Find the distance from A to the centroid of the triangle. A(2,3/2,-4), B(3,-4,2), C(1,3,-7).
Hi,
the centroid has the coordinates $\displaystyle \left(\frac{x_A+x_B+x_C}{3} , \frac{y_A+y_B+y_C}{3}, \frac{z_A+z_B+z_C}{3}\right)$
With your values the centroid is at $\displaystyle \left(2, \frac16 , -3\right)$
Now use the distance formula:
$\displaystyle d=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2+(z_1-z_2)^2}$ that means:
$\displaystyle d=\sqrt{0+\frac{16}{9} + 1}=\sqrt{\frac{25}{9}}=\frac53$
EDIT: I finally found my typos and I believe that now everything is correct.
Hello, johntuan!
Another approach . . .
Find the distance from A to the centroid of the triangle
with vertices: .$\displaystyle A\left(2,\,\frac{3}{2},\,\text{-}4\right),\;B(3,\,\text{-}4,\,2),\;C(1,\,3,\,\text{-}7)$
You are expected to know that the centroid $\displaystyle (G)$ is the common intersection of the medians.
Also, the centroid is $\displaystyle \frac{2}{3}$ the distance from a vertex to the midpoint of the opposite side.
The midpoint of side $\displaystyle BC$ is: .$\displaystyle M\left(\frac{3+1}{2},\:\frac{\text{-}4+3}{2},\:\frac{2-7}{2}\right) \:=\:\left(2,\:\text{-}\frac{1}{2},\:\text{-}\frac{5}{2}\right)$
Distance $\displaystyle AM$ is: .$\displaystyle \overline{AM} \;=\;\sqrt{(2-2)^2 + \left(\frac{3}{2}+\frac{1}{2}\right)^2 + \left(-4+\frac{5}{2}\right)^2} \;=\;\sqrt{0^2 + 2^2 + \left(-\frac{3}{2}\right)^2}$
. . $\displaystyle \overline{AM}\;=\;\sqrt{4 + \frac{9}{4}} \;=\;\sqrt{\frac{25}{4}}\quad\Rightarrow\quad \overline{AM} \;=\;\frac{5}{2}$
Therefore: .$\displaystyle \overline{AG} \;=\;\frac{2}{3}\cdot\overline{AM} \;=\;\frac{2}{3}\cdot\frac{5}{2} \;=\;\boxed{\frac{5}{3}}$
Hi,
I've done this problem recently here:http://www.mathhelpforum.com/math-help/72933-post6.html