The incentre of the triangle ABC,whose vertic | vedclass

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Question

(A) Let the vertices be $A(0,2,1)$,$B(-2,0,0)$,and $C(-2,0,2)$.First,calculate the lengths of the sides of the triangle:$a = BC = \sqrt{(-2 - (-2))^2

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Let the vertices be $A(0,2,1)$,$B(-2,0,0)$,and $C(-2,0,2)$.First,calculate the lengths of the sides of the triangle:$a = BC = \sqrt{(-2 - (-2))^2 + (0 - 0)^2 + (2 - 0)^2} = \sqrt{0 + 0 + 4} = 2$.$b = AC = \sqrt{(-2 - 0)^2 + (0 - 2)^2 + (2 - 1)^2} = \sqrt{4 + 4 + 1} = \sqrt{9} = 3$.$c = AB = \sqrt{(-2 - 0)^2 + (0 - 2)^2 + (0 - 1)^2} = \sqrt{4 + 4 + 1} = \sqrt{9} = 3$.The coordinates of the incentre $I$ are given by $\left(\frac{ax_A + bx_B + cx_C}{a+b+c}, \frac{ay_A + by_B + cy_C}{a+b+c}, \frac{az_A + bz_B + cz_C}{a+b+c}\right)$.$I = \left(\frac{2(0) + 3(-2) + 3(-2)}{2+3+3}, \frac{2(2) + 3(0) + 3(0)}{2+3+3}, \frac{2(1) + 3(0) + 3(2)}{2+3+3}\right)$.$I = \left(\frac{0 - 6 - 6}{8}, \frac{4 + 0 + 0}{8}, \frac{2 + 0 + 6}{8}\right)$.$I = \left(-\frac{12}{8}, \frac{4}{8}, \frac{8}{8}\right) = \left(-\frac{3}{2}, \frac{1}{2}, 1\right)$.

The incentre 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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