4MA0 Higher January 2018 Paper 3 Q24
24
Diagram NOT accurately drawn
\(OABC\) is a parallelogram.
\(\overrightarrow{OA} = 4\mathbf{a}\) \(\overrightarrow{OC} = 3\mathbf{c}\)
The point \(X\) is such that \(CBX\) is a straight line and \(CB : BX = 2 : 3\)
The point \(Y\) is such that \(\overrightarrow{CY} = 2\overrightarrow{AX}\)
Find, in terms of \(\mathbf{a}\) and \(\mathbf{c}\), the vector \(\overrightarrow{OY}\)
Give your answer in its simplest form.
(3)
| Scheme | Marks |
|---|---|
\(\overrightarrow{BX} = \dfrac{3}{2} \times 4\mathbf{a}\;(= 6\mathbf{a})\) or \(\overrightarrow{CX} = \dfrac{5}{2} \times 4\mathbf{a}\;(= 10\mathbf{a})\) or \(\overrightarrow{AX} = \dfrac{3}{2} \times 4\mathbf{a} + 3\mathbf{c}\;(= 6\mathbf{a} + 3\mathbf{c})\) or \(\overrightarrow{OY} = \overrightarrow{OC} + \overrightarrow{CY}\) or \(\overrightarrow{OY} = \overrightarrow{OC} + 2\overrightarrow{AX}\) | M1 |
| \(\overrightarrow{CY}\) (or \(2\overrightarrow{AX}\)) = \(2(-4\mathbf{a} + 3\mathbf{c} + 4\mathbf{a} + 6\mathbf{a})\) (= \(6\mathbf{c} + 12\mathbf{a}\)) | M1 |
| \(9\mathbf{c} + 12\mathbf{a}\) | A1 |
| (3) |
Notes
M1: correct vector, in terms of \(\mathbf{a}\) and/or \(\mathbf{c}\) for \(\overrightarrow{BX}\) or \(\overrightarrow{CX}\) or \(\overrightarrow{AX}\) (need not be simplified)
or \(\overrightarrow{OY} = \overrightarrow{OC} + \overrightarrow{CY}\) or \(\overrightarrow{OY} = \overrightarrow{OC} + 2\overrightarrow{AX}\)
(accept \(3\mathbf{c}\) in place of \(\overrightarrow{OC}\) and their \(\overrightarrow{CY}\) and their \(\overrightarrow{AX}\) if clearly stated in terms of \(\mathbf{a}\) and \(\mathbf{c}\))
M1: correct vector for \(\overrightarrow{CY}\) (or \(2\overrightarrow{AX}\))
(need not be simplified)
implies the award of both method marks
A1: for \(9\mathbf{c} + 12\mathbf{a}\) or \(3(3\mathbf{c} + 4\mathbf{a})\)