Question

Two systems of rectangular axes have the same origin. If a plane cuts them at distance $a,b,$ and $a^{\prime },b^{\prime },c^{\prime }$ respectively from the origin, then $\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=k(\frac{1}{{a^{\prime }}^2}+\frac{1}{{b^{\prime }}^2}+\frac{1}{{c^{\prime }}^2})$, where $k=$

• A. $1$
• B. $2$
• C. $4$
• D. None of these

Answer 1

Answer: (A)

$1$

Let $a,b,c$ be the intercepts when $Ox,Oy,Oz$ are taken as axes:

then the equation of the plane is $\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1$ ...$\left(1\right)$

Also let $a^{\prime },b^{\prime },c^{\prime }$ be the intercepts when $OX,OY,OZ$ are taken as axes;

then in this case equation of the same plane is $\frac{X}{a^{\prime }}+\frac{Y}{b^{\prime }}+\frac{Z}{c^{\prime }}=1$ ...$\left(2\right)$

Now $\left(1\right)$ and $\left(2\right)$ are equation of the same same plane and in both cases, the origin is the same.

Hence length of the perpendicular drawn from the origin to the plane in both the case must be the same,

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

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