Rightarrow \ \frac{1}{a^{2}} + \frac{1}{b^{2}} + \frac{1}{c^{2}} - \frac{1}{a^{'2}} - \frac{1}{b^{'2}} - \frac{1}{c^{'2}} = 0$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.