Higher November 2020 Paper 2 Q19
19 \(OAB\) is a triangle.
\(\overrightarrow{OA} = \mathbf{a} \qquad \overrightarrow{OB} = \mathbf{b}\)
The point \(C\) lies on \(OA\) such that \(OC : CA = 1 : 2\)
The point \(D\) lies on \(OB\) such that \(OD : DB = 1 : 2\)
Using a vector method, prove that \(ABDC\) is a trapezium.
(3)
| Scheme | Marks |
|---|---|
| \(\overrightarrow{AB} = -\mathbf{a} + \mathbf{b}\) or \(\overrightarrow{BA} = \mathbf{a} - \mathbf{b}\) | M1 |
\(\overrightarrow{CD} = \dfrac{1}{3}(-\mathbf{a} + \mathbf{b})\) or \(\overrightarrow{DC} = \dfrac{1}{3}(\mathbf{a} - \mathbf{b})\) oe | M1 |
| Working required Answer: Correct vectors and conclusion including parallel and trapezium | A1 |
| (3) | |
| (3 marks) |
Notes
M1: Correct diagram (condone missing vector labels or arrows – with \(C\) on line segment \(OA\) and \(D\) on line segment \(OB\)) OR for finding \(\overrightarrow{AB}\) or \(\overrightarrow{BA}\) - may be seen as part of later working
M1: Method to find \(\overrightarrow{CD}\) or \(\overrightarrow{DC}\)
A1: eg \(\overrightarrow{AB}\) (\(AB\)) and \(\overrightarrow{CD}\) (\(CD\)) are parallel therefore \(ABDC\) is a trapezium