Spatial Geometry & Kinematic Forces

Vector Algebra, Collinearity & Resultant Navigation

Vectors embody directional quantities governing physical flight, ocean currents, and geometric partitions. Master the Triangle and Polygon laws of addition, parallel scalar conditions $\vec{a} = \lambda \vec{b}$, collinearity proofs, unit vectors $\hat{r}$, and solving ratios through non-parallel vector coefficient comparisons.

1. Fundamental Vector Laws & Criteria

Magnitude & Unit Vector

$\vec{r} = x\mathbf{i} + y\mathbf{j} = \begin{pmatrix} x \\ y \end{pmatrix}$
$|\vec{r}| = \sqrt{x^2 + y^2}$
$\hat{r} = \frac{\vec{r}}{|\vec{r}|} = \frac{x\mathbf{i} + y\mathbf{j}}{\sqrt{x^2 + y^2}}$

$\hat{r}$ has a magnitude of exactly 1 unit and points in the same direction as $\vec{r}$.

Parallel & Collinear

$\vec{u} \parallel \vec{v} \iff \vec{u} = \lambda \vec{v}$
Collinear points $A, B, C$:
$\vec{AB} = k \vec{BC}$

To prove collinearity, state that $\vec{AB} = k\vec{BC}$ and point $B$ is a common vertex.

Non-Parallel Vectors

$h\mathbf{a} + k\mathbf{b} = m\mathbf{a} + n\mathbf{b}$
If $\mathbf{a} \nparallel \mathbf{b}$ and non-zero:
$h = m \quad \text{and} \quad k = n$

The foundational theorem used to solve ratios and unknown constants in Paper 2 geometric diagrams!

2. Progressive SPM Worked Examples with Marking Schemes

Example 1 • SPM Paper 1 Standard [5 Marks]

Given that $\vec{OA} = 3\mathbf{i} - 4\mathbf{j}$ and $\vec{OB} = -5\mathbf{i} + 2\mathbf{j}$, find:

(a) The vector $\vec{AB}$. [2 marks]
(b) The unit vector in the direction of $\vec{AB}$. [3 marks]
Step-by-Step Marking Solution
(a) Finding $\vec{AB}$ using subtraction of position vectors:
$\vec{AB} = \vec{OB} - \vec{OA}$
$= (-5\mathbf{i} + 2\mathbf{j}) - (3\mathbf{i} - 4\mathbf{j})$
$= (-5 - 3)\mathbf{i} + (2 - (-4))\mathbf{j} = -8\mathbf{i} + 6\mathbf{j}$
[K1] $\vec{OB} - \vec{OA}$
[N1] $-8\mathbf{i} + 6\mathbf{j}$
(b) Unit vector:
Magnitude $|\vec{AB}| = \sqrt{(-8)^2 + 6^2} = \sqrt{64 + 36} = \sqrt{100} = 10\text{ units}$. [K1]
$\widehat{AB} = \frac{\vec{AB}}{|\vec{AB}|} = \frac{-8\mathbf{i} + 6\mathbf{j}}{10} = -\frac{4}{5}\mathbf{i} + \frac{3}{5}\mathbf{j}$ (or $\begin{pmatrix} -0.8 \\ 0.6 \end{pmatrix}$).
[K1] magnitude 10
[N1] $-\frac{4}{5}\mathbf{i} + \frac{3}{5}\mathbf{j}$
Example 2 • SPM Paper 2 Section A (Classic Geometric Ratio) [8 Marks]

In $\triangle OPQ$, $\vec{OP} = \mathbf{p}$ and $\vec{OQ} = \mathbf{q}$. Point $R$ lies on $OP$ such that $OR : RP = 2 : 1$. Point $S$ lies on $PQ$ such that $PS : SQ = 3 : 2$. The lines $OS$ and $QR$ intersect at point $T$.

(a) Express $\vec{OS}$ in terms of $\mathbf{p}$ and $\mathbf{q}$. [2 marks]
(b) Express $\vec{QR}$ in terms of $\mathbf{p}$ and $\mathbf{q}$. [2 marks]
(c) Given that $\vec{OT} = \lambda \vec{OS}$ and $\vec{QT} = \mu \vec{QR}$, find the values of $\lambda$ and $\mu$. [4 marks]
Step-by-Step Marking Solution
(a) $\vec{OS}$ expression:
$\vec{PQ} = \vec{OQ} - \vec{OP} = \mathbf{q} - \mathbf{p}$.
$\vec{PS} = \frac{3}{5}\vec{PQ} = \frac{3}{5}(\mathbf{q} - \mathbf{p})$.
$\vec{OS} = \vec{OP} + \vec{PS} = \mathbf{p} + \frac{3}{5}(\mathbf{q} - \mathbf{p}) = \frac{2}{5}\mathbf{p} + \frac{3}{5}\mathbf{q}$.
[K1] ratio path
[N1] $\frac{2}{5}\mathbf{p} + \frac{3}{5}\mathbf{q}$
(b) $\vec{QR}$ expression:
Since $OR : RP = 2 : 1 \implies \vec{OR} = \frac{2}{3}\mathbf{p}$.
$\vec{QR} = \vec{QO} + \vec{OR} = -\mathbf{q} + \frac{2}{3}\mathbf{p} = \frac{2}{3}\mathbf{p} - \mathbf{q}$.
[K1] $\vec{OR} = \frac{2}{3}\mathbf{p}$
[N1] $\frac{2}{3}\mathbf{p} - \mathbf{q}$
(c) Equating coefficients of $\mathbf{p}$ and $\mathbf{q}$:
From $\vec{OT} = \lambda \vec{OS} \implies \vec{OT} = \lambda\left(\frac{2}{5}\mathbf{p} + \frac{3}{5}\mathbf{q}\right) = \frac{2}{5}\lambda\mathbf{p} + \frac{3}{5}\lambda\mathbf{q} \quad \text{--- (1)}$
From another path: $\vec{OT} = \vec{OQ} + \vec{QT} = \mathbf{q} + \mu \vec{QR} = \mathbf{q} + \mu\left(\frac{2}{3}\mathbf{p} - \mathbf{q}\right) = \frac{2}{3}\mu\mathbf{p} + (1 - \mu)\mathbf{q} \quad \text{--- (2)}$ [K1]
Equate coefficients of $\mathbf{p}$: $\frac{2}{5}\lambda = \frac{2}{3}\mu \implies \mu = \frac{3}{5}\lambda \quad \text{--- (3)}$
Equate coefficients of $\mathbf{q}$: $\frac{3}{5}\lambda = 1 - \mu \quad \text{--- (4)}$ [K1]
Substitute (3) into (4): $\frac{3}{5}\lambda = 1 - \frac{3}{5}\lambda \implies \frac{6}{5}\lambda = 1 \implies \lambda = \frac{5}{6}$. [K1, N1]
$\mu = \frac{3}{5}\left(\frac{5}{6}\right) = \frac{1}{2}$. [N1]
Example 3 • KBAT Autonomous Drone Navigation [5 Marks]

An agricultural surveillance drone flies with an air velocity of $\mathbf{v}_{\text{drone}} = (12\mathbf{i} + 16\mathbf{j})\text{ ms}^{-1}$. A persistent crosswind blows across the plantation with velocity $\mathbf{v}_{\text{wind}} = (-4\mathbf{i} + 3\mathbf{j})\text{ ms}^{-1}$.

(a) Find the actual resultant ground velocity vector of the drone, $\mathbf{v}_{\text{ground}}$. [2 marks]
(b) Calculate the ground speed (magnitude) of the drone and its actual direction of motion as a compass bearing. [3 marks]
(a) $\mathbf{v}_{\text{ground}} = \mathbf{v}_{\text{drone}} + \mathbf{v}_{\text{wind}}$
$= (12\mathbf{i} + 16\mathbf{j}) + (-4\mathbf{i} + 3\mathbf{j}) = 8\mathbf{i} + 19\mathbf{j}\text{ ms}^{-1}$.
[K1] vector addition
[N1] $8\mathbf{i} + 19\mathbf{j}\text{ ms}^{-1}$
(b) Ground speed $= |\mathbf{v}_{\text{ground}}| = \sqrt{8^2 + 19^2} = \sqrt{64 + 361} = \sqrt{425} \approx 20.62\text{ ms}^{-1}$. [K1]
Direction relative to the positive $x$-axis (East): $\theta = \tan^{-1}\left(\frac{19}{8}\right) \approx 67.17^\circ$ North of East.
Compass bearing (clockwise from North): $90^\circ - 67.17^\circ = 022.83^\circ$ (or $022^\circ 50'$). [N1] $20.62\text{ ms}^{-1}, 022.83^\circ$