Higher January 2019 Paper 2 Q23
23 \(ABCD\) is a trapezium.
\(\overrightarrow{DC} = 3\overrightarrow{AB}\)
\(\overrightarrow{DA} = \begin{pmatrix} 2 \\ -3 \end{pmatrix}\) \(\overrightarrow{DB} = \begin{pmatrix} -1 \\ 7 \end{pmatrix}\)
Find the exact magnitude of \(\overrightarrow{BC}\)
(5)
| Scheme | Marks |
|---|---|
| e.g. \(\overrightarrow{AB} = \overrightarrow{AD} + \overrightarrow{DB}\) or \(\begin{pmatrix} 2 \\ -3 \end{pmatrix} + \begin{pmatrix} -1 \\ 7 \end{pmatrix}\) | M1 |
| \(\overrightarrow{AB} = \begin{pmatrix} 1 \\ 4 \end{pmatrix}\) | A1 |
| \(\overrightarrow{DC} = 3 \times \begin{pmatrix} 1 \\ 4 \end{pmatrix} \left(= \begin{pmatrix} 3 \\ 12 \end{pmatrix}\right)\) | M1 |
\(\overrightarrow{BC} = \begin{pmatrix} 1 \\ -7 \end{pmatrix} + \begin{pmatrix} 3 \\ 12 \end{pmatrix} \left(= \begin{pmatrix} 4 \\ 5 \end{pmatrix}\right)\) oe or \(\overrightarrow{BC} = \begin{pmatrix} -1 \\ -4 \end{pmatrix} + \begin{pmatrix} 2 \\ -3 \end{pmatrix} + \begin{pmatrix} 3 \\ 12 \end{pmatrix} \left(= \begin{pmatrix} 4 \\ 5 \end{pmatrix}\right)\) oe | M1 |
| \(\sqrt{41}\) cao | A1 |
| (5) | |
| (5 marks) |
Notes
M1: for a correct vector equation for \(\overrightarrow{AB}\)
A1: No isw