Question

A variable plane which remains at a constant distance 3p from the origin cut the coordinate axes at A, B and C. The locus of the centroid of triangle ABC is

• A. $x^{-1} + y^{-1} + z^{-1} = p^{-1}$
• B. $x^{-2} + y^{-2} + z^{-2} = p^{-2}$
• C. $x + y + z = p$
• D. $x^2 + y^2 + z^2 = p^2$

Answer 1

The Correct Option is B

Let equation of the variable plane be $\frac{x}{a} +\frac{y}{b} + \frac{z}{c} = 1$ This meets the coordinate axes at $A\left(a, 0, 0\right), B \left(0, b, 0\right)$ and $C\left(0, 0, c\right)$. Let $P\left(\alpha, \beta, \gamma\right)$ be the centroid of the $\Delta ABC$ .Then $\alpha = \frac{a+0+0}{3}, \beta = \frac{0+b+0}{3}, \gamma = \frac{0+0+c}{3}$ $\therefore\quad a = 3\alpha, b = 3\beta , c = 3\gamma \quad...\left(2\right)$ Plane $\left(1\right)$ is at constant distance $3p$ from the origin, so $3p = \frac{\left|\frac{0}{a}+\frac{0}{b}+\frac{0}{c}-1\right|}{\sqrt{\left(\frac{1}{a}\right)^{2}+\left(\frac{1}{b}\right)^{2}}+\left(\frac{1}{c}\right)^{2}} \Rightarrow \frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}} = \frac{1}{9p^{2}} \quad...\left(3\right)$ From $\left(2\right)$ and $\left(3\right)$, we get $\frac{1}{9\alpha^{2}}+\frac{1}{9\beta ^{2}} + \frac{1}{9\gamma ^{2}} = \frac{1}{9p ^{2}}\quad\Rightarrow \alpha^{-2} + \beta^{-2} + \gamma^{-2} = p^{-2}$ Generalizing $\alpha, \beta, \gamma$, locus of centroid $P \left(\alpha, \beta, \gamma\right)$ is $x^{-2} + y^{-2} + z^{-2} = p^{-2}$