Q143 / 225 · JEE Main 2003 · Mathematics
Two system of rectangular axes have the same origin. If a plane cuts them at distances $\mathrm{a}, \mathrm{b}, \mathrm{c}$ and $\mathrm{a}^{\prime}, \mathrm{b}^{\prime}, \mathrm{c}^{\prime}$ from the origin then
- Correct answerA$\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}-\frac{1}{a^{\prime 2}}-\frac{1}{b^{\prime 2}}-\frac{1}{c^{\prime 2}}=0$
- B$\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{1}{a^{22}}+\frac{1}{b^{22}}+\frac{1}{c^{22}}=0$
- C$\frac{1}{\mathrm{a}^2}+\frac{1}{\mathrm{~b}^2}-\frac{1}{\mathrm{c}^2}+\frac{1}{\mathrm{a}^{\prime 2}}+\frac{1}{\mathrm{~b}^{\prime 2}}-\frac{1}{\mathrm{c}^{\prime 2}}=0$
- D$\frac{1}{a^2}-\frac{1}{b^2}-\frac{1}{c^2}+\frac{1}{a^{\prime 2}}-\frac{1}{b^{\prime 2}}-\frac{1}{c^{\prime 2}}=0$
Solution
Eq. of planes be $\frac{\mathrm{x}}{\mathrm{a}}+\frac{\mathrm{y}}{\mathrm{b}}+\frac{\mathrm{z}}{\mathrm{c}}=1 \& \frac{\mathrm{x}}{\mathrm{a}_1}+\frac{\mathrm{y}}{\mathrm{b}_1}+\frac{\mathrm{z}}{\mathrm{c}_1}=1(\perp \mathrm{r}$ distance on plane from origin is same. $)$
$\left|\frac{-1}{\sqrt{\frac{1}{\mathrm{a}^2}+\frac{1}{\mathrm{~b}^2}+\frac{1}{\mathrm{c}^2}}}\right|=\left|\frac{-1}{\sqrt{\frac{1}{\mathrm{a}_1^2}+\frac{1}{\mathrm{~b}_1^2}+\frac{1}{\mathrm{c}_1^2}}}\right|$
$\therefore \Sigma \frac{1}{a^2}-\Sigma \frac{1}{a_1^2}=0$
Hence, the correct answer is option: 'A'.