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NDA 2018
Exam
NDA
Year
2018
Subject
Mathematics
Chapter
Three Dimensional Geometry

NDA 2018 · Mathematics · Three Dimensional Geometry

A sphere of constant radius r through the origin intersects the...

A sphere of constant radius rr through the origin intersects the coordinate axes in A,B\mathrm{A}, \mathrm{B} and C. What is the locus of the centroid of the triangle ABC ?

Options

  1. Ax2+y2+z2=r2x^2+y^2+z^2=r^2
  2. Bx2+y2+z2=4r2x^2+y^2+z^2=4 r^2
  3. C9(x2+y2+z2)=4r29\left(x^2+y^2+z^2\right)=4 r^2
  4. D3(x2+y2+z2)=2r23\left(x^2+y^2+z^2\right)=2 r^2

Correct answer

C. 9(x2+y2+z2)=4r29\left(x^2+y^2+z^2\right)=4 r^2

Step-by-step solution

Let the sphere passing through points A(a,0,0)A(a, 0,0), B(0,b,0),C(0,0,c)\mathrm{B}(0, \mathrm{ b}, 0), \mathrm{C}(0,0, \mathrm{c}) Equation of sphere is x2+y2+z2−ax−by−cz=0x^2+y^2+z^2-a x-b y-c z=0 radius, r=12a2+b2+c2⇒a2+b2+c2=4r2....(1)\mathrm{r}=\frac{1}{2} \sqrt{\mathrm{a}^2+\mathrm{b}^2+\mathrm{c}^2}\Rightarrow \mathrm{a}^2+\mathrm{b}^2+\mathrm{c}^2=4 \mathrm{r}^2....(1) (Squaring on both sides) Let (α,β,γ)(\alpha, \beta, \gamma) be centroid of sphere. ∴(α,β,γ)=(a+0+03,0+b+03,0+0+c3)=(a3,b3,c3)∴α2+β2+γ2=a29+b29+c29=a2+b2+c29=4r29....from(1)⇒9(α2+β2+γ2)=4r2\begin{aligned} & \therefore(\alpha, \beta, \gamma)=\left(\frac{\mathrm{a}+0+0}{3}, \frac{0+\mathrm{b}+0}{3}, \frac{0+0+\mathrm{c}}{3}\right) \\ & =\left(\frac{\mathrm{a}}{3}, \frac{\mathrm{ b}}{3}, \frac{\mathrm{c}}{3}\right) \\ & \therefore \alpha^2+\beta^2+\gamma^2=\frac{\mathrm{a}^2}{9}+\frac{\mathrm{b}^2}{9}+\frac{\mathrm{c}^2}{9} \\ & =\frac{\mathrm{a}^2+\mathrm{b}^2+\mathrm{c}^2}{9} \\ & =\frac{4 \mathrm{r}^2}{9}....from(1) \\ & \Rightarrow 9\left(\alpha^2+\beta^2+\gamma^2\right)=4 \mathrm{r}^2 \end{aligned} So, Locus is 9(x2+y2+z2)=4r29\left(x^2+y^2+z^2\right)=4 r^2.
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