Answer: \(\begin{pmatrix}-20\\48\end{pmatrix}\); \(\overrightarrow{PQ}=\begin{pmatrix}12\\8\end{pmatrix}+\begin{pmatrix}-5\\-3\end{pmatrix}t\); first time \(t=2.15\) h.

(a) The direction vector is

\(\begin{pmatrix}-5\\12\end{pmatrix}.\)

Its magnitude is

\(\sqrt{(-5)^2+12^2}=13.\)

So the unit vector in this direction is

\(\frac1{13}\begin{pmatrix}-5\\12\end{pmatrix}.\)

The speed is \(52\text{ km h}^{-1}\), so the velocity vector is

\(52\cdot\frac1{13}\begin{pmatrix}-5\\12\end{pmatrix} =4\begin{pmatrix}-5\\12\end{pmatrix} =\begin{pmatrix}-20\\48\end{pmatrix}.\)

(b) Ship \(P\) starts at

\(\begin{pmatrix}0\\0\end{pmatrix}\)

and has velocity

\(\begin{pmatrix}-20\\48\end{pmatrix}.\)

So its position vector after \(t\) hours is

\(\begin{pmatrix}-20\\48\end{pmatrix}t.\)

(c) Ship \(Q\) starts at

\(\begin{pmatrix}12\\8\end{pmatrix}\)

and has velocity

\(\begin{pmatrix}-25\\45\end{pmatrix}.\)

So its position vector after \(t\) hours is

\(\begin{pmatrix}12\\8\end{pmatrix} +\begin{pmatrix}-25\\45\end{pmatrix}t.\)

(d) The displacement vector \(\overrightarrow{PQ}\) is position of \(Q\) minus position of \(P\):

\(\overrightarrow{PQ} =\begin{pmatrix}12\\8\end{pmatrix} +\begin{pmatrix}-25\\45\end{pmatrix}t -\begin{pmatrix}-20\\48\end{pmatrix}t.\)

So

\(\overrightarrow{PQ} =\begin{pmatrix}12\\8\end{pmatrix} +\begin{pmatrix}-5\\-3\end{pmatrix}t.\)

(e) Therefore

\(\overrightarrow{PQ} =\begin{pmatrix}12-5t\\8-3t\end{pmatrix}.\)

Hence

\(PQ=\sqrt{(12-5t)^2+(8-3t)^2}.\)

Expanding,

\(PQ=\sqrt{144-120t+25t^2+64-48t+9t^2}.\)

So

\(PQ=\sqrt{34t^2-168t+208}.\)

(f) When the ships are \(2\) km apart,

\(\sqrt{34t^2-168t+208}=2.\)

Square both sides:

\(34t^2-168t+208=4.\)

So

\(34t^2-168t+204=0.\)

Solving this quadratic gives

\(t=\frac{42\pm\sqrt{30}}{17}.\)

The first time is the smaller value:

\(t=\frac{42-\sqrt{30}}{17}\approx2.15.\)

So the ships are first \(2\) km apart after

\(2.15\text{ h}.\)