Higher November 2021 Paper 3 Q18
18 \(OABC\) is a trapezium.
\(\overrightarrow{OA} = \mathbf{a}\)
\(\overrightarrow{AB} = \mathbf{b}\)
\(\overrightarrow{OC} = 3\mathbf{b}\)
\(D\) is the point on \(OB\) such that \(OD : DB = 2 : 3\)
\(E\) is the point on \(BC\) such that \(BE : EC = 1 : 4\)
Work out the vector \(\overrightarrow{DE}\) in terms of \(\mathbf{a}\) and \(\mathbf{b}\).
Give your answer in its simplest form. (4)
| Answer | Mark | Mark scheme |
|---|---|---|
| \(\dfrac{2}{5}\mathbf{a} + \mathbf{b}\) | P1 | for relationship involving \(D\) eg \(\overrightarrow{OD} = \dfrac{2}{5}\overrightarrow{OB}\) or \(\overrightarrow{DB} = \dfrac{3}{5}\overrightarrow{OB}\) or for relationship involving \(E\) eg \(\overrightarrow{BE} = \dfrac{1}{5}\overrightarrow{BC}\) or \(\overrightarrow{EC} = \dfrac{4}{5}\overrightarrow{BC}\) |
| P1 | for relationship involving \(D\) in terms of \(\mathbf{a}\) and \(\mathbf{b}\) eg \(\overrightarrow{OD} = \dfrac{2}{5}(\mathbf{a} + \mathbf{b})\) or \(\overrightarrow{DB} = \dfrac{3}{5}(\mathbf{a} + \mathbf{b})\) or for relationship involving \(E\) in terms of \(\mathbf{a}\) and \(\mathbf{b}\) eg \(\overrightarrow{BE} = \dfrac{1}{5}(-\mathbf{b} - \mathbf{a} + 3\mathbf{b})\) oe or \(\overrightarrow{EC} = \dfrac{4}{5}(-\mathbf{b} - \mathbf{a} + 3\mathbf{b})\) oe or \(\overrightarrow{BC} = 2\mathbf{b} - \mathbf{a}\) oe or \(\overrightarrow{CB} = \mathbf{a} - 2\mathbf{b}\) oe | |
| P1 | (dep P2) for expression for \(\overrightarrow{DE}\) in terms of \(\mathbf{a}\) and \(\mathbf{b}\) eg \(\overrightarrow{DE} = \dfrac{3}{5}(\mathbf{a} + \mathbf{b}) + \dfrac{1}{5}(-\mathbf{b} - \mathbf{a} + 3\mathbf{b})\) | |
| A1 | for \(\dfrac{2}{5}\mathbf{a} + (1)\mathbf{b}\) or \(\dfrac{1}{5}(2\mathbf{a} + 5\mathbf{b})\) |