A) \[6\]
B) \[9\]
C) \[12\]
D) \[24\]
Correct Answer: A
Solution :
Given plane is \[2x-y+2z=25\] and sphere is \[{{x}^{2}}+{{y}^{2}}+{{z}^{2}}-2x-4y+2z-3=0\] On comparing with \[{{x}^{2}}+{{y}^{2}}+{{z}^{2}}+2ux+2vy+2z+d=0\] We get \[u=-1,v=-2,w=1,\,d=-3\] Centre \[=(1,2,-1)\] and radius \[=\sqrt{1+4+1+3}=3\] Now, perpendicular distance from centre to the plane \[=\left| \frac{2\times 1-1\times 2+2\times (-1)-25}{\sqrt{4+1+4}} \right|=\left| \frac{27}{3} \right|=9\] Required shortest distance = perpendicular distance - radius \[=9-3=6\]
You need to login to perform this action.
You will be redirected in
3 sec
