4400 Higher June 2007 Paper 3 Q16
16 \(PQR\) is a triangle.
\(E\) is the point on \(PR\) such that \(PR = 3PE\).
\(F\) is the point on \(QR\) such that \(QR = 3QF\).
Diagram NOT accurately drawn
\(\overrightarrow{PQ} = \mathbf{a}\), \(\quad \overrightarrow{PE} = \mathbf{b}\).
| Scheme | Marks |
|---|---|
| (i) Answer: \(3\mathbf{b}\) | B1 |
| (ii) Answer: \(3\mathbf{b} - \mathbf{a}\) | B1 |
| (iii) Answer: \(\frac{2}{3}\mathbf{a} + \mathbf{b}\) or \(\mathbf{a} + \frac{1}{3}(3\mathbf{b} - \mathbf{a})\) or \(3\mathbf{b} - \frac{2}{3}(3\mathbf{b} - \mathbf{a})\) oe | B1 |
| (3) |
| Scheme | Marks |
|---|---|
| \(\frac{2}{3}\mathbf{a}\) or \(\frac{2}{3}\overrightarrow{PQ}\) or \(k = \frac{2}{3}\) or \(\mathbf{a} + \frac{1}{3}(3\mathbf{b} - \mathbf{a}) - \mathbf{b}\) or \(\frac{2}{3}\mathbf{a} + \mathbf{b} - \mathbf{b}\) or (a)(iii) − \(\mathbf{b}\) or \(-\mathbf{b} + \mathbf{a} + \frac{1}{3}(3\mathbf{b} - \mathbf{a})\) or \(-\mathbf{b} + \mathbf{a} + \frac{1}{3}\)(a)(ii) or \(2\mathbf{b} - \frac{2}{3}(3\mathbf{b} - \mathbf{a})\) or \(2\mathbf{b} - \frac{2}{3}\)(a)(ii) oe | B2 |
| (2) | |
| (5 marks) |
Notes
B2: for \(\frac{2}{3}\mathbf{a}\) or \(\frac{2}{3}\overrightarrow{PQ}\) or \(k = \frac{2}{3}\) unless clearly obtained by non-vector method
or for expression in terms of \(\mathbf{a}\) and/or \(\mathbf{b}\) (need not be simplified) for \(\overrightarrow{EF}\) either correct or ft from (a)
B1 for correct vector statement with at least 3 terms which includes \(\overrightarrow{EF}\) (or \(\overrightarrow{FE}\)) in terms of capital letters and/or \(\mathbf{a}\), \(\mathbf{b}\)
eg \(\overrightarrow{PQ} = \overrightarrow{PE} + \overrightarrow{EF} + \overrightarrow{FQ}\)
\(\overrightarrow{PF} = \overrightarrow{PE} + \overrightarrow{EF}\) \(\mathbf{a} = \mathbf{b} + \overrightarrow{EF} + \overrightarrow{FQ}\)
If an attempt is crossed out and replaced, mark all attempts, including crossed out one, and award best mark.