Find the lengths of the medians of the triangle with vertices A (0, 0, 6), B (0, 4, 0) and (6, 0, 0).
Find the lengths of the medians of the triangle with vertices A (0, 0, 6), B (0, 4, 0) and (6, 0, 0).
Let AD, BE, and CF be the medians of the given triangle ABC.
Since AD is the median, D is the mid-point of BC.
$\therefore$ Coordinates of point $D=\left(\frac{0+6}{2}, \frac{4+0}{2}, \frac{0+0}{2}\right)=(3,2,0)$
$\mathrm{AD}=\sqrt{(0-3)^{2}+(0-2)^{2}+(6-0)^{2}}=\sqrt{9+4+36}=\sqrt{49}=7$
Since $B E$ is the median, $E$ is the mid-point of $A C$.
$\therefore$ Coordinates of point $\mathrm{E}=\left(\frac{0+6}{2}, \frac{0+0}{2}, \frac{6+0}{2}\right)=(3,0,3)$
$B E=\sqrt{(3-0)^{2}+(0-4)^{2}+(3-0)^{2}}=\sqrt{9+16+9}=\sqrt{34}$
Since $\mathrm{CF}$ is the median, $\mathrm{F}$ is the mid-point of $\mathrm{AB}$.
$\therefore$ Coordinates of point $\mathrm{F}=\left(\frac{0+0}{2}, \frac{0+4}{2}, \frac{6+0}{2}\right)=(0,2,3)$
Length of $\mathrm{CF}=\sqrt{(6-0)^{2}+(0-2)^{2}+(0-3)^{2}}=\sqrt{36+4+9}=\sqrt{49}=7$
Thus, the lengths of the medians of $\triangle \mathrm{ABC}$ are $7, \sqrt{34}$, and 7 .
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