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

For the given triangle ABC. Let AD, BE and CF are the medians

We know that, median divides the line segment into two equal parts, so D is the midpoint of BC, therefore,

Coordinates of point D = \( \dfrac{0+6}{2}, \dfrac{4+0}{2}, \dfrac{0+0}{2}\)

Coordinates of point D = (3,2,0)

AD =\( \sqrt{0– 3)^2 + (0 – 2)^2 + (6– 0)^2} \)

AD = \( \sqrt{9+4+36}\)

AD = \( \sqrt{49}\)

AD = 7
Similarly, E is the midpoint of AC,

Coordinates of point E = \( \dfrac{0+6}{2}, \dfrac{0+0}{2}, \dfrac{0+6}{2}\)

Coordinates of point E = (3,0,3)

AC = \( \sqrt{3– 0)^2 + (0 – 4)^2 + (3– 0)^2} \)

AC = \( \sqrt{9+16+9} \)

AC = \( \sqrt{34}\)

Similarly, F is the midpoint of AB,

Coordinates of point F =\( \dfrac{0+0}{2}, \dfrac{0+4}{2}, \dfrac{6+0}{2}\)

Coordinates of point F = (0,2,3)

CF =\( \sqrt{6– 0)^2 + (0 – 2)^2 + (0– 3)^2} \)

CF = \( \sqrt{36+4+9} \)

CF = \( \sqrt{49}\)

CF = 7

Therefore, the lengths of the medians of the triangle ABC we obtain are,
7, \( \sqrt{34}\), 7