(a) \(m \begin{pmatrix} 2 \\ 1 \end{pmatrix} + n \begin{pmatrix} -1 \\ 2 \end{pmatrix} = \begin{pmatrix} 5 \\ -4 \end{pmatrix}\)
\(2m - n = 5 ...(1)\)
\(m + 2n = -4 ... (2)\)
\((1) \times 2 : 4m - 2n = 10 ... (3)\)
\((2) + (3) : 5m = 6 \implies m = \frac{6}{5}\)
\(2(\frac{6}{5}) - n = 5 \implies n = \frac{12}{5} - 5 = \frac{-13}{5}\)
\(\therefore (m + n) = (\frac{6}{5} + \frac{-13}{5}) = -\frac{7}{5}\).
(b) A(-1, 3) = (-i + 3j)
B(2, -1) = (2i - j)
\(\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA}\)
\(\overrightarrow{AB} = \begin{pmatrix} 2 \\ -1 \end{pmatrix} - \begin{pmatrix} -1 \\ 3 \end{pmatrix}\)
= \(\overrightarrow{AB} = \begin{pmatrix} 3 \\ -4 \end{pmatrix}\)
\(|\overrightarrow{AB}| = \sqrt{3^{2} + (-4)^{2}} = 5\)
Unit vector parallel to AB = \(\frac{1}{5} \begin{pmatrix} 3 \\ -4 \end{pmatrix}\)
\(\overrightarrow{OA} + \overrightarrow{AC} = \overrightarrow{OC}\)
\(\overrightarrow{AC} = \overrightarrow{OC} - \overrightarrow{OA}\)
= \(\begin{pmatrix} 5 \\3 \end{pmatrix} - \begin{pmatrix} -1 \\ 3 \end{pmatrix}\)
= \(\begin{pmatrix} 6 \\ 0 \end{pmatrix}\)
\(|\overrightarrow{AC}| = 6\)
Unit vector parallel to AC = \(\frac{1}{6} \begin{pmatrix} 6 \\ 0 \end{pmatrix}\)
(ii) Let the angle BAC = \(\theta\)
\(\overrightarrow{AB} \cdot \overrightarrow{AC} = |\overrightarrow{AB}||\overrightarrow{AC}| \cos \theta\)
\((3i - 4j) \cdot (6i + 0j) = |3i + 4j||6i + 0j| \cos \theta\)
\(18 = 5 \times 6 \times \cos \theta \implies \cos \theta = \frac{18}{30} = 0.6\)
\(\theta = \cos^{-1} (0.6) = 53.1°\)