Question

A variable plane is at a distance of $6$ units from the origin. If it meets the coordinate axes in $A,B$ and $C$, then the equation of the locus of the centroid of the $\Delta ABC$ is

• A. $\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}=\frac{1}{4}$
• B. $x^2+y^2+z^2=4$
• C. $\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}=1$
• D. $\frac{1}{x^2}+\frac{1}{y^2}-\frac{1}{z^2}=\frac{1}{4}$

Answer 1

Answer: (A)

Let the equation of plane $\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1$
Distance from origin is $6.$
$\therefore 6=\frac{1}{\sqrt{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}}}$
$\Rightarrow \frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=\frac{1}{36}...(i)$
Centroid of plane is $\left(\frac{a}{3},\frac{b}{3},\frac{c}{3}\right)$
$\therefore$ Let $x=\frac{a}{3}$
$\Rightarrow a=3x$
Similarly, $b=3y,c=3z$
On putting the value of $a,b,c$ in Eq. (i), we get
$\frac{1}{9x^2}+\frac{1}{9y^2}+\frac{1}{9z^2}=\frac{1}{36}$
$\Rightarrow \frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}=\frac{1}{4}$