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Math Help - Geometry with Vectors

  1. #1
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    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.

    ---

    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.
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by rualin View Post
    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.

    ---

    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
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  3. #3
    MHF Contributor red_dog's Avatar
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    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}
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  4. #4
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    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.
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  5. #5
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    The medians of a triangle are concurrent at appoint which is \frac{2}{3} the distance from any vertex.
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  6. #6
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    Quote Originally Posted by rualin View Post
    ...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:
    <br />
\overrightarrow{OZ}= \overrightarrow{OB}+\overrightarrow{OC}-\overrightarrow{OB}+\frac{1}{3} \overrightarrow{OA}+\frac13 \overrightarrow{OB}-\frac23 \overrightarrow{OC} . Collect like terms:
    <br />
\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.
    Attached Thumbnails Attached Thumbnails Geometry with Vectors-dreieck_schwerpunkt.gif  
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