Vectors Application 3

**skeske1234** · August 30th 2009, 06:02 PM

Find the length of the median AM in the triangle ABC, for the points A(2,1.5,-4), B (3,-4,2) and C(1,3,-7), then find the distance from A to the centroid of the triangle.

Ok, so this is what I did so far, but I am stuck as to where to go to find the distance from A to the centroid of the triangle, plus my answer for the first part does not seem to be correct with the back.

BC Midpoint=(-2,7,-9)
----------
2
=(-1,7/2,-9/2)

AM=(-3,2,-0.5)
=sqrt (13.5)

that's my answer for part A, above, but the answer in the back says 2.5

For part b, I found the midpoint of each AB, BC, and AC.. but I am not sure how or where to go from here.

**Soroban** · August 30th 2009, 06:47 PM

Hello, skeske1234!

(a) Find the length of the median $AM$ in $\Delta ABC$

with vertices: . $A(2,\tfrac{3}{2},-4),\;B (3,-4,2),\;C(1,3,-7)$

The midpoint is: . $\left(\frac{x_1 {\color{red}+}x_2}{2},\;\frac{y_1 {\color{red}+} y_2}{2},\;\frac{z_1{\color{red}+} z_2}{2}\right)$

$M$ , the midpoint of $BC$ is: . $\left(\frac{3+1}{2},\;\frac{\text{-}4+3}{2},\;\frac{2-7}{2}\right) \;=\;\left(2,\;\text{-}\frac{1}{2},\;\text{-}\frac{5}{2}\right)$

Then: . $AM \;\;=\;\;\sqrt{(2-2)^2 + \left(\text{-}\frac{1}{2} - \frac{3}{2}\right)^2 + \left(\text{-}\frac{5}{2}+4\right)^2} \;\;=\;\;\sqrt{(0)^2 + (\text{-}2)^2 + \left(\frac{3}{2}\right)^2}$

. . . . . . . . $= \;\;\sqrt{0 + 5 + \frac{9}{4}} \;\;=\;\;\sqrt{\frac{25}{4}} \;\;=\;\;\frac{5}{2}$

(b) Find the distance from $A$ to the centroid of the triangle.

The centroid is $\tfrac{2}{3}$ of the way from $A\left(2,\:\tfrac{3}{2},\:-4\right)$ to $M\left(2,\:-\tfrac{1}{2},\:-\tfrac{5}{2}\right)$

Can you find it?

**skeske1234** · August 31st 2009, 06:33 AM

How do you know that the centroid is 2/3 of the way from A to midpoint? Is that the same for all centroids of a triangle in general? so from B to midpoint it would be 2/3 way for centroid, same with C?

**Plato** · August 31st 2009, 06:58 AM

Originally Posted by skeske1234

How do you know that the centroid is 2/3 of the way from A to midpoint? Is that the same for all centroids of a triangle in general? so from B to midpoint it would be 2/3 way for centroid, same with C?

That is a standard theorem: The medians of a triangle are concurrent and the distance from a vertex to that point is 2/3 of the length of the median
That theorem has an easy but messy proof using vectors.

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