Q217 / 270 · NDA 2016 (Phase 1) · Mathematics

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

  • A
    $\frac{2}{\sqrt{29}}$
  • B
    $\frac{4}{\sqrt{29}}$
  • Correct answer
    C
    $\frac{5}{\sqrt{29}}$
  • D
    1

Solution

Plane $P$ touches the sphere $x^2+y^2+z^2=r^2$ then $r=$ Distane between centre of sphere $(0,0,0)$ to plance $P$. $\begin{aligned} & \Rightarrow \mathrm{r}=\left|\frac{5(0)+2(0)-5}{\sqrt{5^2+2^2+(0)^2}}\right| \\ & =\frac{5}{\sqrt{25+4}} \\ & \mathrm{r}=\frac{5}{\sqrt{29}} \end{aligned}$

Hence, the correct answer is option: 'C'.

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