Let (3, 4, -1) and (-1, 2, 3) be the end poin | vedclass

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(C) The diameter $d$ of the sphere is the distance between the two given end points $(3, 4, -1)$ and $(-1, 2, 3)$.Using the distance formula,$d = \sqr

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$3$

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(C) The diameter $d$ of the sphere is the distance between the two given end points $(3, 4, -1)$ and $(-1, 2, 3)$.Using the distance formula,$d = \sqrt{(-1 - 3)^2 + (2 - 4)^2 + (3 - (-1))^2}$.$d = \sqrt{(-4)^2 + (-2)^2 + (4)^2}$.$d = \sqrt{16 + 4 + 16} = \sqrt{36} = 6$.The radius $r$ of the sphere is half of the diameter.$r = \frac{d}{2} = \frac{6}{2} = 3$.Thus,the radius of the sphere is $3$.

Let $(3, 4, -1)$ and $(-1, 2, 3)$ be the end points of a diameter of a sphere. Then the radius of the sphere is equal to

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