Question

Two systems of rectangular axes have the same origin. If a plane cuts them at a distance $a,b,c$ and $a{}^{\prime },b{}^{\prime },c{}^{\prime }$ from the origin then

• A. $a^{-2}+b^{-2}-c^{-2}+a^{\prime }{}^{-2}+b^{\prime }{}^{-2}-c^{\prime }{}^{-2}=0$
• B. $a^{-2}-b^{-2}-c^{-2}+a^{\prime }{}^{-2}-b^{\prime }{}^{-2}-c^{\prime }{}^{-2}=0$
• C. $a^{-2}+b^{-2}+c^{-2}-a^{\prime }{}^{-2}-b^{\prime }{}^{-2}-c^{\prime }{}^{-2}=0$
• D. none of these

Answer 1

The correct option is C $a^{-2}+b^{-2}+c^{-2}-a^{\prime }{}^{-2}-b^{\prime }{}^{-2}-c^{\prime }{}^{-2}=0$

Equation of plane with given intercept are,
$\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1$ and
$\frac{x}{a^{\prime }}+\frac{y}{b^{\prime }}+\frac{z}{c^{\prime }}=1$
Since the perpendicular distance of origin on he plane is same, therefore
$\mid \frac{-1}{\sqrt{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}}}\mid =\mid \frac{-1}{\sqrt{\frac{1}{a^{\prime 2}}+\frac{1}{b^{\prime 2}}+\frac{1}{c^{\prime 2}}}}\mid$

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

$\Rightarrow a^{-2}+b^{-2}+c^{-2}-a^{\prime -2}-b^{\prime -2}-c^{\prime -2}=0$
Hence, option 'C' is correct.