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NDA 2016
Exam
NDA
Year
2016
Subject
Mathematics
Chapter
Three Dimensional Geometry

NDA 2016 · Mathematics · Three Dimensional Geometry

A plane P passes through the line of intersection of the planes 2...

A plane P passes through the line of intersection of the planes 2x−y+3z=2,x+y−z=12 \mathrm{x}-\mathrm{y}+3 \mathrm{z}=2, \mathrm{x}+\mathrm{y}-\mathrm{z}=1 and the point (1,0,1)(1,0,1). If the plane PP touches the sphere x2+y2+z2=r2x^2+y^2+z^2=r^2, then what is rr equal to?

Options

  1. A229\frac{2}{\sqrt{29}}
  2. B429\frac{4}{\sqrt{29}}
  3. C529\frac{5}{\sqrt{29}}
  4. D1

Correct answer

C. 529\frac{5}{\sqrt{29}}

Step-by-step solution

Plane PP touches the sphere x2+y2+z2=r2x^2+y^2+z^2=r^2 then r=r= Distane between centre of sphere (0,0,0)(0,0,0) to plance PP. ⇒r=∣5(0)+2(0)−552+22+(0)2∣=525+4r=529\begin{aligned} & \Rightarrow \mathrm{r}=\left|\frac{5(0)+2(0)-5}{\sqrt{5^2+2^2+(0)^2}}\right| \\ & =\frac{5}{\sqrt{25+4}} \\ & \mathrm{r}=\frac{5}{\sqrt{29}} \end{aligned}
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