Question

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

• A. $\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{1}{a^{\prime 2}}+\frac{1}{b^{\prime 2}}+\frac{1}{c^{\prime 2}}=0$
• B. $\frac{1}{a^2}-\frac{1}{b^2}-\frac{1}{c^2}-\frac{1}{a^{\prime 2}}-\frac{1}{b^{\prime 2}}-\frac{1}{c^{\prime 2}}=0$
• C. $\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}-\frac{1}{a^{\prime 2}}-\frac{1}{b^{\prime 2}}-\frac{1}{c^{\prime 2}}=0$
• D. $\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{1}{a^{\prime 2}}+\frac{1}{b^{\prime 2}}+\frac{1}{c^{\prime 2}}=0$

Answer 1

The correct option is C

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

The planes are:
$\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1and\frac{x}{a^{\prime }}+\frac{y}{b^{\prime }}+\frac{z}{c^{\prime }}=1$
The perpendicular distance of the plane from the origin is the same for both cases.
$\therefore \left|\begin{array}{c}\frac{-1}{\sqrt{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}}}\end{array}\right|=\left|\begin{array}{c}\frac{-1}{\sqrt{\frac{1}{a^{\prime 2}}+\frac{1}{b^{\prime 2}}+\frac{1}{c^{\prime 2}}}}\end{array}\right|\Rightarrow \frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}-\frac{1}{a^{\prime 2}}-\frac{1}{b^{\prime 2}}-\frac{1}{c^{\prime 2}}=0$