Results 1 to 3 of 3
Like Tree2Thanks
  • 1 Post By BobP
  • 1 Post By johng

Math Help - vector problem for the best only

  1. #1
    Newbie
    Joined
    Jul 2013
    From
    algeria
    Posts
    3

    Lightbulb vector problem for the best only

    vector problem for the best only-captures.png
    ps : sorry for my english ; i don't speak english very well, it obvious .So sorry
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    Joined
    Jun 2009
    Posts
    660
    Thanks
    133

    Re: vector problem for the best only

    Rather than show that AF and BD intersect at G, it's easier to show that AF and BD (separately), pass through the midpoint of CE.

    Dealing with AF first, suppose that AF intercects CE at P. We then have to show that P is the midpoint of CE.

    Let \underline{AP}=\lambda\underline{AF}, and let \underline{PE}=\mu\underline{CE} where \lambda \text {  and  } \mu are constants, each less than 1.

    From the triangle ABF we have

    \underline{AF} = \underline{AB}+ \underline{BF} = \underline{AB}+(3/5)\underline{BC},

    and from BEC,

    \underline{CE} = \underline{CB}+ \underline{BE} = \underline{CB}+(1/3)\underline{BA}.

    From the triangle APE,

    \underline{AE} = \underline{AP}+ \underline{PE} = \lambda\underline{AF}+\mu\underline{CE},

    from which, substituting from the earlier equations,

    (2/3)\underline{AB}=\lambda\{ \underline{AB}+(3/5)\underline{BC} \}+\mu \{\underline{CB}+(1/3)\underline{BA} \}.

    Collecting terms,

     \left( \frac{2}{3}-\lambda+\frac{\mu}{3} \right) \underline{AB}= \left( \frac{3\lambda}{5}-\mu \right) \underline{BC}.

    The vectors \underline{AB} \text{ and } \underline{BC} are not parallel, so for equality it's necessary that the two expressions in brackets are each zero.

    Solving simultaneously produces \lambda = 5/6 \text{ and } \mu = 1/2, the latter indicating that P is the midpoint of CE and hence coincident with G.


    Geometrically, the result is obvious. Joining the two points between B and F to the two points on AB gets you two lines parallel with AF, and joining the point between F and C to the midpoint of AC gets you a fourth line also parallel to AF.
    The lines are equally spaced in which case CE will be divided into four parts all of equal length, two to the left of P and two to the right. Therefore, P will be the midpoint of CE.



    The vector routine can be used to show that BD also passes through the point G.
    Thanks from johng
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member
    Joined
    Dec 2012
    From
    Athens, OH, USA
    Posts
    689
    Thanks
    281

    Re: vector problem for the best only

    Hi,
    BobP has given you one solution. Here's a solution in the same spirit, but slightly different:

    vector problem for the best only-mhfgeometry30.png
    Thanks from BobP
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Vector problem
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: May 3rd 2011, 04:39 PM
  2. vector problem
    Posted in the Algebra Forum
    Replies: 2
    Last Post: March 9th 2010, 09:34 AM
  3. vector problem
    Posted in the Geometry Forum
    Replies: 1
    Last Post: September 16th 2009, 05:52 AM
  4. Vector problem
    Posted in the Geometry Forum
    Replies: 6
    Last Post: July 3rd 2009, 07:30 PM
  5. vector problem
    Posted in the Calculus Forum
    Replies: 4
    Last Post: February 21st 2007, 06:08 PM

Search Tags


/mathhelpforum @mathhelpforum