5 Vectors and transformation geometry

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5.1 Vectors

A vector describes a movement with both magnitude (size) and direction. Two vectors are equal only when both properties match, even if they start at different points.

Quantity	What matters
scalar	size only
vector	size and direction
equal vectors	same size and same direction

Read an arrow from its tail to its head: the arrow length represents magnitude and the arrowhead fixes direction. Reversing the arrow produces the negative vector.

Parallel arrows are not automatically equal: they may point oppositely or have different magnitudes.

The vector from $O$ to $A$ is written $\overrightarrow{OA}$ and may be named $\mathbf a$ . A column vector $\begin{pmatrix}x\\y\end{pmatrix}$ means move $x$ horizontally and $y$ vertically.

Component	Positive	Negative
top, $x$	right	left
bottom, $y$	up	down

Order matters: $\overrightarrow{AO}=-\overrightarrow{OA}=-\mathbf a$ . Translate a diagram into notation by naming the start point first and the end point second.

The entries of a column vector are displacements, not the coordinates of its endpoint unless the vector starts at the origin.

Multiplying a vector by a scalar $k$ multiplies every component by $k$ . Its magnitude is multiplied by $|k|$ ; a negative scalar also reverses its direction.

Scalar $k$	Effect on vector
$k>1$	same direction, longer
$0<k<1$	same direction, shorter
$k=0$	zero vector
$k<0$	reversed direction, scaled by $\|k\|$

If $\mathbf a=\begin{pmatrix}3\\-2\end{pmatrix}$ , then $-2\mathbf a=\begin{pmatrix}-6\\4\end{pmatrix}$ .

Do not multiply only one component; the scalar acts on the whole vector.

Add vectors by joining movements head-to-tail or by adding corresponding components. Subtracting a vector means adding its reverse: $\mathbf a-\mathbf b=\mathbf a+(-\mathbf b)$ .

Operation	Column rule
addition	$\begin{pmatrix}x_1\\y_1\end{pmatrix}+\begin{pmatrix}x_2\\y_2\end{pmatrix}=\begin{pmatrix}x_1+x_2\\y_1+y_2\end{pmatrix}$
subtraction	subtract top from top and bottom from bottom
route	$\overrightarrow{AC}=\overrightarrow{AB}+\overrightarrow{BC}$

$\begin{pmatrix}5\\3\end{pmatrix}-\begin{pmatrix}-2\\4\end{pmatrix}=\begin{pmatrix}7\\-1\end{pmatrix}$ : brackets protect the signs.

A route must connect head-to-tail. If an arrow points the wrong way, reverse it and change its sign before combining.

For $\mathbf v=\begin{pmatrix}x\\y\end{pmatrix}$ , its modulus is the non-negative length $|\mathbf v|=\sqrt{x^2+y^2}$ , found using Pythagoras' theorem.

Step	Action
1	identify horizontal and vertical components
2	square both components
3	add the squares
4	take the positive square root

$\left|\begin{pmatrix}9\\12\end{pmatrix}\right|=\sqrt{9^2+12^2}=15$ .

The modulus is a scalar, so it has no direction and cannot be negative. Squaring a negative component makes a positive contribution.

A resultant is the single vector with the same overall effect as two or more successive vectors. Follow a continuous route and add every directed segment.

Route fact	Vector equation
$O\to A\to B\to C$	$\overrightarrow{OC}=\overrightarrow{OA}+\overrightarrow{AB}+\overrightarrow{BC}$
reverse a segment	$\overrightarrow{BA}=-\overrightarrow{AB}$
closed route	vector sum is $\mathbf0$

Choose a route whose start and finish match the required resultant, rewrite every segment in the stated base vectors, then collect the coefficients of each base vector.

Do not add undirected lengths or rely on the visual angle of a diagram; resultant calculations use directed vector equations.

A vector proof translates each geometric condition into an exact vector relation. Equal vectors establish equal directed sides; non-zero scalar multiples establish parallel lines, and a specified fraction locates a division point.

Geometric fact	Vector evidence
midpoint $M$ of $AB$	$\overrightarrow{AM}=\tfrac12\overrightarrow{AB}$
parallel lines	one direction vector is a non-zero scalar multiple of the other
same point by two routes	the two position-vector expressions are equal
collinear points	their connecting vectors are scalar multiples

Example: if $\overrightarrow{OC}=\tfrac13\mathbf a$ and $\overrightarrow{OD}=\tfrac13\mathbf b$ , then $\overrightarrow{CD}=\tfrac13(\mathbf b-\mathbf a)$ . Since $\overrightarrow{AB}=\mathbf b-\mathbf a$ , $CD$ is parallel to $AB$ .

Finishing with an expression is not a proof. State the geometric conclusion justified by the equality or scalar-multiple relationship, and exclude the zero-vector case when asserting a direction.

5.2 Transformation geometry

A rotation turns every point through the same angle about one fixed centre. A complete specification states the centre, angle and direction.

Required part	Meaning
centre	fixed point of the turn
angle	amount of turn
direction	clockwise or anticlockwise

Every point and its image are the same distance from the centre, and the angle between their centre-lines is the rotation angle.

Saying only 'rotation' or giving an angle without a centre is incomplete.

Rotate each vertex about the stated centre, keeping its distance from the centre unchanged, then join the image vertices in the original order.

Step	Action
1	mark the centre
2	trace centre-to-vertex displacement
3	turn that displacement through the given angle
4	plot the image vertex and repeat

For a $180^\circ$ turn about $(a,b)$ , $(x,y)$ maps to $(2a-x,2b-y)$ .

Do not rotate about the origin unless the stated centre is the origin.

By convention, anticlockwise rotations have positive angles and clockwise rotations have negative angles.

Description	Equivalent description
$90^\circ$ clockwise	$-90^\circ$ or $270^\circ$ anticlockwise
$90^\circ$ anticlockwise	$+90^\circ$ or $270^\circ$ clockwise
$180^\circ$	same result in either direction

Keep the centre fixed while deciding direction; the sign describes the turn, not a coordinate sign.

Positive does not mean clockwise, and a full $360^\circ$ change gives the original position.

A reflection maps every point across a mirror line. The mirror line is the perpendicular bisector of the segment joining a point to its image.

Mirror line	Coordinate effect
$x=a$	horizontal distance to $x=a$ changes side
$y=b$	vertical distance to $y=b$ changes side
$y=x$	$(x,y)\mapsto(y,x)$
$y-x=0$	same line as $y=x$

A point on the mirror line stays fixed; paired points lie at equal perpendicular distances on opposite sides.

The mirror line is not usually the line joining a point to its image; it crosses that segment at right angles halfway along.

To reflect a shape, send each vertex along a perpendicular to the mirror line by the same distance to the opposite side. To recover the mirror line, construct perpendicular bisectors of point-image pairs.

Given	Construction
mirror line	measure perpendicular distance for every vertex, copy it across
object and image	join matching vertices, mark midpoints, draw their common perpendicular bisector

Join the reflected vertices in corresponding order and verify that lengths and angles match the original.

Measuring horizontal or vertical distance works only for vertical or horizontal mirror lines; oblique lines require perpendicular distance.

A translation slides every point by the same directed displacement. Its distance and direction are identical for all corresponding point pairs.

Feature	Translation effect
movement	same distance and direction for every point
orientation	unchanged
fixed centre or line	none required

Compare any vertex with its image: the horizontal and vertical changes must match those for every other vertex.

A translation does not turn, flip or resize the shape.

Choose each vertex, apply the same horizontal and vertical displacement, plot its image, then reconnect the vertices in the same order.

Step	Action
1	identify a matching start vertex
2	count horizontal movement
3	count vertical movement
4	repeat exactly for every vertex

Corresponding sides remain parallel and equal, and the image has the same orientation and size.

Do not repeatedly move from the previous image vertex; apply the displacement independently to each original vertex.

The translation vector $\begin{pmatrix}a\\b\end{pmatrix}$ moves each point $a$ units horizontally and $b$ units vertically: $(x,y)\mapsto(x+a,y+b)$ .

Entry	Positive	Negative
top, $a$	right	left
bottom, $b$	up	down

The vector $\begin{pmatrix}-4\\3\end{pmatrix}$ means 4 left and 3 up.

A translation vector uses a column, not coordinate notation; its entries describe a change rather than a location.

Rotations, reflections and translations are rigid transformations: they preserve all lengths and angles, so the image is congruent to the original.

Transformation	Lengths	Angles	Orientation
rotation	preserved	preserved	preserved
translation	preserved	preserved	preserved
reflection	preserved	preserved	reversed

Corresponding side lengths, angle sizes, perimeter and area are unchanged under a rigid transformation.

Congruent does not mean identical position or orientation; an enlargement with scale factor other than 1 is not rigid.

An enlargement is specified by a centre and a positive scale factor $k$ . Each image point lies on the ray from the centre through the original point, at $k$ times the original distance.

Scale factor	Effect
$k>1$	image farther from centre and larger
$k=1$	unchanged
$0<k<1$	image between centre and original, smaller

Lines joining corresponding vertices pass through the centre of enlargement and their distance ratios equal $k$ .

This syllabus specifies positive scale factors only; do not introduce negative enlargements into this objective.

An enlargement preserves corresponding angles and multiplies every corresponding length by the same scale factor $k$ .

Measure	Factor
angle	unchanged
length and perimeter	$k$
area	$k^2$

The image is similar to the original. It is congruent only when $k=1$ .

Preserved angles do not imply preserved lengths; for $k=3$ , every length triples.

For each vertex, draw or imagine a ray from the centre through that vertex and place the image at $k$ times the centre-to-vertex distance.

Step	Action
1	locate the centre
2	form centre-to-vertex rays
3	multiply each displacement by $k$
4	join image vertices in order

With centre $(a,b)$ , $(x,y)$ maps to $(a+k(x-a),\ b+k(y-b))$ .

Multiplying coordinates directly by $k$ works only when the centre is the origin.

Identify the single transformation by comparing size, orientation and point movement, then state every parameter required for that type.

Type	Complete description needs
translation	column vector
reflection	mirror-line equation
rotation	centre, angle and direction
enlargement	centre and positive scale factor

Equal size suggests a rigid transformation; reversed orientation suggests reflection; changed size suggests enlargement. Confirm with corresponding vertices before naming it.

Do not list several transformations when asked for a single one, and do not confuse a centre coordinate $(a,b)$ with a translation vector $\begin{pmatrix}a\\b\end{pmatrix}$ .

5 Vectors and transformation geometry

Exam analysis

Most tested topics

5.1 Vectors

Recognise magnitude and direction in a vector

Read vector and column-vector notation

Multiply a vector by a scalar

Add and subtract vectors

Calculate the modulus of a vector

Find a resultant vector

Prove geometrical facts with vectors

5.2 Transformation geometry

Specify a rotation completely

Rotate a shape about a given point

Use signs for rotation direction

Specify a reflection by its mirror line

Construct mirror lines and reflected shapes

Describe a translation by distance and direction

Translate a shape

Use column vectors for translations

Recognise rigid transformations and congruence

Specify an enlargement completely

Track angle and length effects of enlargement

Enlarge a shape from a centre and scale factor

Give complete transformation descriptions