2020 ACJC Promo Q6

2020 ACJC Promo Q6

7 marks
Promo

The diagram below shows a trapezium \(OACB\), where \(\overrightarrow{OA}=\mathbf{a}\), \(\overrightarrow{OB}=\mathbf{b}\) and \(\overrightarrow{OC}=\mathbf{c}\).

Explain, using the diagram or otherwise, why \(\mathbf{c}=\lambda \mathbf{a}+\mathbf{b}\), for some \(\lambda \in {{\mathbb{R}}^{+}}\).[1]

By considering area of triangles or otherwise, prove that the area of trapezium is

\(\frac{1+\lambda }{2}\left| \mathbf{a}\times \mathbf{b} \right|\).[2]

Show that \({{\left| \mathbf{a}.\mathbf{b} \right|}^{2}}+{{\left| \mathbf{a}\times \mathbf{b} \right|}^{2}}={{\left( \left| \mathbf{a} \right|\left| \mathbf{b} \right| \right)}^{2}}\).[2]

Given further that \(3\mathbf{c}-3\mathbf{b}=\mathbf{a}\), \(\left| \mathbf{a} \right|=2\), \(\left| \mathbf{b} \right|=3\), \(\left| \mathbf{a}.\mathbf{b} \right|=5\), find the area of trapezium \(OACB\) exactly.[2]

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Answer:\(\overrightarrow{OC}=\overrightarrow{OB}+\overrightarrow{BC}=\mathbf{b}+\lambda \mathbf{a}\) for some \(\lambda \in \mathbb{R}'\). Since \(\overrightarrow{BC}\) is in the direction of \(\overrightarrow{OA}\);Area of trapezium \(OACB=\frac{2}{3}\sqrt{11}\)

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