Vector Algebra Question 191
Question: 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]
Options:
A) 1
B) 2
C) 3
D) 4
Show Answer
Answer:
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|} $
$ =\frac{3\sqrt{3}}{\sqrt{1^{2}+1^{2}+1^{2}}}=3 $
Let P be any position of circle, then P lies on plane as well as on sphere.
$ \therefore $ OP = radius of sphere = 5
In $ \Delta OPM $ , we have $ OP^{2}=OM^{2}+PM^{2} $
$ \Rightarrow PM=\sqrt{5^{2}-3^{2}}=4 $ .
BETA
