The incenter of the triangle ABC,whose vertic | vedclass

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(C) Let the vertices be $A(0,2,1)$,$B(-2,0,0)$,and $C(-2,0,2)$.Calculate the side lengths:$a = BC = \sqrt{(-2 - (-2))^2 + (0 - 0)^2 + (2 - 0)^2} = \sq

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$\left(-\frac{3}{2}, \frac{1}{2}, 1\right)$

Vedclass · Answer

(C) Let the vertices be $A(0,2,1)$,$B(-2,0,0)$,and $C(-2,0,2)$.Calculate the side lengths:$a = BC = \sqrt{(-2 - (-2))^2 + (0 - 0)^2 + (2 - 0)^2} = \sqrt{0 + 0 + 4} = 2$$b = AC = \sqrt{(-2 - 0)^2 + (0 - 2)^2 + (2 - 1)^2} = \sqrt{4 + 4 + 1} = 3$$c = AB = \sqrt{(-2 - 0)^2 + (0 - 2)^2 + (0 - 1)^2} = \sqrt{4 + 4 + 1} = 3$The incenter $I$ is given by $\frac{aA + bB + cC}{a + b + c}$.$I = \frac{2(0,2,1) + 3(-2,0,0) + 3(-2,0,2)}{2 + 3 + 3}$$I = \frac{(0,4,2) + (-6,0,0) + (-6,0,6)}{8}$$I = \frac{(-12, 4, 8)}{8} = \left(-\frac{12}{8}, \frac{4}{8}, \frac{8}{8}\right) = \left(-\frac{3}{2}, \frac{1}{2}, 1\right)$.

The incenter of the triangle $ABC$,whose vertices are $A(0,2,1)$,$B(-2,0,0)$,and $C(-2,0,2)$ is

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