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.

Answer: \(\mathbf{v}_A=\binom{\frac{5\sqrt3}{2}}{\frac{15}{2}}\), \(\mathbf{v}_B=\binom{5\sqrt3}{5}\), and the boats collide when \(t=200\) seconds after boat \(A\) starts.

(a) Bearings are measured clockwise from north. Since the \(x\)-direction is east and the \(y\)-direction is north, a speed \(V\) on a bearing \(\alpha\) has velocity vector

\(\binom{V\sin\alpha}{V\cos\alpha}.\)

For boat \(A\), \(V=5\sqrt3\) and \(\alpha=30^\circ\), so

\(\mathbf v_A=\binom{5\sqrt3\sin30^\circ}{5\sqrt3\cos30^\circ} =\binom{\frac{5\sqrt3}{2}}{\frac{15}{2}}.\)

For boat \(B\), \(V=10\) and \(\alpha=60^\circ\), so

\(\mathbf v_B=\binom{10\sin60^\circ}{10\cos60^\circ} =\binom{5\sqrt3}{5}.\)

(b) Let \(t\) be the time in seconds after boat \(A\) starts.

The position vector of boat \(A\) is

\(\mathbf r_A=t\binom{\frac{5\sqrt3}{2}}{\frac{15}{2}}.\)

Boat \(B\) starts after \(100\) seconds from \(\binom0{1000}\). Therefore, for \(t\geqslant100\), its position vector is

\(\mathbf r_B=\binom0{1000}+(t-100)\binom{5\sqrt3}{5}.\)

For a collision, the \(x\)-coordinates and \(y\)-coordinates must be equal at the same time.

Equating the \(x\)-coordinates:

\(\dfrac{5\sqrt3}{2}t=5\sqrt3(t-100).\)

Divide by \(5\sqrt3\):

\(\dfrac t2=t-100.\)

So

\(t=200.\)

Now equate the \(y\)-coordinates:

\(\dfrac{15}{2}t=1000+5(t-100).\)

Simplifying the right-hand side,

\(\dfrac{15}{2}t=5t+500.\)

Multiplying by \(2\),

\(15t=10t+1000,\)

so

\(t=200.\)

Both coordinates give the same time, and \(200\geqslant100\), so boat \(B\) has already started. Therefore the two boats collide \(200\) seconds after boat \(A\) starts.