Vectors in Two Dimensions

Vectors in 0606 escalate quickly: notation and arithmetic (gentle), geometry with position vectors (standard), and relative velocity (the part that fills our enquiry inbox every exam season). The whole topic runs on a handful of rules applied with diagram discipline.

Notation, magnitude, unit vectors

A vector has magnitude and direction. Component forms: $x\mathbf{i} + y\mathbf{j}$ or column vectors. For $\mathbf{v} = 3\mathbf{i} - 4\mathbf{j}$:

• Magnitude: $|\mathbf{v}| = \sqrt{3^2 + (-4)^2} = 5$. Pythagoras, same as coordinate length
• Unit vector: $\mathbf{v}/|\mathbf{v}| = (3\mathbf{i} - 4\mathbf{j})/5$, direction with magnitude $1$

The standard construction the exam loves: a particle moves with speed 26 m/s in the direction of $5\mathbf{i} + 12\mathbf{j}$ $\to$ velocity $= 26 \times (5\mathbf{i} + 12\mathbf{j})/13 = 10\mathbf{i} + 24\mathbf{j}$. Speed $\times$ unit vector. Writing the unit-vector step earns the method mark.

Position vectors and geometry

Position vector: $\overrightarrow{OA} = \mathbf{a}$ locates point $A$ from the origin. The workhorse rule:

$\overrightarrow{AB} = \mathbf{b} - \mathbf{a}$ (destination minus start)

From it: midpoint of $AB$ has position $\frac{1}{2}(\mathbf{a} + \mathbf{b})$; $P$ dividing $AB$ in ratio $\lambda:\mu$ has position $(\mu\mathbf{a} + \lambda\mathbf{b})/(\lambda + \mu)$. Geometry proofs (“show $P$, $Q$, $R$ are collinear”) reduce to showing one vector is a scalar multiple of another. $\overrightarrow{PQ} = k\cdot\overrightarrow{QR}$ with a shared point $\Rightarrow$ collinear, and the conclusion must be stated in words (determine-style command).

In ratio/intersection problems, the method is to express the same point two ways and equate coefficients of $\mathbf{a}$ and $\mathbf{b}$, valid because $\mathbf{a}$ and $\mathbf{b}$ are non-parallel. The line “equating coefficients of $\mathbf{a}$: …” is the visible method that carries the marks.

Velocity and relative velocity

The kinematic backbone: position at time $t$ = initial position + $t \times$ velocity:

$\mathbf{r}(t) = \mathbf{r}_0 + t\mathbf{v}$

Write this for each moving object before doing anything else, it converts the word problem into algebra. Then:

• Interception/collision: set $\mathbf{r}_A(t) = \mathbf{r}_B(t)$; equate $\mathbf{i}$ and $\mathbf{j}$ components; the same $t$ must satisfy both (checking this consistency is often the final mark).
• Relative velocity: velocity of $A$ relative to $B$ $= \mathbf{v}_A - \mathbf{v}_B$; relative position likewise. “When are they closest?” and “show they do not collide” questions live entirely in the relative vector.
• Courses and currents: actual velocity = velocity in still water/air + current/wind vector, draw the triangle, label it, and the geometry (often sine/cosine work) follows.

Worked exam-style question

At 12:00, ship A is at position $(2\mathbf{i} + 3\mathbf{j})$ km moving with velocity $(4\mathbf{i} + \mathbf{j})$ km/h; ship B is at $(14\mathbf{i} - 3\mathbf{j})$ km with velocity $(\mathbf{i} + 2.5\mathbf{j})$ km/h. Show the ships meet, and find when.

$\mathbf{r}_A(t) = (2 + 4t)\mathbf{i} + (3 + t)\mathbf{j}$ (M, position model stated) $\mathbf{r}_B(t) = (14 + t)\mathbf{i} + (-3 + 2.5t)\mathbf{j}$ (M) Equate $\mathbf{i}$: $2 + 4t = 14 + t \to 3t = 12 \to t = 4$ (A) Check $\mathbf{j}$ at $t = 4$: $3 + 4 = 7$; $-3 + 10 = 7$ ✓, same $\mathbf{j}$-component, so the ships meet at 16:00 (B, the verification AND the stated conclusion)

The $\mathbf{j}$-component check is the “show” part, skipping it converts a 5-mark answer into a 3-mark one even with $t = 4$ correct.

Common mistakes in this topic

• $\overrightarrow{AB}$ computed as $\mathbf{a} - \mathbf{b}$ (start minus destination, backwards)
• Speed used as velocity without the unit-vector direction step
• Interception solved on one component only, never verified on the other
• Relative velocity subtracted the wrong way round ($A$ relative to $B$ is $\mathbf{v}_A - \mathbf{v}_B$)
• Diagrams skipped on course/current problems, then the triangle’s angles guessed

Vectors borrow magnitude from coordinate geometry and share its kinematics language with calculus kinematics, though here motion is constant-velocity, there it accelerates. Another classically avoided topic, which makes it A*-differentiating.

Relative velocity is the most-requested rescue topic we teach, and it’s very fixable. Free 1-hour trial class with Teacher Rig: message us on WhatsApp.