Higher June 2022 Paper 1 Q21
21 Five points are connected by vectors.
Not drawn accurately
\(\overrightarrow{FG}\) \(= 2\)\(\overrightarrow{EH}\)
Work out \(\overrightarrow{FE}\) in terms of \(\mathbf{a}\) and \(\mathbf{b}\). [4 marks]
| Answer | Mark | Comments |
|---|---|---|
| \(3\mathbf{a} + \mathbf{b} + \mathbf{a} + 6\mathbf{b}\) or \(4\mathbf{a} + 7\mathbf{b}\) | M1 | \(\overrightarrow{EH}\) may be seen on diagram or as part of a longer vector sum |
| \(2 \times\) their \((4\mathbf{a} + 7\mathbf{b})\) or \(8\mathbf{a} + 14\mathbf{b}\) | M1dep | \(\overrightarrow{FG}\) may be seen on diagram |
| Any correct path from \(F\) to \(E\) eg their \((8\mathbf{a} + 14\mathbf{b}) + (-2\mathbf{a} - 15\mathbf{b})\) or their \((8\mathbf{a} + 14\mathbf{b}) - (2\mathbf{a} + 15\mathbf{b})\) or \((-2\mathbf{a} - 15\mathbf{b}) + (3\mathbf{a} + \mathbf{b}) + (\mathbf{a} + 6\mathbf{b})\) or \(2\mathbf{a} - 8\mathbf{b}\) and their \((8\mathbf{a} + 14\mathbf{b})\) + their \((2\mathbf{a} - 8\mathbf{b}) + (-\mathbf{a} - 6\mathbf{b}) + (-3\mathbf{a} - \mathbf{b})\) or their \((8\mathbf{a} + 14\mathbf{b})\) + their \((2\mathbf{a} - 8\mathbf{b})\) + their \((-4\mathbf{a} - 7\mathbf{b})\) | M1dep | \(\overrightarrow{FG}\) \(+\) \(\overrightarrow{GE}\) \(\overrightarrow{FG}\) \(-\) \(\overrightarrow{EG}\) oe \(\overrightarrow{GE}\) \(+\) \(\overrightarrow{ED}\) \(+\) \(\overrightarrow{DH}\) oe \(\overrightarrow{GH}\) oe \(\overrightarrow{FG}\) \(+\) \(\overrightarrow{GH}\) \(+\) \(\overrightarrow{HD}\) \(+\) \(\overrightarrow{DE}\) oe \(\overrightarrow{FG}\) \(+\) \(\overrightarrow{GH}\) \(+\) \(\overrightarrow{HE}\) |
| \(6\mathbf{a} - \mathbf{b}\) | A1 | SC3 \(-6\mathbf{a} + \mathbf{b}\) or \(\mathbf{b} - 6\mathbf{a}\) |
Additional guidance
| Missing brackets and incorrect addition or subtraction signs can be recovered for all four marks | |
| eg \(8\mathbf{a} + 14\mathbf{b} - 2\mathbf{a} + 15\mathbf{b}\) recovered to \(6\mathbf{a} - \mathbf{b}\) | M1M1M1A1 |
| Condone missing brackets for the third mark | |
| eg \(8\mathbf{a} + 14\mathbf{b} - 2\mathbf{a} + 15\mathbf{b}\) and answer \(6\mathbf{a} + 29\mathbf{b}\) | M1M1M1A0 |
| To receive marks expressions must be in terms of \(\mathbf{a}\) and \(\mathbf{b}\) | |
| Allow subtractions shown in vertical form eg \(\begin{array}{r} 8\mathbf{a} + 14\mathbf{b} \\ -\ \ 2\mathbf{a} + 15\mathbf{b} \end{array}\) | M1M1M1 |