Coordinates of a centroind

Let’s A = [3, 2, 0], B = [1, -2, 4] and C = [1, 1, 1] be 3 points in space. Calculate the coordinates of the centroid of △ABC (the intersection of the medians).

Result

x =  1.667
y =  0.333
z =  1.667

Solution:

x0=3 y0=2 z0=0  x1=1 y1=−2 z1=4  x2=1 y2=1 z2=1  x=x0+x1+x23=3+1+13≐53≐1.6667≐1.667x_{0}=3 \ \\ y_{0}=2 \ \\ z_{0}=0 \ \\ \ \\ x_{1}=1 \ \\ y_{1}=-2 \ \\ z_{1}=4 \ \\ \ \\ x_{2}=1 \ \\ y_{2}=1 \ \\ z_{2}=1 \ \\ \ \\ x=\dfrac{ x_{0}+x_{1}+x_{2} }{ 3 }=\dfrac{ 3+1+1 }{ 3 } \doteq \dfrac{ 5 }{ 3 } \doteq 1.6667 \doteq 1.667
y=y0+y1+y23=2+(−2)+13≐13≐0.3333≐0.333y=\dfrac{ y_{0}+y_{1}+y_{2} }{ 3 }=\dfrac{ 2+(-2)+1 }{ 3 } \doteq \dfrac{ 1 }{ 3 } \doteq 0.3333 \doteq 0.333
z=z0+z1+z23=0+4+13≐53≐1.6667≐1.667z=\dfrac{ z_{0}+z_{1}+z_{2} }{ 3 }=\dfrac{ 0+4+1 }{ 3 } \doteq \dfrac{ 5 }{ 3 } \doteq 1.6667 \doteq 1.667

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