Two systems of rectangular axis have the same origin. If a plane cuts them at distances $a,b,c$ and ${a^\prime },{b^\prime },{c^\prime }$, respectively from the origin, then prove that $\frac{1}{{{a^2}}} + \frac{1}{{{b^2}}} + \frac{1}{{{c^2}}} = \frac{1}{{{a^{\prime 2}}}} + \frac{1}{{{b^{\prime 2}}}} + \frac{1}{{{c^{\prime 2}}}}$.

Consider OX, OY, OZ and ox, oy, oz are two system of rectangular axes.
Let their corresponding equation of plane be

$\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$

………(i)
and $\frac{x}{{{a^\prime }}} + \frac{y}{{{b^\prime }}} + \frac{z}{{{c^\prime }}} = 1$

……..(ii)
Also, the length of perpendicular from origin to Eqs. (i) and (ii) must be same.

$\therefore$ $\frac{{\frac{0}{a} + \frac{0}{b} + \frac{0}{c} - 1}}{{\sqrt {\frac{1}{{{a^2}}} + \frac{1}{{{b^2}}} + \frac{1}{{{c^2}}}} }}$

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

$\Rightarrow$ $\sqrt {\frac{1}{{{a^2}}} + \frac{1}{{{b^2}}} + \frac{1}{{{c^2}}}} = \sqrt {\frac{1}{{{a^2}}} + \frac{1}{{{b^2}}} + \frac{1}{{{c^2}}}}$

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

Long Answer Questions