This mini-lesson covers AQA section J — Vectors, in two and three dimensions: component form (i, j, k), magnitude and direction, unit vectors, addition and scalar multiplication, position vectors and the distance between two points, plus the geometric arguments — parallel vectors, midpoints and ratios — that AQA builds its longer questions from.
Work through each screen, answer the questions as you go (a few are conceptual, most are calculations) and collect ⭐ stars. Everything here is A-level standard, drawn from the AQA 7357 subject content. Press Start when you are ready.
A vector has both magnitude and direction; a scalar has magnitude only. At A-level you work in three dimensions as well as two.
Both notations mean the same thing — AQA uses each, and a column vector is a third way of writing it. A missing component is a zero: 3i + 4j in 3D is 3i + 4j + 0k.
Bold or underlined. In print vectors are bold (a); by hand you must underline them (a̲). Writing a plain "a" for a vector is ambiguous and can cost marks.
The magnitude (or modulus) is the length of the vector — three-dimensional Pythagoras:
a = 3i − 4j + 12k.
|a| = √(3² + (−4)² + 12²) = √(9 + 16 + 144) = √169 = 13.
Note (−4)² = +16 — the squares kill the minus sign, which is why the magnitude is always positive.
Direction: the angle a vector makes with the x-axis is found from
For r = 2i + 3j + 6k: |r| = √(4 + 9 + 36) = 7, so the angle with the x-axis is cos⁻¹(2/7) = 73.4° (1 d.p.).
A unit vector has magnitude exactly 1. To construct one in the direction of a, divide a by its own magnitude:
a = 6i + 6j − 3k. First |a| = √(36 + 36 + 9) = √81 = 9.
So â = (1/9)(6i + 6j − 3k) = (2/3)i + (2/3)j − (1/3)k.
Check: (2/3)² + (2/3)² + (1/3)² = 4/9 + 4/9 + 1/9 = 1 ✓ — magnitude 1, as required.
Typical question: "find λ such that λ(2i − j + 2k) is a unit vector". Since |2i − j + 2k| = √(4 + 1 + 4) = 3, we need λ = 1/3. (Strictly λ = ±1/3, since −1/3 gives a unit vector in the opposite direction.)
Tap a vector, then tap its magnitude. Use |a| = √(x² + y² + z²).
Vector algebra works component by component. Geometrically, addition is the triangle law: place them nose to tail and the resultant runs from the start to the finish.
a = 2i + 3j − k and b = i − j + 4k. Find a + 2b.
2b = 2i − 2j + 8k.
a + 2b = (2 + 2)i + (3 − 2)j + (−1 + 8)k = 4i + j + 7k.
Test for parallel: divide corresponding components. If you get the same λ from every one, they are parallel; if any component disagrees, they are not.
The position vector of a point A is the vector OA from the origin to A. If A = (1, 2, 2), its position vector is a = i + 2j + 2k.
A(1, 2, 2) and B(5, 4, 6).
AB = b − a = (5 − 1, 4 − 2, 6 − 2) = (4, 2, 4).
|AB| = √(4² + 2² + 4²) = √(16 + 4 + 16) = √36 = 6.
The order is everything. AB = b − a, not a − b. Getting it backwards gives you the right length but the opposite direction, which wrecks every subsequent step of a geometry question.
Vectors turn geometry into algebra. Two workhorse results:
A(1, 0, 2) and B(7, 6, 8). Find P, which divides AB in the ratio 1 : 2.
AB = b − a = (6, 6, 6). One third of the way along: p = a + ⅓(6, 6, 6) = (1, 0, 2) + (2, 2, 2) = (3, 2, 4).
Check: AP = (2, 2, 2) and PB = (4, 4, 4) = 2 × AP, so AP : PB = 1 : 2 ✓
Proving three points are collinear: show that AB = λ × BC for some scalar λ. The two vectors are then parallel and share the point B, so all three points lie on one straight line.
AQA explicitly requires you to use vectors in context — which links this section to the mechanics content.
The recurring error: adding magnitudes instead of vectors. Two forces of 5 N do not necessarily give 10 N — if they are perpendicular the resultant is √50 ≈ 7.07 N, and if they are opposite it is 0 N.
Tap an item on the left, then its partner on the right.
Longer AQA vector questions reward a clear method more than clever tricks.
Keep exact surds where you can — √3 rather than 1.73 — unless the question asks for a decimal. Rounding early is a needless way to lose accuracy marks.
Notation: a = xi + yj + zk — A-level vectors are 3D as well as 2D
Magnitude: |a| = √(x² + y² + z²) — always positive
Unit vector: â = a/|a| — divide by the magnitude
Direction: cos θ = x/|a| gives the angle with the x-axis
Parallel: b = λa for some scalar λ
Position vectors: AB = b − a · midpoint = ½(a + b) · collinear = parallel + a common point
In context: velocity and force are vectors; speed is |v|; equilibrium means the resultant is 0
That is AQA section J — Vectors. Press Finish to see your score.
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