Answer: (a)(i) \(\begin{pmatrix}2\\5\end{pmatrix}\); (a)(ii) \(13\); (a)(iii) \(P\) does not pass through the point; (b) \(\mathbf c=4\mathbf b-3\mathbf a\).

Work through the problem step by step, using exact values where possible before giving any final numerical approximation.

(a)(i) The initial position vector is found by putting \(t=0\):

(a)(ii) The velocity vector is the coefficient of \(t\) in the position vector:

So the speed is its magnitude:

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

(a)(iii) For the \(x\)-coordinate to be \(158\),

\(2+12t=158.\)

So

\(12t=156,\)

and

\(t=13.\)

For the \(y\)-coordinate to be \(-48\),

\(5-5t=-48.\)

So

\(-5t=-53,\)

and

\(t=10.6.\)

The two values of \(t\) are different, so \(P\) does not pass through the point with position vector \(\begin{pmatrix}158\\-48\end{pmatrix}\).

(b) Since

\(AB:AC=1:4,\)

we have

\(\overrightarrow{AB}=\frac14\overrightarrow{AC}.\)

Now

\(\overrightarrow{AB}=\mathbf b-\mathbf a\)

and

\(\overrightarrow{AC}=\mathbf c-\mathbf a.\)

Therefore

\(\mathbf b-\mathbf a=\frac14(\mathbf c-\mathbf a).\)

Multiplying by 4 gives

\(4\mathbf b-4\mathbf a=\mathbf c-\mathbf a.\)

Hence

\(\mathbf c=4\mathbf b-3\mathbf a.\)