Question

A variable plane which remains at a constant distance $3p$ from the origin cuts the coordinate axes at $A,B,C$. Show that the locus of the centroid of triangle $ABC$ is $\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}=\frac{1}{p^2}$

Answer 1

Let the equation of the plane be
$\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1$ ...$\left(i\right)$
where $a,b,c$ are variables.
This plane meets $X,Y$ and $Z$ axes at points $A(a,0,0),B(0,b,0)$ and $C(0,0,c)$ respectively. Let $(\alpha ,\beta ,\gamma )$ be the coordinates of the centroid of $\Delta ABC$. Then,
$\alpha =\frac{a+0+0}{3}=\frac{a}{3},\beta =\frac{0+b+0}{3}=\frac{b}{3},$
$\gamma =\frac{0+0+c}{3}=\frac{c}{3}$ ...$\left(ii\right)$
The plane represented by equation $\left(i\right)$ is at a distance $3p$ from the origin.
$\therefore$ $3p$ = Length of perpendicular from $(0,0,0)$ to the plane $\left(i\right)$
$\Rightarrow 3p=\frac{\left|\begin{array}{c}\frac{0}{a}+\frac{0}{b}+\frac{0}{c}-1\end{array}\right|}{\sqrt{{\left(\frac{1}{a}\right)}^2+{\left(\frac{1}{b}\right)}^2+{\left(\frac{1}{c}\right)}^2}}$
$\Rightarrow 3p=\frac{1}{\sqrt{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}}}$
$\Rightarrow \frac{1}{9p^2}=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}$ ...$\left(iii\right)$
From $\left(ii\right)$, we have
$a=3\alpha ,b=3\beta$ and $c=3\gamma$
Substituting the values of $a,b,c$, in $\left(iii\right)$, we obtian
$\Rightarrow \frac{1}{9p^2}=\frac{1}{9{\alpha }^2}+\frac{1}{{\beta }^2}+\frac{1}{9{\gamma }^2}$
$\Rightarrow \frac{1}{p^2}=\frac{1}{{\alpha }^2}+\frac{1}{{\beta }^2}+\frac{1}{{\gamma }^2}$
So, the locus of centroid of triangle is
$\frac{1}{p^2}=\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}$