Question

A variable plane is at a constant distance $p$ from the origin and meets the axes in $A,B$ and $C$. Then locus of the centroid of the tetrahedron $OABC$ is

• A. $x^{-2}+y^{-2}+z^{-2}=16p^{-2}$
• B. $x^{-2}+y^{-2}+z^{-2}=16p^{-1}$
• C. $x^2+y^{-2}+z^{-2}=16$
• D. None of these

Answer 1

Answer: (A)

$x^{-2}+y^{-2}+z^{-2}=16p^{-2}$

Equation of plane at a distance $p$ from origin is given by,
$lx+my+nz=p$, where $l,m,n$ are direction cosine of normal to the plane along distance.

Thus intercept on the axes are,
$A=(\frac{p}{l},0,0),B=(0,\frac{p}{m},0),C=(0,0,\frac{p}{n})$

So centroid of tetrahedron $OABC$ is,
$x=\frac{p}{4l},y=\frac{p}{4m},z=\frac{p}{4n}$

Also we know,
$l^2+m^2+n^2=1$
$\Rightarrow \frac{p^2}{16x^2}+\frac{p^2}{16y^2}+\frac{p^2}{16z^2}=1$

Hence, required locus is
$x^{-2}+y^{-2}+z^{-2}=16p^{-2}$