Higher June 2018 Paper 6 Q10
10 Two vectors, \(\mathbf{a}\) and \(\mathbf{b}\), are shown on the 1 centimetre grid below.
Show that the vector \(\mathbf{a} + 2\mathbf{b}\) has length 7 cm.
You may use the grid below.
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| Correct triangle drawn with \(\mathbf{a} + 2\mathbf{b}\) labelled and with correct arrows or \(\mathbf{a}\) and \(2\mathbf{b}\) labelled and with correct arrows AND length 7cm indicated on diagram  | 3 | M1 for vector \(2\mathbf{b}\) drawn on grid M1 \(\mathbf{a} + k\mathbf{b}\) drawn on grid The two vectors must be joined end to end but arrows may be contradictory. \(k\mathbf{b}\) should be in the direction of \(\mathbf{b}\) | If both methods shown/started, mark the better one For M marks condone missing or incorrect arrows and labels on vectors Mark intent: end of vectors within 2mm of of vertices of relevant square Examples (ignore arrows): M1M1 for \(\mathbf{a} + 2\mathbf{b}\) drawn (3 marks if labelled and 7 cm indicated) M1M1 for \(\mathbf{a} - 2\mathbf{b}\) M1M0 for \(2\mathbf{b}\) or \(-2\mathbf{b}\) M0M1 for \(\mathbf{a} + \mathbf{b}\), \(\mathbf{a} - 1.5\mathbf{b}\) etc |
| OR | |||
| \(\begin{pmatrix} 4 \\ 1 \end{pmatrix} + 2\begin{pmatrix} -2 \\ 3 \end{pmatrix} = \begin{pmatrix} 0 \\ 7 \end{pmatrix}\) with brackets | B1 for \(\begin{pmatrix} 4 \\ 1 \end{pmatrix}\) B1 for \(\begin{pmatrix} -2 \\ 3 \end{pmatrix}\) or \(\begin{pmatrix} -4 \\ 6 \end{pmatrix}\) | For B1 marks, condone missing brackets and fraction lines | |