4MA0 Higher June 2014 Paper 3R Q21
21
Diagram NOT accurately drawn
\(OABC\) is a parallelogram.
\(\overrightarrow{OA} = \mathbf{a}\) and \(\overrightarrow{OC} = \mathbf{c}\)
\(P\) is the point on \(AB\) such that \(AP = \dfrac{1}{4}AB\).
\(Q\) is the point on \(OC\) such that \(OQ = \dfrac{2}{3}OC\).
Find, in terms of \(\mathbf{a}\) and \(\mathbf{c}\), \(\overrightarrow{PQ}\).
Give your answer in its simplest form.
(3)
| Scheme | Marks |
|---|---|
| \((\overrightarrow{PQ} =) -\dfrac{1}{4}\mathbf{c} - \mathbf{a} + \dfrac{2}{3}\mathbf{c}\) or \((\overrightarrow{PQ} =) \dfrac{3}{4}\mathbf{c} - \mathbf{a} - \dfrac{1}{3}\mathbf{c}\) | M2 |
| \(\dfrac{5}{12}\mathbf{c} - \mathbf{a}\) | A1 |
| (3) | |
| (3 marks) |
Notes
M2: Award M1, ignoring signs, for
\(\dfrac{1}{4}\mathbf{c}\) or \(\dfrac{3}{4}\mathbf{c}\) or \(\dfrac{1}{3}\mathbf{c}\) or \(\dfrac{2}{3}\mathbf{c}\)
or \(\overrightarrow{PQ} = \overrightarrow{PA} + \overrightarrow{AO} + \overrightarrow{OQ}\) oe
or \(\overrightarrow{PQ} = \overrightarrow{PB} + \overrightarrow{BC} + \overrightarrow{CQ}\) oe
A1: SC: B2 for \(\mathbf{a} - \dfrac{5}{12}\mathbf{c}\)