Question
Two system of rectangular axes have the same origin. If a plane cuts them at distances $$a,\,b,\,c$$ and $$a',\,b',\,c'$$ from the origin then :
A.
$$\frac{1}{{{a^2}}} + \frac{1}{{{b^2}}} + \frac{1}{{{c^2}}} - \frac{1}{{a{'^2}}} - \frac{1}{{b{'^2}}} - \frac{1}{{c{'^2}}} = 0$$
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.
$$\frac{1}{{{a^2}}} - \frac{1}{{{b^2}}} - \frac{1}{{{c^2}}} + \frac{1}{{a{'^2}}} - \frac{1}{{b{'^2}}} - \frac{1}{{c{'^2}}} = 0$$
Answer :
$$\frac{1}{{{a^2}}} + \frac{1}{{{b^2}}} + \frac{1}{{{c^2}}} - \frac{1}{{a{'^2}}} - \frac{1}{{b{'^2}}} - \frac{1}{{c{'^2}}} = 0$$
Solution :
Equation of planes be $$\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$$ & $$\frac{x}{{a'}} + \frac{y}{{b'}} + \frac{z}{{c'}} = 1$$
( $$ \bot \,\,r$$ distance on plane from origin is same. )
$$\eqalign{
& \left| {\frac{{ - 1}}{{\sqrt {\frac{1}{{{a^2}}} + \frac{1}{{{b^2}}} + \frac{1}{{{c^2}}}} }}} \right| = \left| {\frac{{ - 1}}{{\sqrt {\frac{1}{{a{'^2}}} + \frac{1}{{b{'^2}}} + \frac{1}{{c{'^2}}}} }}} \right| \cr
& = \frac{1}{{{a^2}}} + \frac{1}{{{b^2}}} + \frac{1}{{{c^2}}} - \frac{1}{{a{'^2}}} - \frac{1}{{b{'^2}}} - \frac{1}{{c{'^2}}} = 0 \cr} $$