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NDA 2017
Exam
NDA
Year
2017
Subject
Mathematics
Chapter
Three Dimensional Geometry

NDA 2017 · Mathematics · Three Dimensional Geometry

A variable plane passes through a fixed point (a, b, c) and cuts...

A variable plane passes through a fixed point (a,b,c)(a, b, c) and cuts the axes in A,B\mathrm{A}, \mathrm{B} and C respectively. The locus of the centre of the sphere OABC,O\mathrm{OABC}, \mathrm{O} being the origin, is

Options

  1. Axa+yb+zc=1\frac{\mathrm{x}}{\mathrm{a}}+\frac{\mathrm{y}}{\mathrm{b}}+\frac{\mathrm{z}}{\mathrm{c}}=1
  2. Bax+by+cz=1\frac{\mathrm{a}}{\mathrm{x}}+\frac{\mathrm{b}}{\mathrm{y}}+\frac{\mathrm{c}}{\mathrm{z}}=1
  3. Cax+by+cz=2\frac{\mathrm{a}}{\mathrm{x}}+\frac{\mathrm{b}}{\mathrm{y}}+\frac{\mathrm{c}}{\mathrm{z}}=2
  4. Dxa+yb+zc=2\frac{\mathrm{x}}{\mathrm{a}}+\frac{\mathrm{y}}{\mathrm{b}}+\frac{\mathrm{z}}{\mathrm{c}}=2

Correct answer

C. ax+by+cz=2\frac{\mathrm{a}}{\mathrm{x}}+\frac{\mathrm{b}}{\mathrm{y}}+\frac{\mathrm{c}}{\mathrm{z}}=2

Step-by-step solution


Equation of plane passing through points (p,0,0),(0,q,0)(p, 0,0),(0, q, 0) and (0,0,r)(0,0, r) is xp+yq+zr=1\frac{x}{p}+\frac{y}{q}+\frac{z}{r}=1 Given that this plane passes through (a,b,c)(a, b, c). ∴ap+bq+cr=1\therefore \frac{\mathrm{a}}{\mathrm{p}}+\frac{\mathrm{b}}{\mathrm{q}}+\frac{\mathrm{c}}{\mathrm{r}}=1 Equation of sphere is x2+y2+z2−px−qy−rz=0x^2+y^2+z^2-p x-q y-r z=0. Centre of the sphere =(1,m,n)=(p2,q2,r2)⇒p=2ℓ,q=2m,r=2n∴=(1, m, n)=\left(\frac{\mathrm{p}}{2}, \frac{\mathrm{q}}{2}, \frac{\mathrm{r}}{2}\right)\Rightarrow \mathrm{p}=2 \ell, \mathrm{q}=2 \mathrm{ m}, \mathrm{r}=2 \mathrm{n}\therefore locus of the centre ⇒ax+by+cz=2\Rightarrow \frac{\mathrm{a}}{\mathrm{x}}+\frac{\mathrm{b}}{\mathrm{y}}+\frac{\mathrm{c}}{\mathrm{z}}=2.
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