20. The figure below is triangle \( OPQ \) in which \( OP = \mathbf{p} \) and \( OQ = \mathbf{q} \). Points \( M \) and \( N \) are on \( OQ \) and \( OP \) respectively such that \( ON = NP = 1:2 \) and \( OM: MQ = 3:2 \).

(a) Express the following vectors in terms of \(\mathbf{p}\) and \(\mathbf{q}\):

(i) \(\overrightarrow{PM}\) (2 marks) 
\[
\overrightarrow{PM} = \overrightarrow{PO} + \overrightarrow{OM} \\
= -\mathbf{p} + \frac{3}{5}\mathbf{q}  \\
= \frac{3}{5}\mathbf{q} - \mathbf{p}.
\]

(ii) \(\overrightarrow{QN}\) (1 mark) 
\[
\overrightarrow{QN} = \overrightarrow{QO} + \overrightarrow{ON} \\
= -\mathbf{q} + \tfrac{1}{3}\mathbf{p} \\
= \frac{1}{3}\mathbf{p} - \mathbf{q}.
\]

(iii) \(\overrightarrow{PQ}\) (1 mark) 
\[
\overrightarrow{PQ} = \overrightarrow{PO} + \overrightarrow{OQ}  \\
= -\mathbf{p} + \mathbf{q} \\
= \mathbf{q} - \mathbf{p}.
\]

(b) Lines \( PN \) and \( QM \) intersect at \( X \) such that \( PX = r\mathbf{PM} \) and \( QX = t\mathbf{QN} \). Express \(\overrightarrow{OX}\) in two different ways and find the value of \( r \) and \( t \). (4 marks)

From line \( PX \): 
\[
\overrightarrow{OX} = \overrightarrow{OP} + \overrightarrow{PX} \\
= \mathbf{p} + r\overrightarrow{PM}
= \mathbf{p} + r\left(\frac{3}{5}\mathbf{q} - \mathbf{p}\right) \\
= (1 - r)\mathbf{p} + \frac{3r}{5}\mathbf{q}. \quad (i)
\]

From line \( QX \): 
\[
\overrightarrow{OX} = \overrightarrow{OQ} + \overrightarrow{QX} \\
= \mathbf{q} + t\overrightarrow{QN} \\
= \mathbf{q} + t\left(\frac{1}{3}\mathbf{p} - \mathbf{q}\right) \\
= \frac{t}{3}\mathbf{p} + (1 - t)\mathbf{q}. \quad (ii)
\]

Equate (i) and (ii):
\[
(1 - r)\mathbf{p} + \frac{3r}{5}\mathbf{q} \\ = \frac{t}{3}\mathbf{p} + (1 - t)\mathbf{q}.
\]

Equating components:
\[
\mathbf{p}: \quad 1 - r = \frac{t}{3} \quad \Rightarrow \quad t = 3(1 - r),
\]
\[
\mathbf{q}: \quad \frac{3r}{5} = 1 - t.
\]

Substitute \( t = 3(1 - r) \):
\[
\frac{3r}{5} = 1 - 3(1 - r) = 1 - 3 + 3r = -2 + 3r.
\]

Rearranging:
\[
\frac{3r}{5} - 3r \\  = -2 \quad \Rightarrow \quad \frac{3r - 15r}{5} = -2 \quad \Rightarrow \quad -\frac{12r}{5} = -2.
\]

Solving for \( r \):
\[
r = \frac{5}{6}.
\]

Substitute \( r = \frac{5}{6} \) into \( t = 3(1 - r) \):
\[
t = 3\left(1 - \frac{5}{6}\right) = 3 \cdot \frac{1}{6} = \frac{1}{2}.
\]

Thus, \( r = \frac{5}{6} \) and \( t = \frac{1}{2} \).

(c) Line \( OX \) produced meets \( PQ \) at \( Y \) such that \( PY: YQ = 5:3 \). Using the ratio theorem or otherwise, find \(\overrightarrow{OY}\) in terms of \(\mathbf{p}\) and \(\mathbf{q}\). (2 marks)

By the section formula:
\(
\overrightarrow{OY} = \frac{5\overrightarrow{Q} + 3\overrightarrow{P}}{5 + 3} = \frac{5\mathbf{q} + 3\mathbf{p}}{8}.
\)

Alternatively:
\(
\overrightarrow{OY} = \overrightarrow{OP} + \frac{5}{8}\overrightarrow{PQ} \\
= \mathbf{p} + \frac{5}{8}(\mathbf{q} - \mathbf{p}) \\
= \frac{3}{8}\mathbf{p} + \frac{5}{8}\mathbf{q}.
\)

(a) KCSE 2003, PPI, Q6
Given that:
\(
x = 2\mathbf{i} + \mathbf{j} - 2\mathbf{k}, \quad y = -3\mathbf{i} + 4\mathbf{j} - \mathbf{k}, \quad z = -5\mathbf{i} + 3\mathbf{j} + 2\mathbf{k}
\)
and that:
\(
p = 3x - y + 2z,
\)
find the magnitude of the vector to 3 significant figures. (4 marks)

Solution 
First, substitute the given vectors into \( p \):
\(
p = 3(2\mathbf{i} + \mathbf{j} - 2\mathbf{k}) - (-3\mathbf{i} + 4\mathbf{j} - \mathbf{k}) + 2(-5\mathbf{i} + 3\mathbf{j} + 2\mathbf{k})
\)

Expanding and simplifying:
\(
p = (6\mathbf{i} + 3\mathbf{j} - 6\mathbf{k}) + (3\mathbf{i} - 4\mathbf{j} + \mathbf{k}) + (-10\mathbf{i} + 6\mathbf{j} + 4\mathbf{k})
\)
\(
= (6 + 3 - 10)\mathbf{i} + (3 - 4 + 6)\mathbf{j} + (-6 + 1 + 4)\mathbf{k}
\)
\(
= -\mathbf{i} + 5\mathbf{j} - \mathbf{k}
\)

The magnitude of \( p \) is:
\(
|p| = \sqrt{(-1)^2 + 5^2 + (-1)^2} \\ = \sqrt{1 + 25 + 1} \\  = \sqrt{27}  = 5.1962 \approx 5.20 \text{ (3 s.f)}
\)
b) KCSE 2007, PP2, Q12
Vector \( q \) has a magnitude of 7 and is parallel to vector \( p \). Given that:
\(
p = 3\mathbf{i} - \mathbf{j} + \tfrac{3}{2}\mathbf{k},
\)
express vector \( q \) in terms of \( \mathbf{i}, \mathbf{j}, \) and \( \mathbf{k} \). (2 marks)

Solution  
First, find the magnitude of \( p \):
\(
|p| = \sqrt{3^2 + (-1)^2 + \left(\frac{3}{2}\right)^2} \\ = \sqrt{9 + 1 + 2.25} \\ = \sqrt{12.25} = 3.5
\)

Since \( q \) is parallel to \( p \) and \( |q| = 7 \):
\(
q = \frac{7}{3.5} p = 2p
\)

Substituting \( p \):
\(
q = 2(3\mathbf{i} - \mathbf{j} + \tfrac{3}{2}\mathbf{k}) = 6\mathbf{i} - 2\mathbf{j} + 3\mathbf{k}
\)

Alternatively, the direction can be reversed:
\(
q = -6\mathbf{i} + 2\mathbf{j} - 3\mathbf{k}
\)

9. Given \( \overrightarrow{OA} = 3\mathbf{i} + 4\mathbf{j} - 6\mathbf{k} \) and \( \overrightarrow{OB} = 3\mathbf{i} + 3\mathbf{j} + \mathbf{k} \), point \( P \) divides the line segment \( AB \) in the ratio \( 3: -2 \). Find the coordinates of \( P \). (3 marks)

\(
\begin{pmatrix}
i \\
j \\
k
\end{pmatrix}
=
\frac{-2}{1}
\begin{pmatrix}
3 \\
4 \\
-6
\end{pmatrix}
+
\frac{3}{1}
\begin{pmatrix}
3 \\
3 \\
1
\end{pmatrix}
\)

\(
=
\begin{pmatrix}
-6 \\
-8 \\
12
\end{pmatrix}
+
\begin{pmatrix}
9 \\
9 \\
3
\end{pmatrix}
=
\begin{pmatrix}
3 \\
1 \\
15
\end{pmatrix}
\)

\(
p(3,1,15)
\)

The points \( A \), \( B \), and \( C \) lie on a straight line. The position vectors of \( A \) and \( C \) are given by: \( \overrightarrow{OA} = 2\mathbf{i} + 3\mathbf{j} + 9\mathbf{k} \) and  \( \overrightarrow{OC} = 5\mathbf{i} - 3\mathbf{j} + c\mathbf{k}. \) Point \( B \) divides the line segment \( AC \) internally in the ratio \( 2:1 \). Find the magnitude of the position vector of \( B \) from the origin. 



\(
AB = \tfrac{2}{3}AC
\)

\(
= \tfrac{2}{3} 
\left( 
\begin{pmatrix} 
5 \\ 
-3 \\ 
6 
\end{pmatrix} 
- 
\begin{pmatrix} 
2 \\ 
3 \\ 
9 
\end{pmatrix} 
\right)
\)

\(
= \tfrac{2}{3} 
\begin{pmatrix} 
3 \\ 
-6 \\ 
-3 
\end{pmatrix} 
= 
\begin{pmatrix} 
2 \\ 
-4 \\ 
-2 
\end{pmatrix}
\)

\(
OB = 
\begin{pmatrix} 
2 \\ 
-4 \\ 
-2 
\end{pmatrix} 
+ 
\begin{pmatrix} 
2 \\ 
3 \\ 
9 
\end{pmatrix} 
= 
\begin{pmatrix} 
4 \\ 
1 \\ 
7 
\end{pmatrix}
\)