Question

A plane moves such that its distance from the origin is a constant p. If it intersects the coordinate axes at A, B, C then the locus of the centroid of the triangle ABC is

• A. $\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}=\frac{1}{p^2}$
• B. $\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}=\frac{9}{p^2}$
• C. $\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}=\frac{2}{p^2}$
• D. $\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}=\frac{4}{p^2}$

Answer 1

The correct option is B $\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}=\frac{9}{p^2}$

Let equation of plane be lx + my + nz = p
or $\frac{x}{\frac{p}{l}}+\frac{y}{\frac{p}{m}}+\frac{z}{\frac{p}{n}}=1$
Coordinates of A,B,C are $(\frac{p}{l},0,0),(0,\frac{p}{m},0)$ and
$(0,0,\frac{p}{m})$ respectively.
If centroid of triangle ABC is $(x_1,y_1,z_1)$
Then, $x_1=\frac{p}{3l},y_1=\frac{p}{3m},z_1=\frac{p}{3n}$
$\Rightarrow l^2+m^2+n^2=\frac{p^2}{9x_1^2}+\frac{p^2}{9y_1^2}+\frac{p^2}{9z_1^2}$
Or $\frac{1}{x_1^2}+\frac{1}{y_1^2}+\frac{1}{z_1^2}=\frac{9}{p^2}$
$\therefore$ Required locus is $\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}=\frac{9}{p^2}$