Question

A variable which remains at a constant distance 3p from the origin, cuts the co-ordinate axes A, B, C find the locus of the centroid triangle ABC.

Answer 1

We have,

Let the variable plane at distance 3p from the origin be $ax+by+cz=3p......\left(1\right)$

Where $a,b,c$ are the direction cosines of the normal to the plane.

This meet the Y-axis in A.

Thus $y=z=0$.

$Now,$

$ax=3p$

$x=3\frac{p}{a}$

Similarly,

Coordinate of B and C are,

$B=(0,\frac{3p}{a},0)$

$C=(0,0,\frac{3p}{c})$

Now,

Let the centroid of ABC is $G(x,y,z)$ using the formula of centroid

$(x,y,z)=(\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3},\frac{z_1+z_2+z_3}{3})$

$\Rightarrow G(x,y,z)=(\frac{\frac{3p}{a}+0+0}{3},\frac{0+\frac{3p}{a}+0}{3},\frac{0+0+\frac{3p}{a}}{3})$

$\Rightarrow G(x,y,z)=(\frac{p}{a},\frac{p}{a},\frac{p}{a})$

Now,

$a=\frac{p}{x},y=\frac{p}{b},z=\frac{p}{c}$

We know that,

$a^2+b^2+c^2=1$

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

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

Hence, this is the answer.