1. ## Geometry with Vectors

Let A=(4,2,0), B=(1,3,0) and C=(1,1,3) form a triangle.

Find the coordinates of the point in which the medians of triangle ABC intersect.

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There is probably a very simple yet elegant solution to this but I can't seem to find it. Any help would be greatly appreciated.

2. Originally Posted by rualin
Let A=(4,2,0), B=(1,3,0) and C=(1,1,3) form a triangle.

Find the coordinates of the point in which the medians of triangle ABC intersect.

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There is probably a very simple yet elegant solution to this but I can't seem to find it. Any help would be greatly appreciated.
i suppose you are talking about the point where vectors that perpendicularly bisect the sides of the triangle meet?

find such vectors and equate them. of course, here i am referring to the vector equations of lines. recall that we can represent a line by the equation:

$\vec{r} = \vec{r}_0 + t \vec{v}$

where $\vec{r}_0 = \left< x_0, y_0, z_0 \right>$ is a point the line passes through (in this case, the midpoint of the lines of the sides of the triangle, $\vec{v} = \left< a,b,c \right>$ is a vector describing the direction of the line. in this case,the direction perpendicular to the sides they pass through, $t$ is a parameter, and $\vec{r}$ is just the arbitrary vector $\left< x,y,z \right>$

note that we can write the lines as follows:

$\left< x,y,z \right> = \left< x_0 + at, y_0 + bt, z_0 + ct \right>$

get all you lines in this form and then equate the corresponding components. you will be able to solve for the parameters and hence find the point of intersection

3. I think it's about center of gravity.
In this case you don't need vectors.
The coordinates of the center of gravity G are
$x_G=\frac{x_A+x_B+x_C}{3}, \ y_G=\frac{y_A+y_B+y_C}{3}$

4. I got (2,2,1) using red_dog's suggestion though I don't understand how that magical formula works being that I don't know anything about centers of mass. I couldn't follow through Jhevon's explanation.

It seems to me that I could simply assume the point I care for is a point X=(x,y,z), find the distances from A,B, and C to this point X, add them all up and equate them to zero to get the same solution but I couldn't explain why set them up to zero.

Thanks, Jhevon and red_dog.

5. The medians of a triangle are concurrent at appoint which is $\frac{2}{3}$ the distance from any vertex.

6. Originally Posted by rualin
...I don't understand how that magical formula works ...
Hello,

consider the triangle ABC. You know the coordinates of the vertices. Let $A(x_A, y_A)$ then the vector $\overrightarrow{OA} = [x_A, y_A]$.

As you know: $\overrightarrow{OM_{AB}} = \frac{1}{2}(\overrightarrow{OA}+\overrightarrow{OB })$

Now you are looking for the vector $\overrightarrow{OZ}= \overrightarrow{OB}+\overrightarrow{BC}+\frac{2}{3 } \cdot \overrightarrow{CM_{AB}}$. Now substitute:
$\overrightarrow{BC}=\overrightarrow{OC}-\overrightarrow{OB}$ and
$\overrightarrow{CM_{AB}}=\frac{1}{2} \cdot (\overrightarrow{OA}+\overrightarrow{OB})-\overrightarrow{OC})$
You'll get:
$
\overrightarrow{OZ}= \overrightarrow{OB}+\overrightarrow{OC}-\overrightarrow{OB}+\frac{1}{3} \overrightarrow{OA}+\frac13 \overrightarrow{OB}-\frac23 \overrightarrow{OC}$
. Collect like terms:
$
\overrightarrow{OZ}=\frac13 \cdot (\overrightarrow{OA}+\overrightarrow{OB}+\overrigh tarrow{OC})$

Using the coordinate writing of vectors you have:

$[x_Z, y_Z] = \frac13 \cdot ([x_a, y_A] + [x_B, y_B] + [x_C, y_C])$ which is red_dog's fabulous formula.