JEE Main & Advanced
Mathematics
Vector Algebra
Question Bank
Critical Thinking
question_answer
The radius of the circular section of the sphere \[|\mathbf{r}|\,=5\]by the plane \[\mathbf{r}\,.\,(\mathbf{i}+\mathbf{j}+\mathbf{k})=3\sqrt{3}\] is [DCE 1999]
A)1
B)2
C)3
D)4
Correct Answer:
D
Solution :
The centre of the sphere \[|r|\,=5\] is at the origin and radius\[=5\]. Let M be the foot of perpendicular from O to the given plane. Then OM = length of perpendicular from O to the given plane \[=\frac{|\overrightarrow{OM}\,.\,(i+j+k)-3\sqrt{3}|}{|i+j+k|}\]