Higher January 2020 Paper 2R Q26
26 The diagram shows trapezium \(OACB\).
Diagram NOT accurately drawn
\(\overrightarrow{OA} = 3\mathbf{a}\) \(\overrightarrow{OB} = 6\mathbf{b}\) \(\overrightarrow{AC} = 4\mathbf{b}\)
\(N\) is the point on \(OC\) such that \(ANB\) is a straight line.
Find \(\overrightarrow{ON}\) as a simplified expression in terms of \(\mathbf{a}\) and \(\mathbf{b}\).
(5)
| Scheme | Marks |
|---|---|
| \(\overrightarrow{OC} = 3\mathbf{a} + 4\mathbf{b}\) | B1 |
| \(\overrightarrow{ON} = t(3\mathbf{a} + 4\mathbf{b})\) | M1 |
| \(\overrightarrow{ON} = 3\mathbf{a} + s(-3\mathbf{a} + 6\mathbf{b})\) | M1 |
| \(t(3\mathbf{a} + 4\mathbf{b}) = 3\mathbf{a} + s(-3\mathbf{a} + 6\mathbf{b})\) \(\rightarrow t = 0.6,\; s = 0.4\) | A1 |
| \(\overrightarrow{ON} = 1.8\mathbf{a} + 2.4\mathbf{b}\) oe | A1 |
| (5) | |
| (5 marks) |
Notes
B1: Correct expression for \(\overrightarrow{OC}\)
M1: Correct expressions for \(\overrightarrow{ON}\)
A1: \(t\) or \(s\) value correct
A1: e.g. \(\overrightarrow{ON} = \dfrac{3}{5}(3\mathbf{a} + 4\mathbf{b})\)
Alternative method
| Scheme | Marks |
|---|---|
| \(\overrightarrow{AB} = -3\mathbf{a} + 6\mathbf{b}\) | B1 |
| \(\overrightarrow{AN} = s(-3\mathbf{a} + 6\mathbf{b})\) | M1 |
| \(\overrightarrow{AN} = -3\mathbf{a} + t(3\mathbf{a} + 4\mathbf{b})\) | M1 |
| \(-3\mathbf{a} + t(3\mathbf{a} + 4\mathbf{b}) = s(-3\mathbf{a} + 6\mathbf{b})\) \(\rightarrow t = 0.6,\; s = 0.4 \rightarrow \overrightarrow{AN} = -1.2\mathbf{a} + 2.4\mathbf{b}\) \(\overrightarrow{ON} = 3\mathbf{a} + \overrightarrow{AN}\) | A1 |
| \(\overrightarrow{ON} = 1.8\mathbf{a} + 2.4\mathbf{b}\) oe | A1 |
Notes
B1: Correct expression for \(\overrightarrow{AB}\)
M1: Correct expressions for \(\overrightarrow{AN}\)
A1: \(t\) or \(s\) value correct
A1: e.g. \(\overrightarrow{ON} = \dfrac{3}{5}(3\mathbf{a} + 4\mathbf{b})\)
Alternative method
| Scheme | Marks |
|---|---|
| \(\overrightarrow{OC} = 3\mathbf{a} + 4\mathbf{b}\) | B1 |
| \(ON : NC = 6 : 4\) (i.e. 3 : 2) | M1 |
| \(\overrightarrow{ON} = \dfrac{3}{5}\overrightarrow{OC}\) | M2 |
| \(\overrightarrow{ON} = 1.8\mathbf{a} + 2.4\mathbf{b}\) oe | A1 |
Notes
B1: Correct expression for \(\overrightarrow{OC}\)
A1: e.g. \(\overrightarrow{ON} = \dfrac{3}{5}(3\mathbf{a} + 4\mathbf{b})\)