Question

A variable plane intersects the coordinate axes at $A,B,C$ and is at a constant distance '$p$' from $0(0,0,0)$. Then the locus of the centroid of the tetrahedron $OABC$ is

• A. $\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}=\frac{1}{p^2}$
• B. $\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}=\frac{4}{p^2}$
• C. $\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}=\frac{16}{p^2}$
• D. $\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}=16p^2$

Answer 1

The correct option is C

$\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}=\frac{16}{p^2}$

A variable plane intersects the coordinate axes at $A(a,0,0)$, $B(0,b,0)$, $C(0,0,c)$ centroid of tetrahedron $OABC$ is $(x,y,z)=(\frac{a}{4},\frac{b}{4},\frac{c}{4})$, then equation of plane is $\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1$ perpendicular distance from origin to plane.

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

$\Rightarrow p^2=\frac{1}{\frac{1}{{16x}^2}+\frac{1}{16y^2}+\frac{1}{16z^2}}$

$\Rightarrow \frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}=\frac{16}{p^2}$

Ans: C