POP

E7.2 Vectors in Two Dimensions

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01 — Explanation

What's the difference between a scalar and a vector?

A scalar has magnitude only (mass $=5$ kg, temperature $=25°$ C). A vector has both magnitude and direction (a displacement of $10$ m north-east). On a grid, a translation is a vector, written as a column vector $\binom{x}{y}$ , an arrow $\overrightarrow{AB}$ , or a bold letter $\mathbf{a}$ — where $x$ is the horizontal shift (right if positive) and $y$ is the vertical shift (up if positive).

02 — Explanation

How do I add, subtract and scale vectors?

Operation	Rule
Addition	$\binom{x_1}{y_1}+\binom{x_2}{y_2} = \binom{x_1+x_2}{y_1+y_2}$
Subtraction	$\binom{x_1}{y_1}-\binom{x_2}{y_2} = \binom{x_1-x_2}{y_1-y_2}$
Scalar multiple	$k\binom{x}{y} = \binom{kx}{ky}$

The top row and bottom row are entirely independent — an $x$ -component never interacts with a $y$ -component.

03 — Explanation

What happens when I multiply a vector by a negative scalar?

$k\mathbf{a}$ points in exactly the opposite direction to $\mathbf{a}$ when $k<0$ , not merely a shorter or longer version along the same direction. In particular, $-\mathbf{a} = \binom{-x}{-y}$ is the same length as $\mathbf{a}$ but reversed, so $\overrightarrow{BA} = -\overrightarrow{AB}$ .

04 — Explanation

How do I use vectors to describe a route around a shape?

Any journey between two points can be built by adding vectors "head to tail" along any convenient path, since the combined displacement only depends on the start and end points, not the route taken.

Example: $OABC$ is a parallelogram with $\overrightarrow{OA}=\mathbf{p}$ and $\overrightarrow{OC}=\mathbf{q}$ . Find $\overrightarrow{OB}$ , the diagonal of the parallelogram, in terms of $\mathbf{p}$ and $\mathbf{q}$ .

Since $OABC$ is a parallelogram, $\overrightarrow{CB}=\overrightarrow{OA}=\mathbf{p}$ (opposite sides are equal and parallel). Travelling from $O$ to $B$ via $C$ :

\overrightarrow{OB} = \overrightarrow{OC}+\overrightarrow{CB} = \mathbf{q}+\mathbf{p}

This "travel via a known vertex" technique — choosing whichever route only uses vectors already known — is the foundation for almost every vector-geometry problem at this level, not just the column-vector arithmetic above.

05 — Worked example

Worked Example

Given $\mathbf{a}=\binom{6}{-2}$ , $\mathbf{b}=\binom{-3}{4}$ , $\mathbf{c}=\binom{0}{5}$ , find $3\mathbf{a}-2\mathbf{b}+\mathbf{c}$ as a single column vector.

Solution

3\mathbf{a} = \binom{18}{-6}, \qquad 2\mathbf{b} = \binom{-6}{8}

3\mathbf{a}-2\mathbf{b}+\mathbf{c} = \binom{18}{-6} - \binom{-6}{8} + \binom{0}{5}

Top: $18-(-6)+0 = 24$ . Bottom: $-6-8+5=-9$ .

3\mathbf{a}-2\mathbf{b}+\mathbf{c} = \binom{24}{-9}

Examiner Tip

Subtracting a negative component is the step most often botched under time pressure — $18-(-6)$ must become $18+6$ , not $18-6$ . Rewrite the subtraction as an addition before evaluating if you're prone to this slip.

06 — Worked example

Worked Example 2: reversing the process

Given $\mathbf{u}=\binom{4}{7}$ and $2\mathbf{u}+\mathbf{v}=\binom{9}{10}$ , find $\mathbf{v}$ .

Solution

2\mathbf{u} = \binom{8}{14}

\mathbf{v} = \binom{9}{10}-\binom{8}{14} = \binom{9-8}{10-14} = \binom{1}{-4}

Treating a vector equation exactly like a normal algebraic equation — rearranging to isolate $\mathbf{v}$ before substituting numbers — avoids trying to "read off" an answer from a jumbled combination of vectors.

07 — Explanation

How do I know whether two vectors are parallel?

Two vectors are parallel exactly when one is a scalar multiple of the other — that is, $\mathbf{b} = k\mathbf{a}$ for some number $k$ . Check by comparing the ratio of corresponding components: if $\binom{6}{-2}$ and $\binom{-9}{3}$ are parallel, then $\dfrac{-9}{6}$ should equal $\dfrac{3}{-2}$ , and both do indeed simplify to $-\dfrac32$ , confirming $\binom{-9}{3} = -\dfrac32\binom{6}{-2}$ .