4MA0 Higher June 2015 Paper 4 Q17
17 \(ABCD\) is a parallelogram.
Diagram NOT accurately drawn
\(\overrightarrow{BC} = \begin{pmatrix} 5 \\ -1 \end{pmatrix}\) \(\overrightarrow{DC} = \begin{pmatrix} -2 \\ 3 \end{pmatrix}\)
Find \(\overrightarrow{BD}\) as a column vector.
(2)
| Scheme | Marks |
|---|---|
| \(\begin{pmatrix} 5 \\ -1 \end{pmatrix} + \begin{pmatrix} 2 \\ -3 \end{pmatrix}\) or \(\begin{pmatrix} 5 \\ -1 \end{pmatrix} - \begin{pmatrix} -2 \\ 3 \end{pmatrix}\) | M1 |
| \(\begin{pmatrix} 7 \\ -4 \end{pmatrix}\) | A1 |
| (2) |
Notes
M1: Or for \(\begin{pmatrix} 7 \\ a \end{pmatrix}\) or \(\begin{pmatrix} b \\ -4 \end{pmatrix}\)
A1: Must be written as a vector.