Vectors · Vectors & transformations

Column vectors

A vector has a size and a direction. On a grid it is written as a column vector. The top number is the step across (right is positive), and the bottom number is the step up (up is positive). So $\begin{pmatrix} 4 \\ 3 \end{pmatrix}$ means 4 right and 3 up, and $\begin{pmatrix} -2 \\ 5 \end{pmatrix}$ means 2 left and 5 up.

A column vectorTop: steps across. Bottom: steps up

Adding vectorsHead to tail: add the tops, add the bottoms

To add vectors, add the tops and add the bottoms. On a grid, draw the second starting where the first ends.
To multiply by a number, multiply both parts.
$-\mathbf a$ is $\mathbf a$ turned round, so $\mathbf a - \mathbf b = \mathbf a + (-\mathbf b)$ .

Try it

Column vectorsSet a and b, then pick an operation

31a13b44a + b√32 ≈ 5.66|a + b|

a across = 3a up = 1b across = 1b up = 3

Draw b starting where a ends (head to tail). The sum goes from the start of a to the end of b: add the tops and add the bottoms, 31 + 13 = 44. Its length is √(x² + y²) = √32 ≈ 5.66.

Try each operation. In “k a”, a negative $k$ turns the arrow round.

The length of a vector

The length (or magnitude) of $\begin{pmatrix} x \\ y \end{pmatrix}$ comes from Pythagoras:

\left|\begin{pmatrix} x \\ y \end{pmatrix}\right| = \sqrt{x^2 + y^2}

MagnitudeThe arrow is the hypotenuse: length √(x² + y²)

Position vectors and the vector between two points

The position vector of a point $A$ is $\overrightarrow{OA}$ , the vector from the origin to $A$ . For $A(1, 3)$ it is $\begin{pmatrix} 1 \\ 3 \end{pmatrix}$ . The vector from $A$ to $B$ is “end minus start”:

\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA}

Going backwards reverses the signs: $\overrightarrow{BA} = -\overrightarrow{AB}$ . To get from one point to another, you can go by any route: $\overrightarrow{AC} = \overrightarrow{AB} + \overrightarrow{BC}$ .

From A to BAB = OB − OA: end minus start

Parallel vectors

Two vectors are parallel when one is a number times the other. A negative number means they point in opposite directions.

Parallel vectors2a is twice as long as a; −a points the other way

Midpoints and routes through a shape

The midpoint $M$ of $AB$ is halfway along it, so $\overrightarrow{AM} = \frac12\overrightarrow{AB}$ . Its coordinates are the averages: $M = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)$ .

To find a vector across a shape, go by a route whose pieces you know. For example, $\overrightarrow{AM} = \overrightarrow{AB} + \overrightarrow{BM}$ .

Join the midpoints of two sides of a triangle. The line is parallel to the third side, and half as long. This is the midpoint theorem:

The midpoint theoremMS is parallel to QR and half as long: MS = ½QR

The midpoint theoremDrag P, Q and R

(−1, 1)M, midpoint of PQ(3, 0)S, midpoint of PR4−1vector MS8−2vector QR

M = ((x of P + x of Q) ÷ 2, (y of P + y of Q) ÷ 2) = (−1, 1), and S = (3, 0). MS = (4, −1) is exactly half of QR = (8, −2): MS is parallel to QR and half as long, wherever you drag the corners.

Worked example · WAEC 2017

WAEC 2017 · Paper 2 · Q13 (b)

In $\triangle ABC$ , $\overrightarrow{AB} = \begin{pmatrix} -4 \\ 6 \end{pmatrix}$ and $\overrightarrow{AC} = \begin{pmatrix} 3 \\ -8 \end{pmatrix}$ . If $P$ is the midpoint of $\overline{AB}$ , express $\overrightarrow{CP}$ as a column vector.

Halfway along AB
$\overrightarrow{AP} = \frac12\overrightarrow{AB} = \frac12\begin{pmatrix} -4 \\ 6 \end{pmatrix} = \begin{pmatrix} -2 \\ 3 \end{pmatrix}$ .
Think first. P is the midpoint of AB. What is AP?
A route from C to P
- ${\overrightarrow{CP} = \overrightarrow{CA} + \overrightarrow{AP}}$ .
- ${\overrightarrow{CA} = -\overrightarrow{AC} = \begin{pmatrix} -3 \\ 8 \end{pmatrix}}$ .
Think first. Go from C to A, then A to P.
Add
$\overrightarrow{CP} = \begin{pmatrix} -3 + (-2) \\ 8 + 3 \end{pmatrix} = \begin{pmatrix} -5 \\ 11 \end{pmatrix}$ .
Think first. Add the tops and the bottoms.

Unknowns in a vector equation

A vector equation is two equations at once, one for the tops and one for the bottoms. Solve them together.

Worked example · WAEC 2018

WAEC 2018 · Paper 2 · Q13 (a, b)

The points $R(3, -6)$ , $S(6, -2)$ and $T(p, q)$ are on the $xy$ -plane. If $\frac13\overrightarrow{OR} + \overrightarrow{OS} + \overrightarrow{OT} = \overrightarrow{RS}$ , find the coordinates of $T$ .

If $2\begin{pmatrix} 3 \\ m \end{pmatrix} + n\begin{pmatrix} 1 \\ 2 \end{pmatrix} = \begin{pmatrix} 8 \\ 4 \end{pmatrix}$ , find the values of $m$ and $n$ .

(a) Each vector
- ${\frac13\overrightarrow{OR} = \begin{pmatrix} 1 \\ -2 \end{pmatrix}}$ .
- ${\overrightarrow{OS} = \begin{pmatrix} 6 \\ -2 \end{pmatrix}}$ .
- ${\overrightarrow{RS} = \overrightarrow{OS} - \overrightarrow{OR} = \begin{pmatrix} 3 \\ 4 \end{pmatrix}}$ .
Think first. Write OR, OS and RS as column vectors.
Match tops and bottoms
${p = -4}$ and ${q = 8}$ , so $T$ is ${(-4, 8)}$ .
Think first. 1 + 6 + p = 3 and −2 − 2 + q = 4.
(b) Two equations
- ${6 + n = 8}$ , so ${n = 2}$ .
- Then ${2m + 4 = 4}$ , so ${m = 0}$ .
Think first. Tops: 6 + n = 8. Bottoms: 2m + 2n = 4.

Your turn

WAEC 2014 · Paper 2 · Q13

(a)
If $\mathbf x = \begin{pmatrix} -2 \\ 4 \end{pmatrix}$ and $\mathbf y = \begin{pmatrix} -3 \\ 1 \end{pmatrix}$ , find, correct to 1 decimal place, $|\mathbf x + \mathbf y|$ .
(b)
$P(6, 4)$ , $Q(-2, -2)$ and $R(4, -6)$ are the vertices of triangle $PQR$ . (i) Determine the coordinates of $M$ and $S$ , the midpoints of $\overline{PQ}$ and $\overline{PR}$ respectively. (ii) Find $\overrightarrow{QR}$ and $\overrightarrow{MS}$ . (iii) State the relationship between $\overrightarrow{QR}$ and $\overrightarrow{MS}$ . (iv) Find the equation of $\overline{MS}$ .
Show the answer
(i) $M(2, 1)$ , $S(5, -1)$ ; (ii) $\overrightarrow{QR} = \begin{pmatrix} 6 \\ -4 \end{pmatrix}$ , $\overrightarrow{MS} = \begin{pmatrix} 3 \\ -2 \end{pmatrix}$ ; (iii) $\overrightarrow{QR} = 2\overrightarrow{MS}$ ; (iv) $2x + 3y = 7$

Worked solution (try it first)

(a)

Add the vectors component by component: $\mathbf x + \mathbf y = \begin{pmatrix} -2 + (-3) \\ 4 + 1 \end{pmatrix}$
$= \begin{pmatrix} -5 \\ 5 \end{pmatrix}$ .
Its length is $|\mathbf x + \mathbf y| = \sqrt{(-5)^2 + 5^2}$
$= \sqrt{50}$
$\approx 7.1$ .

(b)(i)

Midpoints average the coordinates: $M = \left(\frac{6 + (-2)}{2}, \frac{4 + (-2)}{2}\right)$
$= (2, 1)$ and $S = \left(\frac{6 + 4}{2}, \frac{4 + (-6)}{2}\right)$
$= (5, -1)$ .

(ii)

A vector from one point to another is "end minus start": $\overrightarrow{QR} = \begin{pmatrix} 4 - (-2) \\ -6 - (-2) \end{pmatrix}$
$= \begin{pmatrix} 6 \\ -4 \end{pmatrix}$ and $\overrightarrow{MS} = \begin{pmatrix} 5 - 2 \\ -1 - 1 \end{pmatrix}$
$= \begin{pmatrix} 3 \\ -2 \end{pmatrix}$ .

(iii)

$\overrightarrow{QR} = 2\overrightarrow{MS}$ : $QR$ is parallel to $MS$ and twice as long.

(iv)

Gradient of $MS = \frac{-1 - 1}{5 - 2} = -\frac23$ .
Through $M(2, 1)$ : $y - 1 = -\frac23(x - 2)$ .
Multiply by 3: $3y - 3 = -2x + 4$ , so $2x + 3y = 7$ .

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