Higher June 2019 Paper 4 Q15
15 The diagram shows triangle OAB and points C and D.
Not to scale
\(\overrightarrow{OA} = 3\mathbf{a}\) and \(\overrightarrow{OB} = 3\mathbf{b}\).
C lies on AB such that AC = 2CB.
D is such that \(\overrightarrow{BD} = 2\mathbf{a} + \mathbf{b}\).
Show, using vectors, that OCD is a straight line. [5]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| Accept any correct justification e.g. two of OC = \(\mathbf{a} + 2\mathbf{b}\) OD = \(2\mathbf{a} + 4\mathbf{b}\) CD = \(\mathbf{a} + 2\mathbf{b}\) and correct conclusion e.g. OD = \(2(\mathbf{a} + 2\mathbf{b})\) = 2OC or OD is a multiple of OC or OC = CD (must be consistent with vectors found) | 5 | B1 for [AB =] \(3\mathbf{b} - 3\mathbf{a}\) oe M1 for each of e.g. OC = \(3\mathbf{a} + \frac{2}{3}(3\mathbf{b} - 3\mathbf{a})\) oe soi by \(\mathbf{a} + 2\mathbf{b}\) OD = \(3\mathbf{b} + 2\mathbf{a} + \mathbf{b}\) oe soi by \(2\mathbf{a} + 4\mathbf{b}\) CD = \(\frac{1}{3}(3\mathbf{b} - 3\mathbf{a}) + 2\mathbf{a} + \mathbf{b}\) oe soi by \(\mathbf{a} + 2\mathbf{b}\) to a maximum of M2 and may be on diagram and condone notation OCD for OD only M1 for [OD =] \(2(\mathbf{a} + 2\mathbf{b})\) or 2OC = OD or OC = CD and must be consistent with vectors found If 0 scored M1 for any correct route leading to OC, CD or OD e.g. OC = OB + BC | |