Answer: (a) \(\frac{40}{41}\mathbf{i}-\frac{9}{41}\mathbf{j}\). (b)(i) \(0\lt k\lt 1\).

Answer: (a) \(\frac{40}{41}\mathbf{i}-\frac{9}{41}\mathbf{j}\). (b)(i) \(0\lt k\lt 1\).

(a) The magnitude of \(40\mathbf{i}-9\mathbf{j}\) is

\(\sqrt{40^2+(-9)^2}=\sqrt{1600+81}=\sqrt{1681}=41\).

So a unit vector in the same direction is

\(\displaystyle \frac{40}{41}\mathbf{i}-\frac{9}{41}\mathbf{j}\).

(b)(i) Since \(R\) lies between \(P\) and \(Q\), the fraction \(\frac{PR}{PQ}=k\) must satisfy

\(0\lt k\lt 1\).

(b)(ii) Moving from \(P\) to \(Q\), the vector \(\overrightarrow{PQ}\) is \(\mathbf{q}-\mathbf{p}\). Since \(\frac{PR}{PQ}=k\),

\(\overrightarrow{PR}=k(\mathbf{q}-\mathbf{p})\).

Therefore the position vector of \(R\) is

\(\mathbf{p}+k(\mathbf{q}-\mathbf{p})\).

Simplifying,

\(\mathbf{r}=(1-k)\mathbf{p}+k\mathbf{q}\).

Thus \(\lambda=1-k\) and \(\mu=k\), so

\(\lambda+\mu=(1-k)+k=1\).