Answer: (b) \(32\mathbf{i}+(16-9\sqrt3)\mathbf{j}\). (c) \(29\mathbf{i}+(16-12\sqrt3t)\mathbf{j}\). (d) \(6\) km.

Answer: (b) \(32\mathbf{i}+(16-9\sqrt3)\mathbf{j}\). (c) \(29\mathbf{i}+(16-12\sqrt3t)\mathbf{j}\). (d) \(6\) km.

(a) A bearing of \(120^{\circ}\) means the direction is \(120^{\circ}\) clockwise from north. The east component is

\(6\sqrt3\sin120^{\circ}=6\sqrt3\cdot\frac{\sqrt3}{2}=9\).

The north component is

\(6\sqrt3\cos120^{\circ}=6\sqrt3\cdot\left(-\frac12\right)=-3\sqrt3\).

So the velocity vector of ship A is

\(9\mathbf{i}-3\sqrt3\mathbf{j}\).

(b) From 09:00 to 12:00 is 3 hours, so the displacement of ship A is

\(3(9\mathbf{i}-3\sqrt3\mathbf{j})=27\mathbf{i}-9\sqrt3\mathbf{j}\).

Its position vector at 12:00 is therefore

\((5\mathbf{i}+16\mathbf{j})+(27\mathbf{i}-9\sqrt3\mathbf{j})\),

which is

\(32\mathbf{i}+(16-9\sqrt3)\mathbf{j}\).

(c) At \(t\) hours after 11:00, ship B has travelled for \(t\) hours from \(Q\). Its position vector is

\(29\mathbf{i}+16\mathbf{j}+t(-12\sqrt3\mathbf{j})\),

so it is

\(29\mathbf{i}+(16-12\sqrt3t)\mathbf{j}\).

(d) At 12:00, ship B has travelled for \(t=1\) hour, so its position vector is

\(29\mathbf{i}+(16-12\sqrt3)\mathbf{j}\).

The displacement from B to A at 12:00 is

\([32-29]\mathbf{i}+[(16-9\sqrt3)-(16-12\sqrt3)]\mathbf{j}\),

which is

\(3\mathbf{i}+3\sqrt3\mathbf{j}\).

Therefore the distance between the ships is

\(\sqrt{3^2+(3\sqrt3)^2}=\sqrt{9+27}=6\).