Answer: (a) \(\overrightarrow{OP}=\frac34\mathbf a+\frac14\mathbf b\); (b) \(\frac1{\sqrt{10}}\binom{3}{1}\).

We start with the main method. Use the measurements and relationships shown in the diagram, then translate them into algebraic or trigonometric equations. Use position vectors and ratios along a line, then compare coefficients of independent vectors.

(a) Since \(AP:PB=1:3\), point \(P\) is one quarter of the way from \(A\) to \(B\).

Therefore

\(\overrightarrow{AP}=\frac14\overrightarrow{AB}=\frac14(\mathbf b-\mathbf a).\)

So

\(\overrightarrow{OP}=\overrightarrow{OA}+\overrightarrow{AP} =\mathbf a+\frac14(\mathbf b-\mathbf a).\)

Hence

\(\overrightarrow{OP}=\frac34\mathbf a+\frac14\mathbf b.\)

(b) The direction vector for \(\mathbf q\) is \(\binom{6}{-3}\), whose magnitude is

\(\sqrt{6^2+(-3)^2}=\sqrt{45}=3\sqrt5.\)

So the unit vector in this direction is

\(\frac{1}{3\sqrt5}\binom{6}{-3}=\binom{2/\sqrt5}{-1/\sqrt5}.\)

Since \(|\mathbf q|=12\sqrt5\),

\(\mathbf q=12\sqrt5\binom{2/\sqrt5}{-1/\sqrt5}=\binom{24}{-12}.\)

For \(\mathbf r\), the direction vector is \(\binom{-5}{5}\), whose magnitude is \(5\sqrt2\). Thus the unit vector is

\(\binom{-1/\sqrt2}{1/\sqrt2}.\)

Since \(|\mathbf r|=15\sqrt2\),

\(\mathbf r=15\sqrt2\binom{-1/\sqrt2}{1/\sqrt2}=\binom{-15}{15}.\)

Therefore

\(\mathbf q+\mathbf r=\binom{24}{-12}+\binom{-15}{15}=\binom{9}{3}.\)

The magnitude of \(\binom{9}{3}\) is

\(\sqrt{9^2+3^2}=\sqrt{90}=3\sqrt{10}.\)

So the required unit vector is

\(\frac{1}{3\sqrt{10}}\binom{9}{3}=\frac1{\sqrt{10}}\binom{3}{1}.\)

This completes the solution and gives the required result.