Question

Consider a plane $\overrightarrow{r}.(2\hat{i}+2\hat{j}-\hat{k})=5$ which cuts a circular section from the sphere $|\overrightarrow{r}|=9$, then the volume of the cone formed by taking circular cross section as base as vertex as centre of sphere is

• A. $\frac{3520\pi }{27}$
• B. $\frac{40}{9}\sqrt{11\pi }9$
• C. $\frac{64}{9}\pi$
• D. $noneofthese$

Answer 1

The correct option is A $\frac{3520\pi }{27}$

Equation of plane

$\overrightarrow{r}\cdot (2\hat{i}+2\hat{j}-\hat{k})=5$

$2x+2y-z-5=0$

$|\overrightarrow{r}|=9$

$x^2+y^2+z^2=81$

radius $=9$

Centre of sphere$=(0,0,0)$

distance of plane from centre of sphere

$=\left|\frac{0+0-0-5}{\sqrt{4+4+1}}\right|=\frac{5}{3}$

So slant height of cone$=$Radius of sphere

$=9$ unit

& height of cone$=5/3$ unit

Let, radius of cone$=$R

$R^2=(9{)}^2-{\left(\frac{3}{3}\right)}^2=81-\frac{29}{9}$

$R^2=\frac{729-29}{9}=\frac{704}{9}$.