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Math Help - vector

  1. #1
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    vector

    The diagram shows non-collinear points O , A , B with P on the line OA such that OP:PA=2:1 and Q on the line AB such that AQ:QB=2:3 . The lines PQ and OB produced meet at the point R . If \vec{OA}=a and \vec{OB}=b

    (1) Show that \vec {PQ}=-\frac{1}{15}a+\frac{2}{5}b

    I managed to prove this .

    (2) Find the position vector of R , relative to O , in terms of b

    need help on this .
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  2. #2
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    Hello hooke
    Quote Originally Posted by hooke View Post
    The diagram shows non-collinear points O , A , B with P on the line OA such that OP:PA=2:1 and Q on the line AB such that AQ:QB=2:3 . The lines PQ and OB produced meet at the point R . If \vec{OA}=a and \vec{OB}=b

    (1) Show that \vec {PQ}=-\frac{1}{15}a+\frac{2}{5}b

    I managed to prove this .

    (2) Find the position vector of R , relative to O , in terms of b

    need help on this .
    The key to this is to realise that (i) \vec{OR} is a multiple of \vec{OB} and (ii) \vec{PR} is a multiple of \vec{PQ}, since the points O,B,R are collinear and P, Q, R are collinear. So, suppose:
    \vec{OR} =m \vec{OB}
    and
    \vec{PR}=n \vec{PQ}
    Therefore:
    \vec{OR}=m\vec{b}
    and
    \vec{PR} = n\left(-\frac{1}{15}\vec a + \frac25\vec b\right)
    But:
    \vec{OR} = \vec{OP}+\vec{PR}

    \Rightarrow m\vec b = \frac23\vec a+n\left(-\frac{1}{15}\vec a + \frac25\vec b\right)
    = \left(\frac23-\frac{1}{15}n\right)\vec a + \frac25n\vec b
    Now the coefficient of \vec a on the LHS is zero. Therefore:
    0 = \frac23-\frac{1}{15}n

    \Rightarrow n = 10
    And if we now compare coefficients of \vec b, we get:
    m = \frac25n = 4
    Hence:
    \vec{OR} = m\vec b = 4\vec b
    Grandad
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