N2011 P1 Q7

N2011 P1 Q7

11 marks
Past Year National Exams
Free

Referred to the origin \(O\), the points \(A\) and \(B\) are such that \(\overrightarrow{OA}=\mathbf{a}\) and \(\overrightarrow{OB}=\mathbf{b}\). The point \(P\) on \(OA\) is such that \(OP:PA=1:2\), and the point \(Q\) on \(OB\) is such that \(OQ:QB=3:2\). The mid-point of \(PQ\) is \(M\) (see diagram).

  1. Find \(\overrightarrow{OM}\) in terms of \(\mathbf{a}\) and \(\mathbf{b}\) and show that the area of triangle \(OPM\) can be written as \(k\left| \mathbf{a}\times \mathbf{b} \right|\), where \(k\) is a constant to be found.[6]
  2. The vectors \(\mathbf{a}\) and \(\mathbf{b}\) are now given by
    \(\mathbf{a}=2p\mathbf{i}-6p\mathbf{j}+3p\mathbf{k}\) and \(\mathbf{b}=\mathbf{i}+\mathbf{j}-2\mathbf{k}\),
    where \(p\) is a positive constant. Given that \(\mathbf{a}\) is a unit vector,
    1. find the exact value of \(p\),[2]
    2. give a geometrical interpretation of \(\left| \mathbf{a}\cdot \mathbf{b} \right|\),[1]
    3. evaluate \(\mathbf{a}\times \mathbf{b}\).[2]

Video Solution:

Default solution

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Answer:(i) k = 1/20; (ii)(a) p = 1/7; (ii)(b) Length of projection of OB onto OA; (ii)(c) a×b = (1/7)(9, 7, 8)

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