Answer: \(\dfrac1{13}\begin{pmatrix}5\\-12\end{pmatrix}\); \(k=-\dfrac32\), \(r=-\dfrac7{10}\); \(\overrightarrow{AB}=3\mathbf q-2\mathbf p\), \(\overrightarrow{AC}=9\mathbf q-6\mathbf p\), \(AB:BC=1:2\).

(a) The magnitude of

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

is

\(\sqrt{5^2+(-12)^2}=\sqrt{25+144}=13.\)

So the unit vector is

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

(b) Equate components:

\(4-2k=-10r\)

and

\(1+3k=5r.\)

From the first equation,

\(r=\frac{k-2}{5}.\)

From the second equation,

\(r=\frac{1+3k}{5}.\)

Therefore

\(k-2=1+3k.\)

So

\(-3=2k,\)

and hence

\(k=-\frac32.\)

Substitute this into \(r=\frac{1+3k}{5}\):

\(r=\frac{1+3\left(-\frac32\right)}5 =\frac{1-\frac92}{5} =-\frac7{10}.\)

(c)(i) Since

\(\overrightarrow{OA}=\mathbf p,\qquad \overrightarrow{OB}=3\mathbf q-\mathbf p,\)

we have

\(\overrightarrow{AB} =\overrightarrow{OB}-\overrightarrow{OA} =3\mathbf q-\mathbf p-\mathbf p =3\mathbf q-2\mathbf p.\)

(c)(ii) Also

\(\overrightarrow{OC}=9\mathbf q-5\mathbf p.\)

Therefore

\(\overrightarrow{AC} =\overrightarrow{OC}-\overrightarrow{OA} =9\mathbf q-5\mathbf p-\mathbf p =9\mathbf q-6\mathbf p.\)

(c)(iii) Now

\(\overrightarrow{AC}=3(3\mathbf q-2\mathbf p)=3\overrightarrow{AB}.\)

So \(\overrightarrow{AB}\) and \(\overrightarrow{AC}\) have the same direction and share the point \(A\). Therefore \(A\), \(B\) and \(C\) lie in a straight line.

(c)(iv) Since

\(\overrightarrow{AC}=3\overrightarrow{AB},\)

we have

\(BC=AC-AB=2AB.\)

Therefore

\(AB:BC=1:2.\)