Results 1 to 10 of 10

Thread: Proof about medians of triangles using vectors

  1. #1
    Newbie
    Joined
    Oct 2009
    Posts
    9

    Proof about medians of triangles using vectors

    Prove the given theorem using vectors.

    The medians of a triangle meet in a point whose distance from each vertex is two-thirds the length of the median from that vertex.
    Last edited by mr fantastic; Oct 27th 2009 at 07:15 PM. Reason: Changed thread title.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Grandad's Avatar
    Joined
    Dec 2008
    From
    South Coast of England
    Posts
    2,570
    Thanks
    1
    Hello kullgirl418
    Quote Originally Posted by kullgirl418 View Post
    Prove the given theorem using vectors.

    The medians of a triangle meet in a point whose distance from each vertex is two-thirds the length of the median from that vertex.
    Continuing from my reply to your previous post, we can generalise the result about the position vector of the mid-point of a line segment, to get the position vector of a point dividing a line segment in a given ratio, and it's this:

    The position vector of the point $\displaystyle R$ that divides the line segment $\displaystyle AB$ in the ratio $\displaystyle \lambda:\mu$ is given by:
    $\displaystyle \vec{r} = \frac{\mu\vec{a}+\lambda\vec{b}}{\lambda+\mu}$
    So suppose the triangle is $\displaystyle ABC, M$ is the mid-point of $\displaystyle BC$ and $\displaystyle G$ divides $\displaystyle AM$ in the ratio $\displaystyle 2:1$. Then:
    $\displaystyle \vec{m} = \tfrac12(\vec{b}+\vec{c})$, as before using the mid-point formula
    and:
    $\displaystyle \vec{g}= \frac{\vec{a}+2\vec{m}}{3}$, using the more general result above

    $\displaystyle =\frac13\Big(\vec{a}+2\times\tfrac12(\vec{b}+\vec{ c})\Big)$

    $\displaystyle =\tfrac13(\vec{a}+\vec{b}+\vec{c})$
    Now you'll see that $\displaystyle G$ is the point that's two-thirds of the way down the median $\displaystyle AM$. I'll leave it to you to show that it's also two-thirds of the way down the other medians as well. (Although this is pretty obvious since the expression for $\displaystyle \vec{g}$ is symmetrical in $\displaystyle \vec{a}, \vec{b}$ and $\displaystyle \vec{c}$.)

    Grandad
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Oct 2009
    Posts
    9
    Thanks this is very helpful! but i am still confused as to how you found vector m and what vector m would look like on the picture?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Grandad's Avatar
    Joined
    Dec 2008
    From
    South Coast of England
    Posts
    2,570
    Thanks
    1
    Hello kullgirl418
    Quote Originally Posted by kullgirl418 View Post
    Thanks this is very helpful! but i am still confused as to how you found vector m and what vector m would look like on the picture?
    $\displaystyle M$ is the mid-point of the side $\displaystyle BC$ of the triangle. So if the position vectors of $\displaystyle B$ and $\displaystyle C$ are $\displaystyle \vec{b}$ and $\displaystyle \vec{c}$, we use the 'mid-point formula' that I showed you in your first posting to get $\displaystyle \vec{m}=\tfrac12(\vec{b}+\vec{c})$.

    It's not terribly helpful to try to imagine what the vector $\displaystyle \vec{m}$ looks like, but if you want to, you'll need to mark a point $\displaystyle O$ somewhere - it doesn't matter where - to represent the origin. Then join $\displaystyle O$ to $\displaystyle M$ - that's the vector $\displaystyle \vec{OM}$, or $\displaystyle \vec{m}$ for short.

    Grandad
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Oct 2009
    Posts
    9
    I am still confused as to how you are labeling B and C as vectors when they are representing points not lines?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Senior Member pankaj's Avatar
    Joined
    Jul 2008
    From
    New Delhi(India)
    Posts
    318
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor
    Grandad's Avatar
    Joined
    Dec 2008
    From
    South Coast of England
    Posts
    2,570
    Thanks
    1

    Position and displacement vectors

    Hello kullgirl418
    Quote Originally Posted by kullgirl418 View Post
    I am still confused as to how you are labeling B and C as vectors when they are representing points not lines?
    It's clear that you're not really understanding vectors at all, so read this carefully.

    I didn't label B and C as vectors - I referred to their position vectors:
    Quote Originally Posted by Grandad View Post
    ... So if the position vectors of $\displaystyle B$ and $\displaystyle C$ are $\displaystyle \vec{b}$ and $\displaystyle \vec{c}$,...
    ...in other words, their positions in relation to some arbitrary origin - usually denoted by O.

    If you're not clear what that phrase 'their positions in relation to some arbitrary origin' means, let me unpack it a little - but you'll have to do the work. So:

    • Draw a triangle ABC - don't make it a special triangle; try not to make it right-angled or isosceles.


    • Mark M as the mid-point of BC.


    • Somewhere else on the paper mark a point O. It's easiest if it's outside the triangle and doesn't lie on any of its sides when they're produced.


    • Join O to A, B, C and M.

    Then the position vectors of the points A, B, C and M in relation to O are the displacement vectors $\displaystyle \vec{OA}, \vec{OB},\vec{OC}$ and $\displaystyle \vec{OM}$. (A displacement vector is a description of how you move from one point to another. So the displacement vector $\displaystyle \vec{OA}$ describes - using whatever language you'd like - how you would move from O to A.)

    To make it simpler to write, we describe these position vectors - these 'movement descriptions' - using lower-case letters:
    $\displaystyle \vec{OA}=\vec{a}$
    $\displaystyle \vec{OB}=\vec{b}$
    $\displaystyle \vec{OC}=\vec{c}$
    $\displaystyle \vec{OM}=\vec{m}$
    The triangle law of addition is then simply two movement descriptions' combined into one. So when we write:
    $\displaystyle \vec{OA}+\vec{AB}=\vec{OB}$
    we're just saying:
    $\displaystyle \underbrace{\text{a movement from O to A }}_{\vec{OA}}$ $\displaystyle \underbrace{\text{ followed by }}_+$ $\displaystyle \underbrace{\text{ a movement from A to B }}_{\vec{AB}}$ $\displaystyle \underbrace{\text{ is the same as }}_=$ $\displaystyle \underbrace{\text{ a movement from O to B}}_{\vec{OB}}$
    So when we use lower-case letters this becomes:
    $\displaystyle \vec{a}+\vec{AB}=\vec{b}$
    in other words
    $\displaystyle \vec{AB}=\vec{b}-\vec{a}$
    which I described in an earlier reply to you as a very important result.

    Finally, let's show you again where the 'mid-point' formula comes from.

    Writing this important result again, but using the points B and M this time:
    $\displaystyle \vec{BM}=\vec{m}-\vec{b}$
    But the movement from B to M is just one-half of the movement from B to C, since M is the mid-point of BC. So
    $\displaystyle \vec{BM}=\tfrac12\vec{BC}$
    Knowing that $\displaystyle \vec{BC}=\vec{c}-\vec{b}$ (there's that result again), we can say:
    $\displaystyle \vec{BM}=\tfrac12\vec{BC}$

    $\displaystyle \Rightarrow \vec{m}-\vec{b}=\tfrac12(\vec{c}-\vec{b})$

    $\displaystyle \Rightarrow \vec{m}=\vec{b}+\tfrac12\vec{c}-\tfrac12\vec{b}$

    $\displaystyle \Rightarrow \vec{m}=\tfrac12(\vec{b}+\vec{c})$
    I hope that helps to make things clearer.

    Grandad
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Senior Member pankaj's Avatar
    Joined
    Jul 2008
    From
    New Delhi(India)
    Posts
    318
    Nice explanation Grandad
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Newbie
    Joined
    Oct 2009
    Posts
    9
    sorry but now i am confused on how you found vector g? because the answer doesn't make sense to me... how do i know that G is two-thirds of the way?
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Newbie
    Joined
    Oct 2009
    Posts
    9
    never mind i understand it! thanks for all of your help!!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 5
    Last Post: May 29th 2011, 03:10 PM
  2. triangles represented using vectors
    Posted in the Calculus Forum
    Replies: 1
    Last Post: Apr 23rd 2010, 05:06 PM
  3. kindly help me for this proof of triangles
    Posted in the Geometry Forum
    Replies: 1
    Last Post: Mar 16th 2010, 10:12 AM
  4. Replies: 1
    Last Post: Dec 27th 2008, 08:38 PM
  5. Vectors and Oblique Triangles
    Posted in the Math Topics Forum
    Replies: 10
    Last Post: Sep 14th 2006, 06:34 PM

Search tags for this page

Search Tags


/mathhelpforum @mathhelpforum