Question

A plane at a constant distance p from the origin meets the coordinate axes in A , B , C. Locus of the centroid of the triangle ABC is
a) x² + y² + z² = p²
​b) x^-2 + y^-2 + z^-2 = 9p^-2
c) x^-1 + y^-1 + z^-1 = 3p^-1
d) none of these

Answer 1

$$\begin{gathered} \mathrm{Let}\mathrm{the}\mathrm{coordinates}\mathrm{of}\mathrm{A},\mathrm{B},\mathrm{C}\mathrm{be}\mathrm{respectively}\left(\mathrm{a},0,0\right);\left(0,\mathrm{b},0\right);\left(0,0,\mathrm{c}\right). \\ \mathrm{Now},\mathrm{the}\mathrm{equation}\mathrm{of}\mathrm{plane}\mathrm{ABC}\mathrm{is}, \\ \frac{\mathrm{x}}{\mathrm{a}}+\frac{\mathrm{y}}{\mathrm{b}}+\frac{\mathrm{z}}{\mathrm{c}}=1 \\ \mathrm{Let}\left(\mathrm{\alpha },\mathrm{\beta },\mathrm{\gamma }\right)\mathrm{be}\mathrm{the}\mathrm{coordinates}\mathrm{of}\mathrm{the}\mathrm{centroid}\mathrm{of}∆\mathrm{ABC}. \\ \mathrm{So},\mathrm{\alpha }=\frac{\mathrm{a}+0+0}{3}=\frac{\mathrm{a}}{3} \\ \mathrm{So},\mathrm{\beta }=\frac{0+\mathrm{b}+0}{3}=\frac{\mathrm{b}}{3} \\ \mathrm{So},\mathrm{\gamma }=\frac{0+0+\mathrm{c}}{3}=\frac{\mathrm{c}}{3} \\ \mathrm{Now},\mathrm{p}=\mathrm{length}\mathrm{of}\perp \left(0,0,0\right)\mathrm{on}\mathrm{the}\mathrm{plane}\mathrm{ABC}, \\ \mathrm{p}=\frac{1}{\sqrt{{\displaystyle \frac{1}{{\mathrm{a}}^2}}+{\displaystyle \frac{1}{{\mathrm{b}}^2}}+{\displaystyle \frac{1}{{\mathrm{c}}^2}}}} \\ \Rightarrow \frac{1}{{\mathrm{a}}^2}+\frac{1}{{\mathrm{b}}^2}+\frac{1}{{\mathrm{c}}^2}=\frac{1}{{\mathrm{p}}^2} \\ \Rightarrow \frac{1}{{\left(3\mathrm{\alpha }\right)}^2}+\frac{1}{{\left(3\mathrm{\beta }\right)}^2}+\frac{1}{{\left(3\mathrm{\gamma }\right)}^2}=\frac{1}{{\mathrm{p}}^2} \\ \Rightarrow \frac{1}{{\mathrm{\alpha }}^2}+\frac{1}{{\mathrm{\beta }}^2}+\frac{1}{{\mathrm{\gamma }}^2}=9{\mathrm{p}}^{-2} \\ \mathrm{So},\mathrm{locus}\mathrm{of}\left(\mathrm{\alpha },\mathrm{\beta },\mathrm{\gamma }\right)\mathrm{is} \\ \frac{1}{{\mathrm{x}}^2}+\frac{1}{{\mathrm{y}}^2}+\frac{1}{{\mathrm{z}}^2}=9{\mathrm{p}}^{-2} \\ \Rightarrow {\mathrm{x}}^{-2}+{\mathrm{y}}^{-2}+{\mathrm{z}}^{-2}=9{\mathrm{p}}^{-2} \end{gathered}$$