Question

A variable plane is at distance, $k$ from the origin and meets the coordinate axes in $A,B,C$. Then, the locus of the centroid of $\Delta ABC$ is

• A. $x^{-2}+y^{-2}+z^{-2}=k^{-2}$
• B. $x^{-2}+y^{-2}+z^{-2}=4k^{-2}$
• C. $x^{-2}+y^{-2}+z^{-2}=16k^{-2}$
• D. $x^{-2}+y^{-2}+z^{-2}=9k^{-2}$

Answer 1

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

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

$a,b,c$ are intercepts on $x,y,z$ axis respectively

Perpendicular distance from origin to plane $=k$

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

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

$centroid(x,y,z)=(\frac{a}{3},\frac{b}{3},\frac{c}{3})$

$\Rightarrow \frac{1}{{\left(3x\right)}^2}+\frac{1}{{\left(3y\right)}^2}+\frac{1}{{\left(3z\right)}^2}=\frac{1}{k^2}$

$\Rightarrow x^{-2}+y^{-2}+z^{-2}=9k^{-2}$