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Mathematics Question on Three Dimensional Geometry

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−1x^{-1} + y^{-1} + z^{-1} = p^{-1}

B

x−2+y−2+z−2=p−2x^{-2} + y^{-2} + z^{-2} = p^{-2}

C

x+y+z=px + y + z = p

D

x2+y2+z2=p2x^2 + y^2 + z^2 = p^2

Answer

x−2+y−2+z−2=p−2x^{-2} + y^{-2} + z^{-2} = p^{-2}

Explanation

Solution

Let equation of the variable plane be xa+yb+zc=1\frac{x}{a} +\frac{y}{b} + \frac{z}{c} = 1 This meets the coordinate axes at A(a,0,0),B(0,b,0)A\left(a, 0, 0\right), B \left(0, b, 0\right) and C(0,0,c)C\left(0, 0, c\right). Let P(α,β,γ)P\left(\alpha, \beta, \gamma\right) be the centroid of the ΔABC\Delta ABC .Then α=a+0+03,β=0+b+03,γ=0+0+c3\alpha = \frac{a+0+0}{3}, \beta = \frac{0+b+0}{3}, \gamma = \frac{0+0+c}{3} ∴a=3α,b=3β,c=3γ...(2)\therefore\quad a = 3\alpha, b = 3\beta , c = 3\gamma \quad...\left(2\right) Plane (1)\left(1\right) is at constant distance 3p3p from the origin, so 3p=∣0a+0b+0c−1∣(1a)2+(1b)2+(1c)2⇒1a2+1b2+1c2=19p2...(3)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 (2)\left(2\right) and (3)\left(3\right), we get 19α2+19β2+19γ2=19p2⇒α−2+β−2+γ−2=p−2\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 α,β,γ\alpha, \beta, \gamma, locus of centroid P(α,β,γ)P \left(\alpha, \beta, \gamma\right) is x−2+y−2+z−2=p−2x^{-2} + y^{-2} + z^{-2} = p^{-2}