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Math Help - Vector question.

  1. #1
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    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; November 3rd 2010 at 02:29 PM. Reason: Re-titled.
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  2. #2
    MHF Contributor Also sprach Zarathustra's Avatar
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    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; November 3rd 2010 at 02:30 PM.
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  3. #3
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    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 A= (x_1, y_1), B= (x_2, y_2) in such a coordinate system. The we can represent vector DA as <x_1, y_1> and vector DB as <x_2, y_2>. Then 2DA= <2x_1, 2y_1>, 3DB= <3x_2, 3y_2> and 2DA+ 3DB= 0 becomes <2x_1+ 3x_2, 2y_1+3y_2>= <0, 0> or 2x_1= -3x_2 and 2y_1= -3y_2.

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

    The result you are asking for, that <-x_1, -y_1>= <(3/5)(x_2- x_1), (3/5)(y_2- y_1)> is the same as -x_1= (3/5)(x_2- x_1)= (3/5)x_2- (3/5)x_1 and -y_1= (3/5)(y_2- y_1). That is, -2x_1= 3x-2 and -2y_1= 3y_2.
    Last edited by mr fantastic; November 3rd 2010 at 02:30 PM.
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  4. #4
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    Hello, lebanon!

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


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

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

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

    Since 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:


    . . \begin{array}{ccccc}<br />
\bullet & ---- & \bullet & --- & \bullet} \\<br />
A && D && B \\ <br />
& \frac{3}{2}\overrightarrow{DB} && \overrightarrow{DB} <br />
\end{array}



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

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