Vectors and Transformations

This lesson covers the algebra and geometry of vectors, combined and inverse transformations, and their role in proofs and similarity arguments.

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Vectors

A vector describes a displacement (movement) — it has magnitude (size) and direction.

Column vector notation: a displacement of a right and b up is written as (a, b).

Magnitude: |(a, b)| = √(a² + b²)

Example: |(3, −4)| = √(9 + 16) = 5

Operations

Operation	Rule	Example
Addition	(a, b) + (c, d) = (a+c, b+d)	(2, 3) + (−1, 5) = (1, 8)
Subtraction	(a, b) − (c, d) = (a−c, b−d)	(4, 1) − (6, −2) = (−2, 3)
Scalar multiplication	k(a, b) = (ka, kb)	3(2, −1) = (6, −3)

Position Vectors

The position vector of a point P is the vector from the origin O to P, written as p or OP.

If A has position vector a and B has position vector b, then: AB = b − a (vector from A to B)

Example: A = (1, 3) and B = (5, −2). AB = (5−1, −2−3) = (4, −5) BA = (1−5, 3−(−2)) = (−4, 5) (= −AB)

Parallel Vectors

Two vectors are parallel if one is a scalar multiple of the other. (2, 6) and (−3, −9) are parallel because (−3, −9) = −1.5 × (2, 6).

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Vectors in Geometric Proofs [H]

Use vectors to prove properties of shapes (e.g. midpoints, parallel lines, collinear points).

Key tools:

• If AB = k × CD, then AB is parallel to CD (and k times its length).
• Points P, Q, R are collinear if PQ = k × PR for some scalar k.
• The midpoint M of AB has position vector: m = (a + b)/2.

Example: OACB is a parallelogram. O is origin; a and b are position vectors of A and B. OC = OA + AC = OA + OB = a + b Midpoint of OC: m = (a + b)/2 Midpoint of AB: (b + a)/2 — these are equal, so diagonals bisect each other. ✓

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Transformations (Review and Extension)

Reflection, Rotation, Translation, Enlargement

(See KS3 lesson for core rules. Here we extend to combined transformations and negative scale factors.)

Negative Scale Factor Enlargement

An enlargement with a negative scale factor places the image on the opposite side of the centre and rotates it through 180°.

Example: Enlarge point (2, 1) by scale factor −3 from centre (0, 0): Image = −3 × (2, 1) = (−6, −3)

Combined Transformations

Apply transformations in order (right to left if written as a composition).

Example: Triangle T has vertices A(1, 2), B(3, 2), C(3, 4). Step 1 — Reflect in the x-axis (y = 0): (x, y) → (x, −y) A′(1, −2), B′(3, −2), C′(3, −4) Step 2 — Rotate 90° anticlockwise about the origin: (x, y) → (−y, x) A″(2, 1), B″(2, 3), C″(4, 3)

Now compare A(1, 2)→A″(2, 1), B(3, 2)→B″(2, 3), C(3, 4)→C″(4, 3). The mapping (x, y) → (y, x) is a reflection in the line y = x. So the single equivalent transformation is a reflection in y = x.

Invariant Points