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Math Help - Vectors Application 3

  1. #1
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    Vectors Application 3

    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.
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  2. #2
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    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?

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  3. #3
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    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?
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  4. #4
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    Quote Originally Posted by skeske1234 View Post
    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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