(a) Given that \(p = (4i – 3j)\) and \(q = (-i + 5j)\), find r such that \(|r| = 15\) and is in the direction \((2p + 3q)\).
(b) 
Forces of magnitude 8N, 6N and 4N act at the point P, as shown in the above diagram. Find the : (i) magnitude ; (ii) direction of the resultant force.
Explanation
(a) \(p = (4i - 3j) ; q = (-i + 5j)\)
\(2p + 3q = \begin{pmatrix} 8i - 3i \\ -6j + 15j \end{pmatrix}\)
= \(\begin{pmatrix} 5i \\ 9j \end{pmatrix}\)
Let \(r = k \begin{pmatrix} 5i \\ 9j \end{pmatrix} = \begin{pmatrix} 5ki \\ 9kj \end{pmatrix}\)
\(|r| = \sqrt{(5k)^{2} + (9k)^{2}} = 15\)
\(\sqrt{25k^{2} + 81k^{2}} = 15 \implies 25k^{2} + 81k^{2} = 225\)
\(106k^{2} = 225 \implies k^{2} = \frac{225}{106} = 2.123\)
\(k = \sqrt{2.123} = 1.457\)
\(\therefore r = \begin{pmatrix} 5i \times 1.457 \\ 9j \times 1.457 \end{pmatrix}\)
= \(\begin{pmatrix} 7.285i \\ 13.113j \end{pmatrix}\)
(b)(i) 
Resolve the forces along the x and y- axis.
Let \(F_{1} = \begin{pmatrix} 8 \cos 120° \\ 8 \sin 120° \end{pmatrix}\)
\(F_{2} = \begin{pmatrix} 6 \cos 90° \\ 6 \sin 90° \end{pmatrix}\)
\(F_{3} = \begin{pmatrix} 4 \cos 30° \\ 4 \sin 30° \end{pmatrix}\)
\(R = F_{1} + F_{2} + F_{3}\)
= \(\begin{pmatrix} 8 \cos 120° \\ 8 \sin 120° \end{pmatrix} + \begin{pmatrix} 6 \cos 90° \\ 6 \sin 90° \end{pmatrix} + \begin{pmatrix} 4 \cos 30° \\ 4 \sin 30° \end{pmatrix}\)
= \(\begin{pmatrix} -8 \cos 60° \\ 8 \sin 60 \end{pmatrix} + \begin{pmatrix} 6 \cos 90° \\ 6 \sin 90° \end{pmatrix} + \begin{pmatrix} 4 \cos 30° \\ 4 \sin 30° \end{pmatrix}\)
= \(\begin{pmatrix} -8 \times 0.5 \\ 8 \times 0.866 \end{pmatrix} + \begin{pmatrix} 6 \times 0 \\ 6 \times 1 \end{pmatrix} + \begin{pmatrix} 4 \times 0.866 \\ 4 \times 0.5 \end{pmatrix}\)
= \(\begin{pmatrix} -4 + 0 + 3.464 \\ 6.928 + 6 + 2 \end{pmatrix}\)
= \(\begin{pmatrix} -0.536 \\ 14.928 \end{pmatrix}\)
\(|R| = \sqrt{(-0.536)^{2} + (14.928)^{2}} = \sqrt{0.2873 + 222.8452}\)
= \(\sqrt{223.1325} = 14.94N\)
(ii)Direction \(\theta\)
\(\tan \theta = \frac{14.928}{-0.536} = -27.851\)
\(\theta = -87.9° = 92.1°\)
The direction is N 02.1° W or a bearing of 267.9°.