Question

Find the lengths of the medians of the triangle with vertices $A(0,0,6),B(0,4,0)$ and $(6,0,0).$

Answer 1

Let $AD,BE$ and $CF$ be the medians of the given $△ABC$.
Since $AD$ is the median, $D$ is the mid-point of $BC$.

$\therefore$ coordinates of point $D=$ $(\frac{0+6}{2},\frac{4+0}{2},\frac{0+0}{2})=(3,2,0)$

$AD=$ $\sqrt{{(0-3)}^2+{(0-2)}^2+{(6-0)}^2}=\sqrt{9+4+36}=\sqrt{49}=7$

Since $BE$ is the median, $E$ is the mid-point of $AC$.

$\therefore$ coordinates of point $E=$ $(\frac{0+6}{2},\frac{0+0}{2},\frac{6+0}{2})=(3,0,3)$

$BE=$ $\sqrt{{(3-0)}^2+{(0-4)}^2+{(3-0)}^2}=\sqrt{9+16+9}=\sqrt{34}$

Since $CF$ is the median, $F$ is the mid point of $AB$.

$\therefore$ coordinates of point $F=$ $(\frac{0+0}{2},\frac{0+4}{2},\frac{6+0}{2})=(0,2,3)$

Length of $CF=$ $\sqrt{{(6-0)}^2+{(0-2)}^2+{(0-3)}^2}=\sqrt{36+4+9}=\sqrt{49}=7$

Thus the lengths of the medians of $△ABC$ are $7,\sqrt{34}$ and $7$ units.