Results 1 to 10 of 10

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

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

    =\tfrac13(\vec{a}+\vec{b}+\vec{c})
    Now you'll see that G is the point that's two-thirds of the way down the median 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 \vec{g} is symmetrical in \vec{a}, \vec{b} and \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?
    M is the mid-point of the side BC of the triangle. So if the position vectors of B and C are \vec{b} and \vec{c}, we use the 'mid-point formula' that I showed you in your first posting to get \vec{m}=\tfrac12(\vec{b}+\vec{c}).

    It's not terribly helpful to try to imagine what the vector \vec{m} looks like, but if you want to, you'll need to mark a point O somewhere - it doesn't matter where - to represent the origin. Then join O to M - that's the vector \vec{OM}, or \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
    317
    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 B and C are \vec{b} and \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 \vec{OA}, \vec{OB},\vec{OC} and \vec{OM}. (A displacement vector is a description of how you move from one point to another. So the displacement vector \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:
    \vec{OA}=\vec{a}
    \vec{OB}=\vec{b}
    \vec{OC}=\vec{c}
    \vec{OM}=\vec{m}
    The triangle law of addition is then simply two movement descriptions' combined into one. So when we write:
    \vec{OA}+\vec{AB}=\vec{OB}
    we're just saying:
    \underbrace{\text{a movement from O to A }}_{\vec{OA}} \underbrace{\text{ followed by }}_+ \underbrace{\text{ a movement from A to B }}_{\vec{AB}} \underbrace{\text{ is the same as }}_= \underbrace{\text{ a movement from O to B}}_{\vec{OB}}
    So when we use lower-case letters this becomes:
    \vec{a}+\vec{AB}=\vec{b}
    in other words
    \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:
    \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
    \vec{BM}=\tfrac12\vec{BC}
    Knowing that \vec{BC}=\vec{c}-\vec{b} (there's that result again), we can say:
    \vec{BM}=\tfrac12\vec{BC}

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

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

    \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
    317
    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: April 23rd 2010, 05:06 PM
  3. kindly help me for this proof of triangles
    Posted in the Geometry Forum
    Replies: 1
    Last Post: March 16th 2010, 10:12 AM
  4. Replies: 1
    Last Post: December 27th 2008, 08:38 PM
  5. Vectors and Oblique Triangles
    Posted in the Math Topics Forum
    Replies: 10
    Last Post: September 14th 2006, 06:34 PM

Search Tags


/mathhelpforum @mathhelpforum