Higher November 2024 Paper 1 Q15
15 \(OXYZ\) is a parallelogram.
\(\overrightarrow{OY} = \mathbf{a}\) and \(\overrightarrow{OZ} = \mathbf{b}\)
\(M\) is the point on \(OY\) such that \(OM : MY = 1 : 3\)
\(N\) is the point on \(OZ\) such that \(ON : NZ = 1 : 2\)
Work out the ratio \(XN : MN\)
You must show all your working. (4)
| Answer | Mark | Mark scheme |
|---|---|---|
| 4 : 1 | P1 | for \(\overrightarrow{OM} = \dfrac{1}{4}\mathbf{a}\) or \(\overrightarrow{MO} = -\dfrac{1}{4}\mathbf{a}\) or \(\overrightarrow{ON} = \dfrac{1}{3}\mathbf{b}\) or \(\overrightarrow{NO} = -\dfrac{1}{3}\mathbf{b}\) \(\overrightarrow{OX} = \mathbf{a} - \mathbf{b}\) or \(\overrightarrow{XO} = -\mathbf{a} + \mathbf{b}\) or \(\overrightarrow{ZY} = \mathbf{a} - \mathbf{b}\) or \(\overrightarrow{YZ} = -\mathbf{a} + \mathbf{b}\) |
| P1 | for \(\overrightarrow{XN} = \mathbf{b} - \mathbf{a} + \dfrac{1}{3}\mathbf{b}\ \left(= \dfrac{4}{3}\mathbf{b} - \mathbf{a}\right)\) oe or \(\overrightarrow{MN} = \dfrac{1}{3}\mathbf{b} - \dfrac{1}{4}\mathbf{a}\) oe or \(\overrightarrow{XM} = \mathbf{b} - \mathbf{a} + \dfrac{1}{4}\mathbf{a}\ \left(= \mathbf{b} - \dfrac{3}{4}\mathbf{a}\right)\) oe | |
| P1 | for \(\overrightarrow{XN} = \mathbf{b} - \mathbf{a} + \dfrac{1}{3}\mathbf{b}\) oe and \(\overrightarrow{MN} = \dfrac{1}{3}\mathbf{b} - \dfrac{1}{4}\mathbf{a}\) or \(\dfrac{4}{3}\mathbf{b} - \mathbf{a} - \mathbf{b} + \dfrac{3}{4}\mathbf{a}\) oe | |
| A1 | for 4 : 1 oe |
Additional guidance
Implies 1st P1
A correct answer with no supportive working gets 0 marks