Question

A point $P$ moves on a plane $\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1$ and $O$ be the origin. A plane through $P$ and perpendicular to $OP$ meets the coordinate axes at $A,B$ and $C$ respectively. Let $G$ be the centroid of the triangle $ABC$. If the planes through $A,B$ and $C$ parallel to the planes $x=0,y=0$ and $z=0$, respectively, intersect at $Q$, then

• A. If $OA=OB=OC$, then $OG=\frac{1}{3}OP$
• B. Locus of $Q$ is
  $\frac{1}{ax}+\frac{1}{by}+\frac{1}{cz}=\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}$
• C. Locus of $Q$ is
  $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}$
• D. If $OA=OB=OC$, then $OG=OP$

Answer 1

Answer: (B), (D)

If $OA=OB=OC$, then $OG=OP$

Let coordinates of $P$ be $(h,k,l)$
$\Rightarrow \frac{h}{a}+\frac{k}{b}+\frac{l}{c}=1\cdots \left(1\right)OP=\sqrt{h^2+k^2+l^2}$

Direction cosines of $OP$ is
$\frac{h}{\sqrt{h^2+k^2+l^2}},\frac{k}{\sqrt{h^2+k^2+l^2}},\frac{l}{\sqrt{h^2+k^2+l^2}}$
Equation of the plane passing through $P$ and perpendicular to $OP$ is
$\frac{hx}{\sqrt{h^2+k^2+l^2}}+\frac{ky}{\sqrt{h^2+k^2+l^2}}+\frac{lz}{\sqrt{h^2+k^2+l^2}}=\sqrt{h^2+k^2+l^2}$

$\Rightarrow hx+ky+lz=h^2+k^2+l^2$

$\therefore A(\frac{h^2+k^2+l^2}{h},0,0)B(0,\frac{h^2+k^2+l^2}{k},0)C(0,0,\frac{h^2+k^2+l^2}{l})OA=OB=OC\Rightarrow h=k=l$
Centroid, $G(\frac{h^2+k^2+l^2}{3h},\frac{h^2+k^2+l^2}{3k},\frac{h^2+k^2+l^2}{3l})\Rightarrow G(h,k,l)\Rightarrow OG=OP$

If coordinates of $Q$ is $(e,f,g)$, then
$e=\frac{h^2+k^2+l^2}{h}\Rightarrow \frac{h}{a}=\frac{h^2+k^2+l^2}{ae},f=\frac{h^2+k^2+l^2}{k}\Rightarrow \frac{k}{b}=\frac{h^2+k^2+l^2}{bf},g=\frac{h^2+k^2+l^2}{l}\Rightarrow \frac{l}{c}=\frac{h^2+k^2+l^2}{cg}$

Now, $\frac{1}{e^2}+\frac{1}{f^2}+\frac{1}{g^2}=\frac{1}{h^2+k^2+l^2}$
From $\left(1\right)$,
$\Rightarrow \frac{h^2+k^2+l^2}{ae}+\frac{h^2+k^2+l^2}{bf}+\frac{h^2+k^2+l^2}{cg}=1\Rightarrow \frac{1}{ae}+\frac{1}{bf}+\frac{1}{cg}=\frac{1}{h^2+k^2+l^2}\Rightarrow \frac{1}{ae}+\frac{1}{bf}+\frac{1}{cg}=\frac{1}{e^2}+\frac{1}{f^2}+\frac{1}{g^2}$

$\therefore$ Locus of $Q$ is $\frac{1}{ax}+\frac{1}{by}+\frac{1}{cz}=\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}$