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Thread: Vectors help please!!

  1. #1
    Feb 2009

    Vectors help please!!

    In the triangle OAB, OA = a and OB = b
    The point P is the point on AB such that AP = 2PB and Q is a point such that OP = 3QP

    Okay, this is a long question, I just couldn't do the last part.

    The point R is on OA so that RQ is parallel to AB. Show that OR = 2/3 a (Hint: Let QR = cBA, where c is a scalar quantity)

    I figured that OR = OQ + QR, and
    OQ = 2/9 a + 4/9 b

    now I am stuck.... any help will be appreciated. Thanks
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  2. #2
    Super Member

    May 2006
    Lexington, MA (USA)
    Hello, anonymous_maths!

    In $\displaystyle \Delta OAB,\;\overrightarrow{OA} = \vec a,\;\overrightarrow{OB} = \vec b$

    $\displaystyle P$ is on $\displaystyle \overrightarrow{AB}$ such that:.$\displaystyle \overrightarrow{AP} \:=\: 2\!\cdot\!\overrightarrow{PB}$

    . . and $\displaystyle Q$ is a point such that: .$\displaystyle \overrightarrow{OP} \:=\: 3\!\cdot\!\overrightarrow{QP}$

    Point $\displaystyle R$ is on $\displaystyle \overrightarrow{OA}$ so that $\displaystyle \overrightarrow{RQ} \parallel \overrightarrow{AB}$

    Show that: .$\displaystyle \overrightarrow{OR} \:=\:\tfrac{2}{3}\vec a$

    As you said, we have: .$\displaystyle \overrightarrow{OR} \;=\;\overrightarrow{OQ} + \overrightarrow{QR}$ .[1]

    Since $\displaystyle \Delta OQR \sim \Delta OPA\!:\;\;\overrightarrow{OQ} \:=\:\tfrac{2}{3}\,\overrightarrow{OP}\:\text{ and }\:\overrightarrow{QR} \:=\:\tfrac{2}{3}\,\overrightarrow{PA}$

    Substitute into [1]: .$\displaystyle \overrightarrow{OR} \;=\;\tfrac{2}{3}\,\overrightarrow{OP} + \tfrac{2}{3}\,\overrightarrow{PA} \;=\;\tfrac{2}{3}\left(\overrightarrow{OP} + \overrightarrow{PA}\right) \;=\;\tfrac{2}{3}\,\overrightarrow{OA}$

    . . Therefore: .$\displaystyle \overrightarrow{OR} \;=\;\tfrac{2}{3}\vec a$

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