Question

The radius of the sphere $(\hat{r}-2\hat{i}+3\hat{j}-\hat{k}).(\hat{r}+3\hat{i}-\hat{j}+2\hat{k})=0$ is

• A. $5$
• B. $5\sqrt{2}$
• C. $5/\sqrt{2}$
• D. $2\sqrt{5}$

Answer 1

Answer: (C)

$5/\sqrt{2}$

In the equation of sphere, $R^2-2R.C+d=0$ where $C$ is the centre and $R$ is any point on it.

Let Radius be $x$

$d=C^2-(x{)}^2$

Similarly, $(\overline{r}-\overline{a}).(\overline{r}-\overline{b})=0$,

Here, $\overline{C}=\frac{\overline{a}+\overline{b}}{2}$ and $d=\overline{a}.\overline{b}$
$\therefore d=-6-3-2=-11$ and $C=(\frac{-1}{2},-1,\frac{-1}{2})$
$\Rightarrow C^2=\frac{1}{4}+1+\frac{1}{4}=1.5$
$\Rightarrow (x{)}^2=1.5+11=12.5$
$\Rightarrow x=\frac{5}{\sqrt{2}}$