A) 2
B) \[\sqrt{2}\]
C) 3
D) 1
Correct Answer: D
Solution :
Perpendicular distance to centre \[\left( \frac{1}{2},0,-\frac{1}{2} \right)\] from \[x+2y-z=4\]is, \[P=\frac{\left| \frac{1}{2}+\frac{1}{2}-4 \right|}{\sqrt{6}}=\sqrt{\frac{3}{2}}\] and radius of sphere \[R=\sqrt{\frac{5}{2}}\], So, \[r=\sqrt{{{R}^{2}}-{{P}^{2}}}=\sqrt{\frac{5}{2}-\frac{3}{2}}=1\].You need to login to perform this action.
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