4MA0 Higher June 2016 Paper 4R Q19
19
Diagram NOT accurately drawn
\(AB\) is parallel to \(DC\)
\(DC = 2AB\)
\(M\) is the midpoint of \(BC\)
\(\overrightarrow{AD} = 2\mathbf{b}\)
\(\overrightarrow{AB} = 4\mathbf{a}\)
Give your answer in its simplest form. (2)
\(N\) is the point such that \(DCN\) is a straight line and \(DC : CN = 2 : 1\)
| Scheme | Marks |
|---|---|
| \(\overrightarrow{BC} = -4\mathbf{a} + 2\mathbf{b} + 8\mathbf{a}\) (\(= 4\mathbf{a} + 2\mathbf{b}\)) | M1 |
| \(2\mathbf{a} + \mathbf{b}\) | A1 |
| (2) |
Notes
M1: A correct method to find \(\overrightarrow{BC}\) in terms of \(\mathbf{a}\) and \(\mathbf{b}\)
| Scheme | Marks |
|---|---|
| \(\overrightarrow{AM} = 4\mathbf{a} + \text{‘}2\mathbf{a} + \mathbf{b}\text{’}\) (\(= 6\mathbf{a} + \mathbf{b}\)) and \(\overrightarrow{AN} = 2\mathbf{b} + 8\mathbf{a} + 4\mathbf{a}\) (\(= 12\mathbf{a} + 2\mathbf{b}\)) or \(\overrightarrow{AM} = 4\mathbf{a} + \text{‘}2\mathbf{a} + \mathbf{b}\text{’}\) (\(= 6\mathbf{a} + \mathbf{b}\)) and \(\overrightarrow{MN} = \text{‘}\mathbf{b} + 2\mathbf{a}\text{’} + 4\mathbf{a}\) (\(= 6\mathbf{a} + \mathbf{b}\)) or \(\overrightarrow{AN} = 2\mathbf{b} + 8\mathbf{a} + 4\mathbf{a}\) (\(= 12\mathbf{a} + 2\mathbf{b}\)) and \(\overrightarrow{MN} = \text{‘}\mathbf{b} + 2\mathbf{a}\text{’} + 4\mathbf{a}\) (\(= 6\mathbf{a} + \mathbf{b}\)) oe | M1ft |
| Show | A1 |
| (2) |
Notes
M1ft: Correct vectors for \(\overrightarrow{AM}\) and \(\overrightarrow{AN}\) or for \(\overrightarrow{AM}\) and \(\overrightarrow{MN}\) or for \(\overrightarrow{AN}\) and \(\overrightarrow{MN}\) (need not be simplified) ft their \(\overrightarrow{BM}\) from (a)
A1: For \(\overrightarrow{AN} = 2\overrightarrow{AM}\) or \(\overrightarrow{AM} = \overrightarrow{MN}\) or \(\overrightarrow{AN} = 2\overrightarrow{MN}\) oe and there is a common point. oe