Question

The radius of the circular section of the sphere $\left|\overrightarrow{r}\right|=5$ by the plane $\overrightarrow{r}.(\overrightarrow{i}+\overrightarrow{j}+\overrightarrow{k})=3\sqrt{3}$ is

• A. $16$
• B. $8$
• C. $4$
• D. None of these\

Answer 1

The correct option is C $4$

The sphere $\left|\overrightarrow{r}\right|=5$ has centre at the origin and radius $5$.

Distance of the the plane $\overrightarrow{r}.(\overrightarrow{i}+\overrightarrow{j}+\overrightarrow{k})=3\sqrt{3}$ from the origin.

$=\frac{3\sqrt{3}}{|\overrightarrow{i}+\overrightarrow{j}+\overrightarrow{k}|}=\frac{3\sqrt{3}}{\sqrt{1^2+1^2+1^2}}=3$

Thus, in figure, $OP=5,ON=3.$

$\therefore {NP}^2={OP}^2-{ON}^2={\left(5\right)}^2-{\left(3\right)}^2=16\Rightarrow NP=4$

Hence, the radius of the circular section $=NP=4$.