Question

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

Answer 1

Let $A$ be the foot of the perpendicular from the centre $O$ to the plane $\overrightarrow{r}\cdot (\hat{i}+\hat{j}+\hat{k})-3\sqrt{3}=0$
Then $OA=\left|\frac{\overrightarrow{0}\cdot (\hat{i}+\hat{j}+\hat{k})-3\sqrt{3}}{|\hat{i}+\hat{j}+\hat{k}|}\right|$
$\Rightarrow OA=\frac{3\sqrt{3}}{\sqrt{3}}=3$

If $P$ is any point on the circle, then $P$ lies on the plane as well as on the sphere.
Therefore, $OP=\text{radius of the sphere}=5$
Now, $A{P}^2=O{P}^2-O{A}^2=5^2-3^2$
$\Rightarrow A{P}^2=16$
$\Rightarrow AP=4$

Hence, the radius of the circular section formed is $4$