Foundation June 2018 Paper 3 Q30
30 \(\mathbf{a} = \begin{pmatrix} 6 \\ -10 \end{pmatrix} \qquad \mathbf{b} = \begin{pmatrix} -1 \\ 2 \end{pmatrix} \qquad \mathbf{c} = \begin{pmatrix} -4 \\ 7 \end{pmatrix}\)
(a) Work out \(\qquad \mathbf{a} + \mathbf{b} + \mathbf{c}\) [2 marks]
(b) Show that \(\qquad \mathbf{a} + 2\mathbf{c} = k\mathbf{b}\), where \(k\) is an integer. [2 marks]
| Answer | Mark | Comments |
|---|---|---|
| \(\begin{pmatrix} 1 \\ -1 \end{pmatrix}\) | B2 | B1 for 1 correct value in correct position Condone a divisor line |
| Answer | Mark | Comments |
|---|---|---|
| \(\begin{pmatrix} 6 \\ -10 \end{pmatrix} + \begin{pmatrix} 2 \times -4 \\ 2 \times 7 \end{pmatrix}\) or \(\begin{pmatrix} 6 \\ -10 \end{pmatrix} + \begin{pmatrix} -8 \\ 14 \end{pmatrix}\) or \(\begin{pmatrix} -2 \\ 4 \end{pmatrix}\) | M1 | oe |
| \(\begin{pmatrix} -2 \\ 4 \end{pmatrix} = 2\begin{pmatrix} -1 \\ 2 \end{pmatrix}\) or \(\begin{pmatrix} -2 \\ 4 \end{pmatrix}\) and k = 2 or \(2\mathbf{b} = \begin{pmatrix} -2 \\ 4 \end{pmatrix}\) | A1 | oe |
Additional guidance
| Condone vectors written as coordinates, eg (–1, 2) is half of (–2, 4) | |
| Must see \(\begin{pmatrix} -2 \\ 4 \end{pmatrix}\) or (–2, 4) to award the A mark | |
| Condone missing brackets and divisor lines | |
| \(\begin{pmatrix} -2 \\ 4 \end{pmatrix}\) seen and \(\mathbf{a} + 2\mathbf{c}\) is \(2\mathbf{b}\) | M1A1 |
| \(\begin{pmatrix} -2 \\ 4 \end{pmatrix} \div 2 = \begin{pmatrix} -1 \\ 2 \end{pmatrix}\) | M1A1 |
| \(\begin{pmatrix} 6 \\ -10 \end{pmatrix} + 2\begin{pmatrix} -4 \\ 7 \end{pmatrix}\) | M0 |