OCR section 1.10 extends vectors into three dimensions. You need magnitude, unit vectors, position vectors, the distance between two points in 3D, and how to use vectors to prove geometric facts.
One spec detail matters: OCR asks for the direction of a vector in two dimensions only. Work through each screen, answer the questions as you go and collect ⭐ stars. Press Start when you're ready.
A vector has magnitude and direction; a scalar has magnitude only. Write vectors as a (bold, or underlined by hand) or as AB with an arrow over it.
|3i − 4j| = √(9 + 16) = √25 = 5.
|2i + 3j − 6k| = √(4 + 9 + 36) = √49 = 7.
A unit vector along 6i + 8j: its magnitude is √(36 + 64) = 10, so the unit vector is (6i + 8j)/10 = 0.6i + 0.8j. Check: √(0.36 + 0.64) = 1. ✓
Note the signs vanish: magnitude is always positive — the components are squared. |3i − 4j| and |3i + 4j| are both 5.
Add and subtract vectors component by component. Multiplying by a scalar k stretches the vector by a factor k (and reverses it if k is negative).
a + b = (4 − 2)i + (−1 + 5)j + (2 + 1)k = 2i + 4j + 3k.
|a + b| = √(2² + 4² + 3²) = √(4 + 16 + 9) = √29 = 5.39 (3 s.f.).
Parallel vectors: a is parallel to b exactly when a = kb for some scalar k. So 6i + 9j is parallel to 2i + 3j (with k = 3), but 6i + 9j is not parallel to 2i + 4j — no single k works for both components.
The position vector of a point A is OA, measured from the origin. The vector from A to B is then:
A is (1, 2, 3) and B is (4, 6, 15).
AB = (4 − 1)i + (6 − 2)j + (15 − 3)k = 3i + 4j + 12k.
Distance AB = √(3² + 4² + 12²) = √(9 + 16 + 144) = √169 = 13.
Order matters for the vector, not the distance. BA = −AB points the other way, but |BA| = |AB| = 13 either way.
In two dimensions you can quote a vector’s direction as the angle it makes with the positive x-direction.
tan θ = 4/3 = 1.3333 ⇒ θ = 53.1° (1 d.p.) above the positive x-direction.
Check: the vector points up and to the right (both components positive), so an angle in the first quadrant is right. ✓
Spec detail worth knowing: OCR extends vectors to three dimensions excluding the direction of a vector in three dimensions. So you will be asked for a 3D magnitude or distance, but the ‘angle it makes’ questions are 2D only.
Calculator trap: arctan always returns an angle between −90° and 90°. For −3i − 4j the calculator still says 53.1°, but the vector points into the third quadrant — the true direction is 53.1° + 180° = 233.1°. Sketch it.
Vectors prove geometric facts without coordinates.
Show that AB and BC are parallel (one is a scalar multiple of the other) and that they share the point B.
Example: A(1, 2), B(3, 6), C(6, 12). AB = 2i + 4j; BC = 3i + 6j = 1.5(2i + 4j). Parallel, sharing B ⇒ A, B and C are collinear. ✓
The midpoint M of AB has position vector (a + b)/2.
The point P dividing AB in the ratio 1:2 has position vector a + ⅓(b − a).
Parallel is not the same as collinear. Two parallel vectors can sit on completely separate lines. You need a common point as well.
Tap a vector, then tap the bin matching its magnitude.
Tap a description on the left, then the correct expression.
Vector vs scalar: a vector has magnitude and direction; a scalar has magnitude only
Magnitude: |xi + yj + zk| = √(x² + y² + z²) — always positive
Unit vector: a/|a| — same direction, length 1
Position vectors: AB = b − a; distance AB = |AB|
Parallel: a = kb; collinear also needs a shared point
Direction: tan θ = y/x — but OCR asks for direction in 2D only; sketch it to fix the quadrant
Midpoint: (a + b)/2
That is OCR 1.10 — and vectors are the natural language of the Mechanics papers. Press Finish to see your score.
You've worked through Vectors for OCR A-level Mathematics A. 🎉
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