In the diagram, $OACB$ is a trapezium where $AC$ is parallel to $OB$. The line $OA$ is

Question

In the diagram, $OACB$ is a trapezium where $AC$ is parallel to $OB$. The line $OA$ is produced to the point $D$ such that $\frac{OA}{AD} = \frac{1}{2}$. Given that $\overrightarrow{OA} = \mathbf{a}$ and $\overrightarrow{OB} = \mathbf{b}$, express, as simply as possible, in terms of $\mathbf{a}$ and/or $\mathbf{b}$, $\overrightarrow{BD}$, $\overrightarrow{OC}$. Given that $\overrightarrow{OE} = 3\mathbf{a} + 2\mathbf{b}$, state the name of the quadrilateral $ODEB$, explain why $O$, $C$ and $E$ lie in a straight line. Find, giving your answers as fractions in the simplest form, $\frac{\text{area of}\hspace{0.5em} \Delta ADC}{\text{area of}\hspace{0.5em} \Delta ODB}$, $\frac{\text{area of}\hspace{0.5em} \Delta ADC}{\text{area of quadrilateral}\hspace{0.5em} ODEB}$.

Accepted 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) $\dfrac4{27}$

Vector Geometry of Parallelograms and Trapezia

Vector Geometry of Parallelograms and Trapezia

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