The circumradius of the triangle formed by th | vedclass

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(D) Let the vertices be $A(2, -1, 1)$,$B(1, -3, -5)$,and $C(3, -4, -4)$.First,calculate the side lengths:$AB = \sqrt{(1-2)^2 + (-3-(-1))^2 + (-5-1)^2}

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$\frac{\sqrt{41}}{2}$

Vedclass · Answer

(D) Let the vertices be $A(2, -1, 1)$,$B(1, -3, -5)$,and $C(3, -4, -4)$.First,calculate the side lengths:$AB = \sqrt{(1-2)^2 + (-3-(-1))^2 + (-5-1)^2} = \sqrt{(-1)^2 + (-2)^2 + (-6)^2} = \sqrt{1 + 4 + 36} = \sqrt{41}$.$BC = \sqrt{(3-1)^2 + (-4-(-3))^2 + (-4-(-5))^2} = \sqrt{2^2 + (-1)^2 + 1^2} = \sqrt{4 + 1 + 1} = \sqrt{6}$.$AC = \sqrt{(3-2)^2 + (-4-(-1))^2 + (-4-1)^2} = \sqrt{1^2 + (-3)^2 + (-5)^2} = \sqrt{1 + 9 + 25} = \sqrt{35}$.Observe that $AB^2 = 41$ and $BC^2 + AC^2 = 6 + 35 = 41$.Since $AB^2 = BC^2 + AC^2$,the triangle is a right-angled triangle with the right angle at $C$.For a right-angled triangle,the circumradius $R$ is half of the hypotenuse.Here,the hypotenuse is $AB = \sqrt{41}$.Therefore,$R = \frac{AB}{2} = \frac{\sqrt{41}}{2}$.

The circumradius of the triangle formed by the points $A(2, -1, 1)$,$B(1, -3, -5)$,and $C(3, -4, -4)$ is:

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