(a) The position vectors of points L and M are (5i + 6j) and (13i + 4j) respectively. If point K lies on LM such that LK : KM is 2 : 3, find the position vector of K.
(b) Three poles are situated at points A, B and C on the same horizontal plane such that \(AB = (8km, 060°)\) and \(BC = (12km, 130°)\). Calculate,
(i) |AC|, correct to three significant figures ; (ii) the bearing of C from A, correct to the nearest degree.
(a) The position vector of L, \(\overrightarrow{OL} = 5i + 6j\)
The coordinates of L = (5, 6)
The position vector of M, \(\overrightarrow{OM} = 13i + 4j\)
The coordinates of M = (13, 4).
Let \(\overrightarrow{OK} = xi + yj\) be the position vector of K.
\(\overrightarrow{LK} = (x - 5)i + (y - 6)j\), \(\overrightarrow{KM} = (13 - x)i + (4 - y)j\)
\(LK : KM = 2 : 3 ; \frac{LK}{KM} = \frac{2}{3}\)
\(3[(x - 5)i + (y - 6)j] = 2[(13 - x)i + (4 - y)j]\)
Equating by components,
\(i : 3x - 15 = 26 - 2x \implies 5x = 41\)
\(x = \frac{41}{5}\)
\(j : 3y - 18 = 8 - 2y \implies 5y = 26\)
\(y = \frac{26}{5}\)
The position vector of K is \(\overrightarrow{OK} = \frac{41}{5} i + \frac{26}{5} j\)
(b) \(\overrightarrow{AC} = \overrightarrow{AB} + \overrightarrow{BC}\)
= \(\begin{pmatrix} 8 \cos 60° \\ 8 \sin 60° \end{pmatrix} + \begin{pmatrix} 12 \cos 130° \\ 12 \sin 130° \end{pmatrix}\)
= \(\begin{pmatrix} 8 \times \frac{1}{2} \\ 8 \times \frac{\sqrt{3}}{2} \end{pmatrix} + \begin{pmatric} -12 \times 0.6428 \\ 12 \times 0.7660 \end{pmatrix}\)
= \(\begin{pmatrix} 4 \\ 4\sqrt{3} \end{pmatrix} + \begin{pmatrix} -7.7136 \\ 9.9192 \end{pmatrix} = \begin{pmatrix} -3.7136 \\ 16.12 \end{pmatrix}\)
\(|AC| = \sqrt{(-3.7136)^{2} + (16.12)^{2}} = \sqrt{13.794 + 259.9}\)
= \(\sqrt{273.694} = 16.544 \approxeq 16.5 km\)
\(AB = (8 \cos 60 i + 8 \sin 60 j) ; AC = (-3.7136 i + 16.12 j)\)
(ii) 
Let the angle between \(\overrightarrow{AB}\) and \(\overrightarrow{AC}\) be \(\theta\).
Using sine rule,
\(\frac{\sin B}{AC} = \frac{\sin A}{BC}\)
\(\frac{\sin 110°}{16.5} = \frac{\sin \theta}{12}\)
\(\sin \theta = \frac{12 \sin 110°}{16.5}\)
\(\sin \theta = 0.6834\)
\(\theta = \sin^{-1} (0.6834) = 43.11°\)
Bearing of C from A = 60° + 43.11° = 103.11° \(\approxeq\) 103° (nearest degree)
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