Answer: \(\begin{pmatrix}30\\10\end{pmatrix}+t\begin{pmatrix}-8\\6\end{pmatrix}\); speed \(13\text{ m s}^{-1}\); distance \(100\sqrt2\).

The direction vector for \(P\) is \(\begin{pmatrix}-4\\3\end{pmatrix}\), whose magnitude is

\(\sqrt{(-4)^2+3^2}=5.\)

Since the speed is \(10\), the velocity vector of \(P\) is

\(10\cdot\frac1{5}\begin{pmatrix}-4\\3\end{pmatrix} =\begin{pmatrix}-8\\6\end{pmatrix}.\)

Therefore the position vector of \(P\) after \(t\) seconds is

\(\begin{pmatrix}30\\10\end{pmatrix} +t\begin{pmatrix}-8\\6\end{pmatrix}.\)

The velocity vector of \(Q\) is \(\begin{pmatrix}5\\12\end{pmatrix}\), so its speed is

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

When \(t=10\),

\(P=\begin{pmatrix}30\\10\end{pmatrix} +10\begin{pmatrix}-8\\6\end{pmatrix} =\begin{pmatrix}-50\\70\end{pmatrix},\)

and

\(Q=\begin{pmatrix}-80\\90\end{pmatrix} +10\begin{pmatrix}5\\12\end{pmatrix} =\begin{pmatrix}-30\\210\end{pmatrix}.\)

The difference in position is

\(\begin{pmatrix}-30\\210\end{pmatrix} -\begin{pmatrix}-50\\70\end{pmatrix} =\begin{pmatrix}20\\140\end{pmatrix}.\)

So the distance is

\(\sqrt{20^2+140^2} =\sqrt{20000} =100\sqrt2.\)