Answer: \(\begin{pmatrix}10\\-24\end{pmatrix}\), \(\begin{pmatrix}3+10t\\6-24t\end{pmatrix}\), \(d^2=5t^2+40t+80\), and the particles do not collide.

(a) The direction vector is

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

Its magnitude is

\(\sqrt{5^2+(-12)^2}=\sqrt{169}=13\).

So a unit vector in this direction is

\(\dfrac1{13}\begin{pmatrix}5\\-12\end{pmatrix}\).

The speed is \(26\), so the velocity vector of \(P\) is

\(26\cdot\dfrac1{13}\begin{pmatrix}5\\-12\end{pmatrix}=2\begin{pmatrix}5\\-12\end{pmatrix}=\begin{pmatrix}10\\-24\end{pmatrix}\).

(b) At \(t=0\), \(P\) has position vector \(\begin{pmatrix}3\\6\end{pmatrix}\). Therefore at time \(t\),

\(\mathbf{r}_P=\begin{pmatrix}3\\6\end{pmatrix}+t\begin{pmatrix}10\\-24\end{pmatrix}=\begin{pmatrix}3+10t\\6-24t\end{pmatrix}\).

(c) The position vector of \(Q\) is

\(\mathbf{r}_Q=\begin{pmatrix}8t-5\\2-25t\end{pmatrix}\).

Hence the displacement from \(Q\) to \(P\) is

\(\mathbf{r}_P-\mathbf{r}_Q=\begin{pmatrix}3+10t-(8t-5)\\6-24t-(2-25t)\end{pmatrix}=\begin{pmatrix}2t+8\\t+4\end{pmatrix}\).

So

\(d^2=(2t+8)^2+(t+4)^2\).

Expanding,

\(d^2=4t^2+32t+64+t^2+8t+16=5t^2+40t+80\).

Therefore \(m=5\), \(n=40\), \(r=80\).

(d) For a collision, \(d=0\), so \(d^2=0\). But

\(d^2=5t^2+40t+80=5(t^2+8t+16)=5(t+4)^2\).

This is zero only when \(t=-4\), which is not possible for the time after \(t=0\). Hence \(P\) and \(Q\) do not collide.