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    Edexcel C4 textbook, excercise 5k question 2. >_>;

    The position vectors of the points A and B are relative to an origin O are 5i+4j+k, -i+j-2k respectively. Find the position vector of the point P which lies on AB produced such that AP=2BP.

    I thought of drawing a diagram, so:



    This means that:

    \vec{OP} = \mathbf{a} + \frac{2}{3}(\vec{AB}) \\

\vec{OP} = \mathbf{a} + \frac{2}{3}(\mathbf{b} - \mathbf{a})

    Now what?
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    (Original post by wanderlust.xx)
    Edexcel C4 textbook, excercise 5k question 2. >_>;

    The position vectors of the points A and B are relative to an origin O are 5i+4j+k, -i+j-2k respectively. Find the position vector of the point P which lies on AB produced such that AP=2BP.

    I thought of drawing a diagram, so:



    This means that:

    \vec{OP} = \mathbf{a} + \frac{2}{3}(\vec{AB})

    Now what?
    You also have the equation AP + PB = AB, and you can solve that simultaneously with the other equation you got, I think.

    EDIT:

    Forget that.

    You know the vector OA and you can work out the vector AB (It's the same as OB - OA) and then you can just plug those right into the equation that you got

    Actually, I think there are two answers.

    Your first equation could also be: OP = a + 2(AB)
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    (Original post by tazarooni89)
    You also have the equation AP + PB = AB, and you can solve that simultaneously with the other equation you got, I think.

    EDIT:

    Forget that.

    You know the vector OA and you can work out the vector AB (It's the same as OB - OA) and then you can just plug those right into the equation that you got

    Actually, I think there are two answers.

    Your first equation could also be: OP = a + 2(AB)
    P is on AB produced and so the order of the letters will be A, B, P. It follows that B is the midpoint of AP. Hence vector AB = vector BP. Etc.
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    Maths Buster actually right. First you've to find out AB = b-a

    Then use the condition AP = 2BP. i.e. AP = 2AB.

    Therefore you can find out OP.
 
 
 
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