Higher June 2018 Paper 3 Q11
11 \(\mathbf{a} = \begin{pmatrix} 6 \\ -10 \end{pmatrix} \qquad \mathbf{b} = \begin{pmatrix} -1 \\ 2 \end{pmatrix} \qquad \mathbf{c} = \begin{pmatrix} -4 \\ 7 \end{pmatrix}\)
(a) Work out \(\quad \mathbf{a} + \mathbf{b} + \mathbf{c}\) [2 marks]
(b) Show that \(\quad \mathbf{a} + 2\mathbf{c} \quad\) is parallel to \(\mathbf{b}\) [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 |
|---|---|---|
| \(\begin{pmatrix} -2 \\ 4 \end{pmatrix}\) seen | M1 | |
| Valid reason | A1 | eg \(\begin{pmatrix} -2 \\ 4 \end{pmatrix} = 2 \times \begin{pmatrix} -1 \\ 2 \end{pmatrix}\) \(\begin{pmatrix} -2 \\ 4 \end{pmatrix} = 2\mathbf{b}\) \(\begin{pmatrix} -2 \\ 4 \end{pmatrix}\) is a multiple of \(\begin{pmatrix} -1 \\ 2 \end{pmatrix}\) \(\mathbf{a} + 2\mathbf{c}\) is a multiple of \(\mathbf{b}\) \(2\mathbf{b} = \mathbf{a} + 2\mathbf{c}\) |
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 / or divisor lines | |
| \(\begin{pmatrix} -2 \\ 4 \end{pmatrix}\) seen and both gradient \(-2\) | M1A1 |
| \(\begin{pmatrix} -2 \\ 4 \end{pmatrix}\) seen and double so parallel | M1A1 |
| \(\begin{pmatrix} -2 \\ 4 \end{pmatrix}\) seen and half so parallel | M1A1 |
| \(\begin{pmatrix} -2 \\ 4 \end{pmatrix}\) seen and \(\mathbf{a} + 2\mathbf{c}\) is \(2\mathbf{b}\) | M1A1 |
| \(\begin{pmatrix} -2 \\ 4 \end{pmatrix}\) seen and \(\mathbf{b} = \tfrac{1}{2}\mathbf{a} + 2\mathbf{c}\) | M1A0 |
| \(\begin{pmatrix} -2 \\ 4 \end{pmatrix}\) seen and both have same ratio | M1A0 |
| \(\dfrac{-2}{4}\) and \(\dfrac{-1}{2}\) both equal \(-0.5\) | M1A0 |