Higher June 2024 Paper 2 Q23
23
Not drawn accurately
Prove that \(DEF\) is a straight line. [4 marks]
| Answer | Mark | Comments |
|---|---|---|
| One correct expression eg \((\overrightarrow{DE} =)\ 6\mathbf{a} + \mathbf{b} + 2\mathbf{a} - 5\mathbf{b}\) or \((\overrightarrow{DF} =)\ 6\mathbf{a} + \mathbf{b} + 4\mathbf{a} - 6\mathbf{b}\) or \((\overrightarrow{EF} =)\ -2\mathbf{a} + 5\mathbf{b} + 4\mathbf{a} - 6\mathbf{b}\) | M1 | oe eg \((\overrightarrow{ED} =)\ -6\mathbf{a} - \mathbf{b} - 2\mathbf{a} + 5\mathbf{b}\) or \((\overrightarrow{FD} =)\ -6\mathbf{a} - \mathbf{b} - 4\mathbf{a} + 6\mathbf{b}\) or \((\overrightarrow{FE} =)\ 2\mathbf{a} - 5\mathbf{b} - 4\mathbf{a} + 6\mathbf{b}\) accept unprocessed brackets eg \((\overrightarrow{EF} =)\ -(2\mathbf{a} - 5\mathbf{b}) + 4\mathbf{a} - 6\mathbf{b}\) |
| Two correct expressions from \(\overrightarrow{DE}\) \(\overrightarrow{DF}\) \(\overrightarrow{EF}\) | M1dep | oe eg \(\overrightarrow{DE}\) and \(\overrightarrow{FD}\) accept unprocessed brackets |
| Two fully simplified expressions from \((\overrightarrow{DE} =)\ 8\mathbf{a} - 4\mathbf{b} \quad (\overrightarrow{DF} =)\ 10\mathbf{a} - 5\mathbf{b}\) \((\overrightarrow{EF} =)\ 2\mathbf{a} - \mathbf{b}\) | A1 | oe eg \((\overrightarrow{DE} =)\ 8\mathbf{a} - 4\mathbf{b}\) and \((\overrightarrow{FD} =)\ -10\mathbf{a} + 5\mathbf{b}\) |
| Two fully simplified expressions from \((\overrightarrow{DE} =)\ 8\mathbf{a} - 4\mathbf{b}\) \((\overrightarrow{DF} =)\ 10\mathbf{a} - 5\mathbf{b}\) \((\overrightarrow{EF} =)\ 2\mathbf{a} - \mathbf{b}\) and valid indication that the vectors are parallel | A1 | eg \((\overrightarrow{DE} =)\ 8\mathbf{a} - 4\mathbf{b}\) and \((\overrightarrow{FE} =)\ -2\mathbf{a} + \mathbf{b}\) and \(8\mathbf{a} - 4\mathbf{b} = -4(-2\mathbf{a} + \mathbf{b})\) or \((\overrightarrow{DF} =)\ 10\mathbf{a} - 5\mathbf{b}\) and \((\overrightarrow{EF} =)\ 2\mathbf{a} - \mathbf{b}\) and \(\overrightarrow{DF} = 5\overrightarrow{EF}\) |
Additional guidance
| Condone absence of vector notation | |
| Condone eg \(\overrightarrow{DCE}\) or \(D\) to \(E\) for \(\overrightarrow{DE}\) | |
| If the only two correct expressions are eg \(\overrightarrow{DE}\) and \(\overrightarrow{ED}\) the maximum possible mark is M1 | |
| Only combining the three given vectors | Zero |
| \(\overrightarrow{DF} = \overrightarrow{DE} + \overrightarrow{EF}\) is not a valid indication | |
| Stating eg \(\overrightarrow{DF}\) is a (scalar) multiple of \(\overrightarrow{EF}\) is not enough for the final A1 |