Question

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

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

Answer 1

The correct option is B $\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}=\frac{16}{p^2}$

Let points are : $A(a,0,0),B(0,b,0)$ and $C=(0,0,c)$
So, equation of plane using intercept form :
$\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1\cdots \left(i\right)$
Converting it into normal form :
$\Rightarrow \frac{x}{a\sqrt{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}}}+\frac{y}{b\sqrt{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}}}+\frac{z}{c\sqrt{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}}}=\frac{1}{\sqrt{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}}}$
So, perpendicular distance $p=\frac{1}{\sqrt{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}}}\cdots \left(ii\right)$
Now, centroid of tetrahedron $OABC:$
$(x,y,z)\equiv (\frac{a}{4},\frac{b}{4},\frac{c}{4})\Rightarrow a=4x,b=4y,c=4z$
putting the above values in $\left(ii\right),$ we get
$\frac{1}{16x^2}+\frac{1}{16y^2}+\frac{1}{16z^2}=\frac{1}{p^2}\Rightarrow \frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}=\frac{16}{p^2}$