4MA0 Higher January 2013 Paper 3 Q20
20 The diagram shows a parallelogram, \(PQRS\).
\(M\) is the midpoint of \(PS\).
\(\overrightarrow{PM} = \mathbf{a}\quad \overrightarrow{PQ} = \mathbf{b}\)
Diagram NOT accurately drawn
(a) Find, in terms of \(\mathbf{a}\) and/or \(\mathbf{b}\),
(i) \(\overrightarrow{PS}\)
(ii) \(\overrightarrow{PR}\)
(iii) \(\overrightarrow{MQ}\)
(3)\(N\) is the point on \(MQ\) such that \(MN = \dfrac{1}{3}MQ\)
(b) Use a vector method to prove that \(PNR\) is a straight line. (2)
| Scheme | Marks |
|---|---|
| (i) \(2\mathbf{a}\) oe | B1 |
| (ii) \(2\mathbf{a} + \mathbf{b}\) oe | B1 |
| (iii) \(-\mathbf{a} + \mathbf{b}\) oe | B1 |
| (3) |
| Scheme | Marks |
|---|---|
| \(\overrightarrow{PN} = \mathbf{a} + \tfrac{1}{3}(\text{“}{-\mathbf{a} + \mathbf{b}}\text{”})\) \(\overrightarrow{PN} = \tfrac{2}{3}\mathbf{a} + \tfrac{1}{3}\mathbf{b}\ \{= \tfrac{1}{3}(2\mathbf{a} + \mathbf{b})\}\) | M1ft |
| stating \(\overrightarrow{PN} = \overrightarrow{PR}/3\) | A1 |
| (2) |
Notes
M1ft: from (a)(iii) i.e. a valid path from \(P\) to \(N\), or \(N\) to \(P\), using lower case letters
A1: Arrows not necessary. Dependent on M1
Alt
| Scheme | Marks |
|---|---|
| \(\overrightarrow{NR} = \tfrac{2}{3}(\text{“}{-\mathbf{a} + \mathbf{b}}\text{”}) + 2\mathbf{a}\) \(\overrightarrow{NR} = \tfrac{4}{3}\mathbf{a} + \tfrac{2}{3}\mathbf{b}\ \{= \tfrac{2}{3}(2\mathbf{a} + \mathbf{b})\}\) | M1ft |
| stating \(\overrightarrow{NR} = 2\overrightarrow{PR}/3\) | A1 |
Notes
M1ft: from (a)(iii) i.e. a valid path from \(N\) to \(R\), or \(R\) to \(N\), using lower case letters
A1: Arrows not necessary. Dependent on M1
NB: If both \(\overrightarrow{PN}\) and \(\overrightarrow{NR}\) worked out correctly, award M1A1 for stating \(2PN = NR\) or stating or showing \(PN + NR = PR\)