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Thread: Vector

  1. #1
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    Vector

    In ∆OAB, P is the mid-point of AB and Q is the point on OP such that Q = 3/4P. Given that $\displaystyle \overrightarrow {OA} = \underline{a} $ and $\displaystyle \overrightarrow {OB} = \underline{b}.$

    The point R on OB is such that OR = kOB, where 0<k<1.

    Find, in terms of a, b and k, the vector $\displaystyle \overrightarrow{AR}.$


    I found this: $\displaystyle \overrightarrow{AR} = - \underline{a} + k\underline{b}$

    Given that AQR is a straight line:
    Find the ratio in which Q divides AR and the value of k.

    I did:
    $\displaystyle
    \overrightarrow{AR} = \overrightarrow{AQ} + \overrightarrow{QR}$

    $\displaystyle \overrightarrow{QR} = -\frac{3}{8} \underline{a} + (k - \frac{3}{8})\underline{b} $

    Iím confused to find the ratio & the value of k.

    Is it like:
    $\displaystyle
    \overrightarrow{AR} = \overrightarrow{AQ} : \overrightarrow{QR}
    =(\frac{3}{8} \underline{b} - \frac{5}{8} \underline{a}) : (k - \frac{3}{8})\underline{b} - \frac{3}{8}\underline{a}$

    ThenÖ?
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  2. #2
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    earboth's Avatar
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    Quote Originally Posted by geton View Post
    In ∆OAB, P is the mid-point of AB and Q is the point on OP such that Q = 3/4P. Given that $\displaystyle \overrightarrow {OA} = \underline{a} $ and $\displaystyle \overrightarrow {OB} = \underline{b}.$

    The point R on OB is such that OR = kOB, where 0<k<1.

    Find, in terms of a, b and k, the vector $\displaystyle \overrightarrow{AR}.$


    I found this: $\displaystyle \overrightarrow{AR} = - \underline{a} + k\underline{b}$

    Given that AQR is a straight line:
    Find the ratio in which Q divides AR and the value of k.

    I did:
    $\displaystyle
    \overrightarrow{AR} = \overrightarrow{AQ} + \overrightarrow{QR}$

    $\displaystyle \overrightarrow{QR} = -\frac{3}{8} \underline{a} + (k - \frac{3}{8})\underline{b} $

    I’m confused to find the ratio & the value of k.

    Is it like:
    $\displaystyle
    \overrightarrow{AR} = \overrightarrow{AQ} : \overrightarrow{QR}
    =(\frac{3}{8} \underline{b} - \frac{5}{8} \underline{a}) : (k - \frac{3}{8})\underline{b} - \frac{3}{8}\underline{a}$

    Then…?
    1. $\displaystyle \overrightarrow{AQ}=-\vec a+\frac34\left(\vec a+\frac12\left(\vec b - \vec a\right) \right) = -\frac58 \vec a + \frac38 \vec b$

    2. $\displaystyle \overrightarrow{AR} = r \cdot (\overrightarrow{AQ}) = -\frac58 r \cdot \vec a + \frac38 r \cdot \vec b$

    3. $\displaystyle \overrightarrow{AR} = - \vec{a} + k\vec{b}$

    4. Compare the coefficients at 2. and 3.:

    $\displaystyle -\frac58 r = -1~\implies~r = \frac85$

    $\displaystyle \frac38 r = k~\implies~\frac38 \cdot \frac85 = k~\implies~k = \frac35$

    5. Calculate the ratio.
    Attached Thumbnails Attached Thumbnails Vector-vektorverhaeltnis.png  
    Last edited by earboth; Nov 10th 2008 at 11:16 PM. Reason: removed a typo
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  3. #3
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    Quote Originally Posted by earboth View Post
    1. $\displaystyle \overrightarrow{AQ}=-\vec a+\frac34\left(\vec a+\frac12\left(\vec b - \vec a\right) \right) = -\frac58 \vec a + \frac38 \vec b$

    2. $\displaystyle \overrightarrow{AR} = r \cdot (\overrightarrow{AQ}) = -\frac58 r \cdot \vec a + \frac38 r \cdot \vec b$

    3. $\displaystyle \overrightarrow{AR} = - \vec{a} + k\vec{b}$

    4. Compare the coefficients at 2. and 3.:

    $\displaystyle -\frac58 r = -1~\implies~r = \frac85$

    $\displaystyle \frac38 r = k~\implies~\frac38 \cdot \frac85 = k~\implies~k = \frac35$

    5. Calculate the ratio.
    Thank you so much
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