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Let (3, 4, ?1) and (?1, 2, 3) are the end points of a diameter of sphere. Then the radius of the sphere is equal to                                                               [Orissa JEE 2003]

A)            1

B)            2

C)            3

D)            9

Correct Answer: C

Solution :

                   \[d=\sqrt{{{(-1-3)}^{2}}+{{(2-4)}^{2}}+{{(3+1)}^{2}}}\]                          = \[\sqrt{16+4+16}\] \[=\sqrt{36}=6\]                               So, radius \[r=\frac{d}{2}=\frac{6}{2}=3.\]