A variable plane which remains at a constant distance 3p from the origin cuts the coordinate axes at A, B, C. Show that the locus of the centroid of $\triangle ABC$ is
$$\frac {1}{x^2}+ \frac {1}{y^2} + \frac {1}{z^2} = \frac {1}{p^2}$$
My work:
As it is given that there are equal intercepts cut at A , B & C so we get (a,0,0),(0,b,0),(0,0,c)
So this plane is at a distance 3p from (0,0,0)
I understood this much ,Any help would be appreciated
Best Answer
I'll give you some hints!
Hint 1: Start by assuming the equation of plane be $\dfrac{x}{a} + \dfrac{y}{b} + \dfrac{z}{c} = 1$ where $a, b$ and $c$ are intercepts on plane on $x, y$ and $z$ axis respectively.
Hint 2: Now find the distance of the plane from the origin.
Hint 3: Centroid of triangle will be $\left(\dfrac{a}{3}, \dfrac{b}{3}, \dfrac{c}{3}\right)$ let it be $(x,y, z)$
Now establish the relationship.