Answer: (a) \(\overrightarrow{OR}=-3\mathbf{i}+15\mathbf{j}\). (b) \(\overrightarrow{OP}=16\mathbf{i}-24\mathbf{j}\).

(a) The vector \(-\mathbf i+5\mathbf j\) has magnitude

\(\sqrt{(-1)^2+5^2}=\sqrt{26}.\)

A vector in the same direction with magnitude \(3\sqrt{26}\) must be \(3\) times this vector. Therefore

\(\overrightarrow{OR}=3(-\mathbf i+5\mathbf j)=-3\mathbf i+15\mathbf j.\)

(b) Since \(\overrightarrow{OP}\) is in the same direction as \(2\mathbf i-3\mathbf j\), write

\(\overrightarrow{OP}=k(2\mathbf i-3\mathbf j).\)

Then

\(\overrightarrow{OP}=2k\mathbf i-3k\mathbf j.\)

Using \(\overrightarrow{PQ}=\overrightarrow{OQ}-\overrightarrow{OP}\),

\(\overrightarrow{PQ}=(10\mathbf i+6\mathbf j)-(2k\mathbf i-3k\mathbf j).\)

So

\(\overrightarrow{PQ}=(10-2k)\mathbf i+(6+3k)\mathbf j.\)

This is in the same direction as \(-\mathbf i+5\mathbf j\), so the ratio of components must be the same. Write

\((10-2k,\,6+3k)=\lambda(-1,5).\)

From the first component,

\(10-2k=-\lambda,\)

so

\(\lambda=2k-10.\)

Using the second component,

\(6+3k=5\lambda=5(2k-10).\)

Therefore

\(6+3k=10k-50.\)

So

\(56=7k,\qquad k=8.\)

Hence

\(\overrightarrow{OP}=8(2\mathbf i-3\mathbf j)=16\mathbf i-24\mathbf j.\)

Answer: \(AC:AB=1:3\).

The vector from \(A\) to \(B\) is

\(\overrightarrow{AB}=\binom{7}{4}-\binom{1}{7}=\binom{6}{-3}.\)

A point on the line segment \(AB\) can be written as

\(\overrightarrow{OA}+t\overrightarrow{AB}=\binom{1}{7}+t\binom{6}{-3}=\binom{1+6t}{7-3t},\)

where \(0\leqslant t\leqslant1\). The value of \(t\) represents the fraction of the way from \(A\) to \(B\).

Since \(C\) has position vector

\(\overrightarrow{OC}=k\binom12=\binom{k}{2k},\)

we equate components:

\(k=1+6t,\qquad 2k=7-3t.\)

Substitute \(k=1+6t\) into the second equation:

\(2(1+6t)=7-3t.\)

Thus

\(2+12t=7-3t,\)

so

\(15t=5,\qquad t=\dfrac13.\)

Therefore \(C\) is one third of the way from \(A\) to \(B\). Hence

\(AC:AB=\dfrac13:1=1:3.\)