(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.
Explanation
(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 \ 4sqrt{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)