Question

A variable plane remains at constant distance $p$ from the origin. If it meets coordinates axes at points $A,B,C$ then the locus of the centroid of $△ABC$ is

• A. $x^{-2}+y^{-2}+z^{-2}=9p^{-2}$
• B. $x^{-3}+y^{-3}+z^{-3}=9p^{-3}$
• C. $x^2+y^2+z^2=9p^2$
• D. $x^3+y^3+z^3=9p^3$

Answer 1

The correct option is A $x^{-2}+y^{-2}+z^{-2}=9p^{-2}$

Let the variable plane be $\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1$

$\therefore A=(a,0,0),B=(0,b,0),C=(0,0,c)$

Let $G(\alpha ,\beta ,\gamma )$ be the centroid of $△ABC$

$\therefore \alpha =\frac{a}{3},\beta =\frac{b}{3},\gamma =\frac{c}{3}$ .........$\left(1\right)$

Also given that, distance of plane from origin is $p$

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

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

$\Rightarrow \frac{1}{{\alpha }^2}+\frac{1}{{\beta }^2}+\frac{1}{{\gamma }^2}=\frac{9}{p^2}$ using $\left(1\right)$

Hence, required locus of $G(\alpha ,\beta ,\gamma )$ is,

$\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}=\frac{9}{p^2}$

i.e. $x^{-2}+y^{-2}+z^{-2}=9p^{-2}$