Higher June 2017 Paper 1 Q19
19
\(OABC\) is a parallelogram.
\(\overrightarrow{OA} = \mathbf{a}\) and \(\overrightarrow{OC} = \mathbf{c}\)
\(X\) is the midpoint of the line \(AC\).
\(OCD\) is a straight line so that \(OC : CD = k : 1\)
Given that \(\overrightarrow{XD} = 3\mathbf{c} - \dfrac{1}{2}\mathbf{a}\)
find the value of \(k\). (4)
| Answer | Mark | Notes |
|---|---|---|
| \(\dfrac{2}{5}\) | P1 | for first step to solve the problem e.g. \(\overrightarrow{AC} = -\mathbf{a} + \mathbf{c}\) or \(\overrightarrow{OX} = \dfrac{1}{2}\mathbf{a} + \dfrac{1}{2}\mathbf{c}\) or demonstrates the location of \(D\) and \(X\) on the diagram |
| P1 | for a correct vector statement using \(\overrightarrow{CD}\) eg \(\overrightarrow{CD} = \overrightarrow{CX} + \overrightarrow{XD}\) or \(\overrightarrow{CD} = \overrightarrow{OD} - \overrightarrow{OC}\) or \(\overrightarrow{OD} = \dfrac{7}{2}\mathbf{c}\) or \(\overrightarrow{CD} = 2.5\mathbf{c}\) oe | |
| P1 | for a correct equation or ratio using \(k\) eg equating \(\overrightarrow{XD} = 3\mathbf{c} - \dfrac{1}{2}\mathbf{a} = \dfrac{1}{2}(-\mathbf{a} + \mathbf{c}) + \dfrac{1}{k}\mathbf{c}\) or \(\dfrac{\overrightarrow{OD}}{\overrightarrow{OC}} = \dfrac{k + 1}{k}\) or \(k = \dfrac{1}{2.5}\) or using a ratio approach eg \((\overrightarrow{OC} : \overrightarrow{CD}) = k : 1 = 1 : 2.5\) | |
| A1 | cao |