Higher January 2023 Paper 1R Q22
22 The diagram shows triangle \(OAB\) with \(OA\) extended to \(E\)
Diagram NOT accurately drawn
\(\overrightarrow{OA} = \mathbf{a}\) \(\overrightarrow{OB} = \mathbf{b}\)
\(M\) is the point on \(OB\) such that \(OM : MB = 4 : 1\)
\(N\) is the point on \(AB\) such that \(AN : NB = 3 : 2\)
\(OA : AE = 5 : 3\)
(a) Find an expression for \(\overrightarrow{ON}\) in terms of \(\mathbf{a}\) and \(\mathbf{b}\)
Give your answer in its simplest form. (2)
Give your answer in its simplest form. (2)
(b) Use a vector method to show that \(MNE\) is a straight line. (3)
| Scheme | Marks |
|---|---|
| \(\overrightarrow{ON} = \mathbf{b} + \dfrac{2}{5}(\mathbf{a} - \mathbf{b})\) oe or \(\overrightarrow{ON} = \mathbf{a} + \dfrac{3}{5}(\mathbf{b} - \mathbf{a})\) oe | M1 |
Correct answer scores full marks (unless from obvious incorrect working) Answer: \(\dfrac{2}{5}\mathbf{a} + \dfrac{3}{5}\mathbf{b}\) | A1 |
| (2) |
Notes
A1: oe eg \(\dfrac{1}{5}(2\mathbf{a} + 3\mathbf{b})\) but must be one term in \(\mathbf{a}\) and one in \(\mathbf{b}\)
| Scheme | Marks |
|---|---|
\(\overrightarrow{ME} = \dfrac{8}{5}\mathbf{a} - \dfrac{4}{5}\mathbf{b}\) \(\overrightarrow{NE} = \dfrac{6}{5}\mathbf{a} - \dfrac{3}{5}\mathbf{b}\) (all oe but simplified) \(\overrightarrow{MN} = \dfrac{2}{5}\mathbf{a} - \dfrac{1}{5}\mathbf{b}\) | M1ft |
\(\overrightarrow{ME} = \dfrac{8}{5}\mathbf{a} - \dfrac{4}{5}\mathbf{b}\) \(\overrightarrow{NE} = \dfrac{6}{5}\mathbf{a} - \dfrac{3}{5}\mathbf{b}\) (all oe but simplified) \(\overrightarrow{MN} = \dfrac{2}{5}\mathbf{a} - \dfrac{1}{5}\mathbf{b}\) | M1 |
| Evidence of a vector method needed Answer: shown | A1 |
| (3) | |
| (5 marks) |
Notes
M1ft: for one of \(\overrightarrow{ME}\), \(\overrightarrow{NE}\) or \(\overrightarrow{MN}\) or one of \(\overrightarrow{EM}\), \(\overrightarrow{EN}\) or \(\overrightarrow{NM}\)
ft (dep on M1 in (a)) their expression for \(\overrightarrow{ON}\) for this mark only
\(\left[\overrightarrow{ME} = \overrightarrow{ON} + \dfrac{6}{5}\mathbf{a} - \dfrac{7}{5}\mathbf{b}\right.\)
\(\left.\overrightarrow{MN} = \overrightarrow{ON} - \dfrac{4}{5}\mathbf{b},\; \overrightarrow{NE} = -\overrightarrow{ON} + \dfrac{8}{5}\mathbf{a}\right]\)
(corrected from the printed mark scheme, which shows \(\dfrac{11}{3}\mathbf{a}\) here: \(\overrightarrow{OE} = \dfrac{8}{5}\mathbf{a}\))
M1: for two of \(\overrightarrow{ME}\), \(\overrightarrow{NE}\) or \(\overrightarrow{MN}\) or two of \(\overrightarrow{EM}\), \(\overrightarrow{EN}\) or \(\overrightarrow{NM}\) must be correct
A1: eg \(\overrightarrow{ME} = 4 \times \overrightarrow{MN}\) or \(\overrightarrow{NE} = 3 \times \overrightarrow{MN}\) or \(\overrightarrow{ME} = \dfrac{4}{3} \times \overrightarrow{NE}\)
or showing they are multiples of the same vector eg \(\overrightarrow{MN} = \dfrac{1}{5}(2\mathbf{a} - \mathbf{b})\) and \(\overrightarrow{NE} = \dfrac{3}{5}(2\mathbf{a} - \mathbf{b})\)