Results 1 to 4 of 4

Thread: Vector question.

  1. #1
    Junior Member
    Joined
    Dec 2008
    Posts
    68

    Vector question.

    A,B,C and D are points of the plane
    1)show that :vector 2DA+vector 3DB = zero vector
    then AD vector = 3/5 AB vector
    Last edited by mr fantastic; Nov 3rd 2010 at 02:29 PM. Reason: Re-titled.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor Also sprach Zarathustra's Avatar
    Joined
    Dec 2009
    From
    Russia
    Posts
    1,506
    Thanks
    1
    Quote Originally Posted by lebanon View Post
    A,B,C and D are points of the plane
    1)show that :vector 2DA+vector 3DB = zero vector
    then AD vector = 3/5 AB vector

    There some information is missing.
    Last edited by mr fantastic; Nov 3rd 2010 at 02:30 PM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor

    Joined
    Apr 2005
    Posts
    19,769
    Thanks
    3027
    Quote Originally Posted by lebanon View Post
    A,B,C and D are points of the plane
    1)show that :vector 2DA+vector 3DB = zero vector
    then AD vector = 3/5 AB vector
    Why even mention "C"? We can always set up a coordinate system so that one point, say D, is the origin. Write $\displaystyle A= (x_1, y_1)$, $\displaystyle B= (x_2, y_2)$ in such a coordinate system. The we can represent vector DA as $\displaystyle <x_1, y_1>$ and vector DB as $\displaystyle <x_2, y_2>$. Then $\displaystyle 2DA= <2x_1, 2y_1>$, $\displaystyle 3DB= <3x_2, 3y_2>$ and 2DA+ 3DB= 0 becomes $\displaystyle <2x_1+ 3x_2, 2y_1+3y_2>= <0, 0>$ or $\displaystyle 2x_1= -3x_2$ and $\displaystyle 2y_1= -3y_2$.

    In that same system, $\displaystyle AB= <x_2- x_1, y_2- y_1>$ while $\displaystyle AD= <-x_1, -y_1>$.

    The result you are asking for, that $\displaystyle <-x_1, -y_1>= <(3/5)(x_2- x_1), (3/5)(y_2- y_1)>$ is the same as $\displaystyle -x_1= (3/5)(x_2- x_1)= (3/5)x_2- (3/5)x_1$ and $\displaystyle -y_1= (3/5)(y_2- y_1)$. That is, $\displaystyle -2x_1= 3x-2$ and $\displaystyle -2y_1= 3y_2$.
    Last edited by mr fantastic; Nov 3rd 2010 at 02:30 PM.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    12,028
    Thanks
    848
    Hello, lebanon!

    A strange problem . . . Point $\displaystyle \,C$ is unnecessary.


    $\displaystyle \,A,B,D\text{ are points on the plane.}$

    $\displaystyle \text{1) Show that if: }\:2\overrightarrow{DA} + 3\overrightarrow{DB} \:=\: \vec 0 $

    . . . $\displaystyle \text{then: }\;\overrightarrow{AD} \:=\:\frac{3}{5}\overrightarrow{AB}$

    Since $\displaystyle 2\overrightarrow{DA} + 3\overrightarrow{DB} \:=\:0\quad\Rightarrow\quad \overrightarrow{DA} \:=\:-\frac{3}{2}\overrightarrow{DB} \quad\Rightarrow \quad \overrightarrow{AD} \:=\:\frac{3}{2}\overrightarrow{DB} $
    . . the diagram must look like this:


    . . $\displaystyle \begin{array}{ccccc}
    \bullet & ---- & \bullet & --- & \bullet} \\
    A && D && B \\
    & \frac{3}{2}\overrightarrow{DB} && \overrightarrow{DB}
    \end{array}$



    Since $\displaystyle \overrightarrow{AB} \:=\:\frac{3}{2}\overrightarrow{DB} + \overrightarrow{DB} \:=\:\frac{5}{2}\overrightarrow{DB}$

    . . then: .$\displaystyle \dfrac{\overrightarrow{AD}}{\overrightarrow{AB}} \;=\;\dfrac{\frac{3}{2}\overrightarrow{DB}}{\frac{ 5}{2}\overrightarrow{DB}} \quad\Rightarrow\quad \dfrac{\overrightarrow{AD}}{\overrightarrow{AB}} \;=\;\dfrac{3}{5} $


    Therefore: .$\displaystyle \overrightarrow{AD} \;=\;\frac{3}{5}\overrightarrow{AB}$
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. vector question
    Posted in the Pre-Calculus Forum
    Replies: 8
    Last Post: Sep 16th 2010, 05:07 AM
  2. Vector question
    Posted in the Algebra Forum
    Replies: 1
    Last Post: Jan 21st 2010, 12:48 AM
  3. another vector question
    Posted in the Algebra Forum
    Replies: 2
    Last Post: Jan 20th 2010, 07:01 AM
  4. vector question help pls
    Posted in the Geometry Forum
    Replies: 1
    Last Post: Aug 16th 2009, 07:55 AM
  5. Vector question
    Posted in the Calculus Forum
    Replies: 4
    Last Post: Mar 25th 2009, 10:09 AM

Search Tags


/mathhelpforum @mathhelpforum