Question: Given $\mathbf{p} = \begin{pmatrix} 3 \\ -4 \end{pmatrix}$, $\mathbf{q} = \begin{pmatrix} -15 \\ 8 \end{pmatrix}$ and $\mathbf{r} = \begin{pmatrix} 2 \\ 4 \end{pmatrix}$, find $|\mathbf{p}|$, $\;2\mathbf{p}+\mathbf{q}$ and $\;\mathbf{p}+2\mathbf{q}-3\mathbf{r}$.

Step 1. Magnitude is Pythagoras on the two components:

$$|\mathbf{p}| = \sqrt{3^2 + (-4)^2} = \sqrt{25} = 5$$

Step 2. Scalar multiples first, then add component by component:

$$2\mathbf{p}+\mathbf{q} = \begin{pmatrix} 6 \\ -8 \end{pmatrix} + \begin{pmatrix} -15 \\ 8 \end{pmatrix} = \begin{pmatrix} -9 \\ 0 \end{pmatrix}$$

Step 3. Same routine with three terms:

$$\mathbf{p}+2\mathbf{q}-3\mathbf{r} = \begin{pmatrix} 3 \\ -4 \end{pmatrix} + \begin{pmatrix} -30 \\ 16 \end{pmatrix} - \begin{pmatrix} 6 \\ 12 \end{pmatrix} = \begin{pmatrix} -33 \\ 0 \end{pmatrix}$$

Step 4. Read the answers geometrically: both results have $y = 0$, so they are horizontal vectors — pointing in the negative $x$-direction.

⚠ Watch out: Squaring kills the minus sign, so $|\mathbf{p}| = \sqrt{3^2 + (-4)^2}$, never $\sqrt{3^2 - 4^2}$. A magnitude is a positive number — writing $|\mathbf{p}|$ as a column vector loses the mark.

Question: $\overrightarrow{PQ} = \begin{pmatrix} -9 \\ 12 \end{pmatrix}$ and $\overrightarrow{RS} = \frac{4}{3}\overrightarrow{PQ}$. Find $|\overrightarrow{PQ}|$, express $\overrightarrow{RS}$ as a column vector, write down $|\overrightarrow{RS}|$, and — given $P(5, -8)$ — find the coordinates of $Q$.

Step 1. Magnitude:

$$|\overrightarrow{PQ}| = \sqrt{(-9)^2 + 12^2} = \sqrt{225} = 15$$

Step 2. Scalar multiple, component by component:

$$\overrightarrow{RS} = \dfrac{4}{3}\begin{pmatrix} -9 \\ 12 \end{pmatrix} = \begin{pmatrix} -12 \\ 16 \end{pmatrix}$$

Step 3. No need to recompute — lengths scale by the same factor:

$$|\overrightarrow{RS}| = \dfrac{4}{3} \times 15 = 20$$

Step 4. Coordinates come from position vectors (vectors from the origin):

$$\overrightarrow{OQ} = \overrightarrow{OP} + \overrightarrow{PQ} = \begin{pmatrix} 5 \\ -8 \end{pmatrix} + \begin{pmatrix} -9 \\ 12 \end{pmatrix} = \begin{pmatrix} -4 \\ 4 \end{pmatrix} \;\Rightarrow\; Q(-4,\ 4)$$

⚠ Watch out: A point's coordinates and its position vector carry the same numbers, but only $\overrightarrow{OQ}$ is a vector — write $Q(-4, 4)$ as the final coordinate answer. And the route $O \to P \to Q$ is the standard trick for locating any unknown point.

Question: $\overrightarrow{AB} = \begin{pmatrix} 3 \\ 1 \end{pmatrix}$. $D$ is $(-2, 1)$ and $E$ is $(h, 4)$. Express $\overrightarrow{DE}$ as a column vector. Find $h$ if $\overrightarrow{DE}$ is parallel to $\overrightarrow{AB}$; find the two possible $h$ if instead $|\overrightarrow{DE}| = |\overrightarrow{AB}|$.

Step 1. Vector between two points = end minus start:

$$\overrightarrow{DE} = \begin{pmatrix} h - (-2) \\ 4 - 1 \end{pmatrix} = \begin{pmatrix} h+2 \\ 3 \end{pmatrix}$$

Step 2. Parallel means one is a multiple of the other: $\overrightarrow{DE} = k\begin{pmatrix} 3 \\ 1 \end{pmatrix}$.

The $y$-components give $3 = k(1)$, so $k = 3$; then the $x$-components:

$$h + 2 = 3(3) = 9 \;\Rightarrow\; h = 7$$

Step 3. For equal magnitudes, $|\overrightarrow{AB}| = \sqrt{3^2+1^2} = \sqrt{10}$:

$$(h+2)^2 + 3^2 = 10$$

Step 4. Solve the quadratic:

$$(h+2)^2 = 1 \;\Rightarrow\; h + 2 = \pm 1 \;\Rightarrow\; h = -1 \text{ or } h = -3$$

⚠ Watch out: Parallel and equal-length are different conditions — parallel fixes the direction (one value of $h$), equal magnitude fixes only the length (two values, from the $\pm$ square root). Keep both roots in Step 4; dropping the negative one is the standard slip.

Question: $\overrightarrow{OA} = 4\mathbf{a}$ and $\overrightarrow{OB} = 3\mathbf{b}$. $D$ lies on $AB$ with $AD = \frac{1}{4}AB$. Express $\overrightarrow{AB}$, $\overrightarrow{AD}$ and $\overrightarrow{OD}$ as simply as possible in terms of $\mathbf{a}$ and $\mathbf{b}$.

[diagram in original sheet: triangle OAB with D on AB]

Step 1. Travel $A \to O \to B$ — reversing an arrow flips its sign:

$$\overrightarrow{AB} = \overrightarrow{AO} + \overrightarrow{OB} = -4\mathbf{a} + 3\mathbf{b}$$

Step 2. $D$ divides $AB$ with $AD = \frac{1}{4}AB$, same direction:

$$\overrightarrow{AD} = \dfrac{1}{4}\overrightarrow{AB} = \dfrac{1}{4}(-4\mathbf{a} + 3\mathbf{b}) = -\mathbf{a} + \dfrac{3}{4}\mathbf{b}$$

Step 3. Route to $D$ through a point you know:

$$\overrightarrow{OD} = \overrightarrow{OA} + \overrightarrow{AD} = 4\mathbf{a} + \left(-\mathbf{a} + \dfrac{3}{4}\mathbf{b}\right) = 3\mathbf{a} + \dfrac{3}{4}\mathbf{b}$$

Step 4. Every unknown vector is found the same way: pick a path along arrows you already know, then simplify the $\mathbf{a}$'s and $\mathbf{b}$'s separately.

⚠ Watch out: $\overrightarrow{AO} = -\overrightarrow{OA}$ — direction matters, and the sign flip is where most marks die. "As simply as possible" means collect like terms fully: $4\mathbf{a} - \mathbf{a} + \frac{3}{4}\mathbf{b}$ is unfinished.

Question: $ABCDEF$ is a regular hexagon with $\overrightarrow{AB} = \mathbf{a}$ and $\overrightarrow{AF} = \mathbf{b}$. Express $\overrightarrow{BF}$, $\overrightarrow{FC}$ and $\overrightarrow{BC}$ as simply as possible in terms of $\mathbf{a}$ and/or $\mathbf{b}$.

[diagram in original sheet: regular hexagon labelled A–F anticlockwise]

Step 1. $\overrightarrow{BF}$ by the triangle route $B \to A \to F$:

$$\overrightarrow{BF} = \overrightarrow{BA} + \overrightarrow{AF} = -\mathbf{a} + \mathbf{b} = \mathbf{b} - \mathbf{a}$$

Step 2. Use the hexagon's structure for $\overrightarrow{FC}$: in a regular hexagon the long diagonal $FC$ is parallel to the side $AB$ and twice its length:

$$\overrightarrow{FC} = 2\mathbf{a}$$

Step 3. Chain the two answers for $\overrightarrow{BC}$, going $B \to F \to C$:

$$\overrightarrow{BC} = \overrightarrow{BF} + \overrightarrow{FC} = (\mathbf{b} - \mathbf{a}) + 2\mathbf{a} = \mathbf{a} + \mathbf{b}$$

Step 4. Sanity-check with symmetry: $\overrightarrow{BC}$ should point "between" the directions of $\mathbf{a}$ and $\mathbf{b}$ — and $\mathbf{a} + \mathbf{b}$ does exactly that.

⚠ Watch out: The hexagon facts doing the work: each long diagonal through the centre is parallel to a side and equal to $2 \times$ that side. Don't try to force every answer through $\mathbf{b}-\mathbf{a}$ paths — spotting the parallel diagonal ($\overrightarrow{FC} = 2\overrightarrow{AB}$) is the intended shortcut.

Question: From the same figure ($\overrightarrow{OA} = 4\mathbf{a}$, $\overrightarrow{OB} = 3\mathbf{b}$, $D$ on $AB$ with $\overrightarrow{OD} = 3\mathbf{a} + \frac{3}{4}\mathbf{b}$), the point $E$ satisfies $3\overrightarrow{AE} = \overrightarrow{OB}$. Find $\overrightarrow{DE}$ and show that $O$, $D$ and $E$ lie on a straight line.

Step 1. First $\overrightarrow{AE} = \dfrac{1}{3}\overrightarrow{OB} = \mathbf{b}$. Route $D \to A \to E$:

$$\overrightarrow{DE} = -\overrightarrow{AD} + \overrightarrow{AE} = -\left(-\mathbf{a} + \dfrac{3}{4}\mathbf{b}\right) + \mathbf{b} = \mathbf{a} + \dfrac{1}{4}\mathbf{b}$$

Step 2. Line up the two vectors that share the point $D$:

$$\overrightarrow{OD} = 3\mathbf{a} + \dfrac{3}{4}\mathbf{b}, \qquad \overrightarrow{DE} = \mathbf{a} + \dfrac{1}{4}\mathbf{b}$$

Step 3. Spot the multiple:

$$\overrightarrow{OD} = 3\,\overrightarrow{DE}$$

Step 4. Conclude with both ingredients stated: $\overrightarrow{OD}$ is a multiple of $\overrightarrow{DE}$ (so they are parallel), and they share the common point $D$ — therefore $O$, $D$, $E$ are collinear. (Bonus: the multiple also says $OD = 3\,DE$ in length.)

⚠ Watch out: Parallel alone is not collinear — the two vectors must also pass through a common point; say so explicitly in the conclusion. Any pair works ($\overrightarrow{OD}$ with $\overrightarrow{OE}$, etc.) as long as both use two of the three points.

Question: $WXYZ$ is a trapezium with $WX$ // $ZY$ and $WX : ZY = 3 : 4$. $P$ on $XZ$ satisfies $ZP : PX = 1 : 3$. Given $\overrightarrow{WX} = 9\mathbf{a}$ and $\overrightarrow{WZ} = \mathbf{b}$, express $\overrightarrow{ZX}$ and $\overrightarrow{WP}$, and show $XY$ is parallel to $WP$.

[diagram in original sheet: trapezium WXYZ with P on the diagonal ZX]

Step 1. Since $WX : ZY = 3:4$ and the sides are parallel, $\overrightarrow{ZY} = 12\mathbf{a}$. The diagonal:

$$\overrightarrow{ZX} = \overrightarrow{ZW} + \overrightarrow{WX} = -\mathbf{b} + 9\mathbf{a}$$

Step 2. $ZP : PX = 1:3$ puts $P$ a quarter of the way along $ZX$:

$$\overrightarrow{WP} = \overrightarrow{WZ} + \dfrac{1}{4}\overrightarrow{ZX} = \mathbf{b} + \dfrac{1}{4}(-\mathbf{b} + 9\mathbf{a}) = \dfrac{3}{4}\mathbf{b} + \dfrac{9}{4}\mathbf{a} = \dfrac{3}{4}(\mathbf{b} + 3\mathbf{a})$$

Step 3. Now the target line:

$$\overrightarrow{XY} = \overrightarrow{XW} + \overrightarrow{WZ} + \overrightarrow{ZY} = -9\mathbf{a} + \mathbf{b} + 12\mathbf{a} = 3\mathbf{a} + \mathbf{b}$$

Step 4. Compare:

$$\overrightarrow{WP} = \dfrac{3}{4}(3\mathbf{a} + \mathbf{b}) = \dfrac{3}{4}\overrightarrow{XY} \;\Rightarrow\; WP \text{ // } XY$$

⚠ Watch out: The factored form $\frac{3}{4}(\mathbf{b}+3\mathbf{a})$ is what makes the multiple visible — always factor before comparing. A ratio $ZP:PX = 1:3$ means $ZP = \frac{1}{4}ZX$ (out of $1+3 = 4$ parts), not $\frac{1}{3}$.

Question: $ABCD$ is a parallelogram; $X$ lies on $CD$ with $CX : XD = 2 : 1$; $AD$ produced meets $BX$ produced at $Y$. Find $\dfrac{\text{area } \triangle BCX}{\text{area } \triangle DYX}$ and $\dfrac{\text{area } \triangle AXY}{\text{area } \triangle BAX}$.

[diagram in original sheet: parallelogram with BX extended to meet AD extended at Y]

Step 1. $\triangle BCX$ and $\triangle DYX$ are similar ($BC$ // $DY$, vertically opposite angles at $X$), with ratio $CX : XD = 2 : 1$:

$$\dfrac{\text{area } \triangle BCX}{\text{area } \triangle DYX} = \left(\dfrac{2}{1}\right)^2 = 4$$

Step 2. The similarity also gives $BX : XY = 2 : 1$ along the line $BY$.

Step 3. $\triangle AXY$ and $\triangle BAX$ are not similar — but their bases $XY$ and $BX$ lie on the same straight line, so they share the perpendicular height from $A$:

$$\dfrac{\text{area } \triangle AXY}{\text{area } \triangle BAX} = \dfrac{XY}{BX} = \dfrac{1}{2}$$

Step 4. That is the whole toolkit: similar triangles → square the length ratio; same-height triangles → plain ratio of bases. Every vector-area question is one of these two (or a chain of them).

⚠ Watch out: Decide which rule applies before writing anything — squaring a same-height ratio (or not squaring a similar one) is the classic error. The angles decide: parallel lines create the similar pair, a shared vertex over one straight line creates the same-height pair.