IB Maths AA HL Topic 3 — Geometry & Trigonometry Paper 1 & 2 ~5 min read HL only

Adding & Subtracting Vectors

Add corresponding components — that’s all there is to it. The sum is called the resultant. Geometrically, place vectors nose-to-tail and the resultant goes from start to finish. Subtraction is just adding the reverse: ab = a + (−b).

📘 What you need to know

The component rule

Component-wise addition & subtraction a ± b  =  (a₁ ± b₁,   a₂ ± b₂,   a₃ ± b₃)

In base vector form, that means collecting i, j, and k terms separately.

(3i − 2j + 5k) + (i + 4jk) = 4i + 2j + 4k
If a term is missing from a base form vector, treat it as 0. i − 4k means the j coefficient is 0 — handy for clean addition.

The geometric picture

Addition (nose-to-tail)
a + b
place tail of b at head of a; resultant goes from start of a to end of b
Subtraction (reverse + add)
ab = a + (−b)
flip b‘s direction, then add nose-to-tail
Two vectors with the same start? Their sum is the diagonal of the parallelogram they form. Their difference (ab) is the vector from the head of b to the head of a.

🧭 Recipe — add or subtract two vectors

  1. Write both in the same form (column or base vector).
  2. Fill in zeros for any missing components.
  3. Operate on each component separately (top with top, middle with middle, bottom with bottom).
  4. Watch the signs — subtraction flips every component of the second vector.
  5. Convert back to the form the question asks for.

Worked examples

WE 1

Add two column vectors

Given u = 4−31 and v = 25−7, find u + v.

Add corresponding components x: 4 + 2 = 6 y: −3 + 5 = 2 z: 1 + (−7) = −6 u + v = 62−6
WE 2

Subtract two column vectors

Given p = −148 and q = 3−25, find pq.

Subtract corresponding components x: −1 − 3 = −4 y: 4 − (−2) = 4 + 2 = 6 z: 8 − 5 = 3 p − q = −463 subtracting a negative becomes addition — easy place to drop a sign
WE 3

Subtract base vectors with a missing term

Given u = 6i + 4k and v = 5i − 3j + 2k, find uv in base vector form.

Step 1: Fill in the missing j term in u (coefficient 0) u = 6i + 0j + 4k Step 2: Subtract like terms i: 6 − 5 = 1 j: 0 − (−3) = 3 k: 4 − 2 = 2 u − v = i + 3j + 2k
WE 4

Multi-step combination with scalars

Given a = (2, 1, −3), b = (−1, 4, 0), and c = (5, −2, 6), find 2a + bc.

Step 1: Compute 2a (scalar multiplication first) 2a = (4, 2, −6) Step 2: Add b 2a + b = (4 + (−1), 2 + 4, −6 + 0) = (3, 6, −6) Step 3: Subtract c (3 − 5, 6 − (−2), −6 − 6) = (−2, 8, −12) 2a + b − c = (−2, 8, −12) do scalar multiples first, then add and subtract componentwise
WE 5

Find unknown components from a known sum

Given a = (3, p, 5) and b = (−1, 4, q), and that a + b = (2, 7, −1), find the values of p and q.

Equate components of a + b with the given resultant Step 1: x-check (no unknowns here) 3 + (−1) = 2 ✓ Step 2: y-component p + 4 = 7 → p = 3 Step 3: z-component 5 + q = −1 → q = −6 p = 3, q = −6
WE 6

Resultant velocity of boat in current

A boat sails at velocity v = 6i + 2j m/s in still water. A current pushes it with velocity w = −i + 3j m/s. Find the boat’s resultant velocity.

Step 1: Resultant velocity = v + w v + w = (6i + 2j) + (−i + 3j) Step 2: Collect like terms i: 6 + (−1) = 5 j: 2 + 3 = 5 resultant velocity = 5i + 5j m/s the resultant tells you the boat’s actual motion over the ground

💡 Top tips

⚠ Common mistakes

Next note: Position & Displacement Vectors. The position vector OA places a point relative to the origin, while displacement vectors AB = ba link any two points — the bridge between coordinates and vectors.

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