Vector Geometry of Parallelograms and Trapezia

Vector Geometry of Parallelograms and Trapezia

Secondary 4

In the diagram, \(OACB\) is a trapezium where \(AC\) is parallel to \(OB\). The line \(OA\) is produced to the point \(D\) such that \(\frac{OA}{AD} = \frac{1}{2}\).

  1. Given that \(\overrightarrow{OA} = \mathbf{a}\) and \(\overrightarrow{OB} = \mathbf{b}\), express, as simply as possible, in terms of \(\mathbf{a}\) and/or \(\mathbf{b}\),
    1. \(\overrightarrow{BD}\),
    2. \(\overrightarrow{OC}\).
  2. Given that \(\overrightarrow{OE} = 3\mathbf{a} + 2\mathbf{b}\),
    1. state the name of the quadrilateral \(ODEB\),
    2. explain why \(O\), \(C\) and \(E\) lie in a straight line.
  3. Find, giving your answers as fractions in the simplest form,
    1. \(\frac{\text{area of}\hspace{0.5em} \Delta ADC}{\text{area of}\hspace{0.5em} \Delta ODB}\),
    2. \(\frac{\text{area of}\hspace{0.5em} \Delta ADC}{\text{area of quadrilateral}\hspace{0.5em} ODEB}\).

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Answer:(a)(i) \(3\mathbf a-\mathbf b\) (ii) \(\mathbf a+\dfrac23\mathbf b\) (b)(i) Trapezium (ii) \(\overrightarrow{OE}=3\overrightarrow{OC}\), so \(O,C,E\) are collinear. (c)(i) \(\dfrac49\) (ii) \(\dfrac4{27}\)

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