First slide

Introduction to 3-D Geometry

Question

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

Easy

A

$7, 7, \sqrt{34}$
B

$7, 8, \sqrt{34}$
C

$7, 9, \sqrt{34}$
D

None of these

Solution

ABC is a triangle with vertices A(0,0,6), B (0,4,0) and C (6,0,0).
Let points D, E and F are the mid-points of BC, AC and AB, respectively.
So, AD , BE and CF will be the medians of the
coordinates of point $D = (\frac{0 + 6}{2}, \frac{4 + 0}{2}, \frac{0 + 0}{2}) = (3, 2, 0)$
Coordinates of point $E = (\frac{0 + 6}{2}, \frac{0 + 0}{2}, \frac{6 + 0}{2}) = (3, 0, 3)$
and coordinates of point $F = (\frac{0 + 0}{2}, \frac{0 + 4}{2}, \frac{6 + 0}{2})$
$= (0, 2, 3)$
Now, length of median AD = Distance between A and D
$\begin{aligned} AD = \sqrt{(0 - 3)^{2} + (0 - 2)^{2} + (6 - 0)^{2}} \\ = \sqrt{9 + 4 + 36} = \sqrt{49} = 7 \\ Similarly, BE = \sqrt{(0 - 3)^{2} + (4 - 0)^{2} + (0 - 3)^{2}} \\ = \sqrt{9 + 16 + 9} = \sqrt{34} \\ and CF = \sqrt{(6 - 0)^{2} + (0 - 2)^{2} + (0 - 3)^{2}} \\ = \sqrt{36 + 4 + 9} = \sqrt{49} = 7 \end{aligned}$

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