Higher January 2021 Paper 1R Q13
13 Here are two vectors.
\(\overrightarrow{AB} = \begin{pmatrix} 5 \\ 3 \end{pmatrix} \qquad \overrightarrow{CB} = \begin{pmatrix} -2 \\ 4 \end{pmatrix}\)
Find, as a column vector, \(\overrightarrow{AC}\)
(2)
| Scheme | Marks |
|---|---|
| e.g. \(\begin{pmatrix} 5 \\ 3 \end{pmatrix} - \begin{pmatrix} -2 \\ 4 \end{pmatrix}\) or \(\begin{pmatrix} 5 \\ 3 \end{pmatrix} + \begin{pmatrix} 2 \\ -4 \end{pmatrix}\) | M1 |
| Correct answer scores full marks (unless from obvious incorrect working) Answer: \(\begin{pmatrix} 7 \\ -1 \end{pmatrix}\) | A1 |
| (2) | |
| (2 marks) |
Notes
M1: or for \(\begin{pmatrix} 7 \\ a \end{pmatrix}\) where \(a \neq -1\) or \(\begin{pmatrix} b \\ -1 \end{pmatrix}\) where \(b \neq 7\)