IB Maths AI HL Vector Properties Paper 1 & 2 Resultant & nose-to-tail ~7 min read

Adding & Subtracting Vectors

Vectors combine component by component: add the i coefficients, then the j coefficients, then the k coefficients. The result is called the resultant — itself a vector. Geometrically, addition is “nose-to-tail”: the resultant goes from the start of the first vector to the end of the last. Subtraction reverses the second vector before adding: a − b = a + (−b). Multiplying by a scalar simply scales every component.

📘 What you need to know

Component-wise rule: (a₁, a₂, a₃)^T ± (b₁, b₂, b₃)^T = (a₁ ± b₁, a₂ ± b₂, a₃ ± b₃)^T.
Resultant = sum of two or more vectors. It’s also a vector, with the same component-wise rule.
Subtraction = adding the reverse: a − b = a + (−b). The minus sign flips every component.
Scalar multiplication: k(a₁, a₂, a₃)^T = (ka₁, ka₂, ka₃)^T. Distributes over sums.
Geometric “nose-to-tail”: place the start of b at the end of a; the resultant a + b goes from the start of a to the end of b.
Parallelogram rule: if a and b share a starting point, the resultant a + b is the diagonal of the parallelogram they form.

Adding component-by-component

For two vectors written in column form, line them up and add (or subtract) the entries row by row — same for any number of vectors. In base-vector form, collect like terms in i, j, and k separately. If one vector is in column form and the other in base-vector form, convert one to match before combining. Scalar multipliers distribute: k(a + b) = ka + kb.

Vector a (teal) plus vector b (orange) drawn nose-to-tail gives the resultant a + b (blue). The faint dashed lines complete the parallelogram — the diagonal reaches the same point, confirming the equivalent parallelogram rule.

Adding vectors (column form) (a₁, a₂, a₃)^T ± (b₁, b₂, b₃)^T = (a₁ ± b₁, a₂ ± b₂, a₃ ± b₃)^T in base-vector form: collect i, j, k coefficients separately

Scalar multiplication and solving vector equations

Multiplying a vector by a scalar scales every component. This combines with addition and subtraction in expressions like 2a − 3b: compute the multiples first, then combine. You’ll also see equations of the form kx + a = b — solve them exactly like a regular linear equation, rearranging to isolate x, then performing the component-wise arithmetic.

Strategy: pick one form (column or base-vector) and stick with it for the whole calculation. Mixing forms in the middle is where most arithmetic slips happen.

🧭 Recipe — combine vectors

Convert to one form: if some vectors are in column form and others in i–j–k form, write all of them as columns (usually easier).
Apply any scalar multipliers: scale each entry of the vector by the scalar before combining.
Add or subtract row by row: compute each component of the result independently.
Convert back if needed: if the question wants base-vector form, write each row as the coefficient of i, j, k.
For vector equations like kx + a = b, rearrange: x = (b − a) / k, then compute.

Worked examples

WE 1

Column addition and subtraction

Given a = (3, −1, 4)^T and b = (2, 5, −3)^T, find a + b and a − b.

add row by row a + b = (3+2, −1+5, 4+(−3)) = (5, 4, 1)^T subtract row by row a − b = (3−2, −1−5, 4−(−3)) = (1, −6, 7)^T a + b = (5, 4, 1)^T; a − b = (1, −6, 7)^T watch the sign on the third row of a − b: 4 − (−3) = +7.

WE 2

Adding in base-vector form

Given p = 4i − 2j + 3k and q = 2i + 5j − 6k, find p + q in base-vector form and column form.

collect i, j, k coefficients i: 4 + 2 = 6 j: −2 + 5 = 3 k: 3 + (−6) = −3 write both forms p + q = 6i + 3j − 3k = (6, 3, −3)^T three independent component sums — line up under each base vector.

WE 3

Mixed notation with a scalar multiple

Given u = (3, 0, −2)^T and v = 4i − j + 5k, find 2u − v.

convert v to column form v = (4, −1, 5)^T scalar multiply: 2u 2u = (6, 0, −4)^T subtract row by row 2u − v = (6−4, 0−(−1), −4−5) = (2, 1, −9)^T 2u − v = (2, 1, −9)^T = 2i + j − 9k middle row: 0 − (−1) = +1. Negative signs are the most common slip.

WE 4

Linear combination with two scalar multipliers

Given a = (2, −1, 3)^T and b = (1, 4, −2)^T, find 3a − 2b.

scalar multiples first 3a = (6, −3, 9)^T 2b = (2, 8, −4)^T subtract row by row 3a − 2b = (6−2, −3−8, 9−(−4)) = (4, −11, 13)^T 3a − 2b = (4, −11, 13)^T always compute the scalar multiples before combining — distributing first prevents sign errors.

WE 5

Three vectors with mixed signs

Given m = i − 2j + k, n = 3j − 2k, and p = −2i + j + 4k, find m − n + p.

write all three as columns m = (1, −2, 1)^T n = (0, 3, −2)^T p = (−2, 1, 4)^T combine row by row i: 1 − 0 + (−2) = −1 j: −2 − 3 + 1 = −4 k: 1 − (−2) + 4 = 7 m − n + p = −i − 4j + 7k = (−1, −4, 7)^T n has no i term — that’s a 0 in the top row. k coefficient: 1 − (−2) + 4 = 7.

WE 6

Solve a vector equation for x

Given a = (3, −1, 2)^T and b = (−1, 5, 8)^T, find the vector x that satisfies 2x + a = b.

rearrange like a normal equation 2x = b − a compute b − a row by row b − a = (−1−3, 5−(−1), 8−2) = (−4, 6, 6)^T divide each component by 2 x = (−2, 3, 3)^T x = −2i + 3j + 3k = (−2, 3, 3)^T check: 2(−2,3,3)^T + (3,−1,2)^T = (−4+3, 6−1, 6+2)^T = (−1, 5, 8)^T = b ✓

💡 Top tips

Pick one form and commit — convert any mixed-form question into all-columns or all-base-vector before doing arithmetic.
Compute scalar multiples first: write out 2u and 3v as columns before subtracting, to keep signs clean.
Missing terms = zeros — if a vector has no i component, put 0 in the first row of its column.
Verify by reverse-checking: add your result back to test the original equation, especially for “solve for x” questions.
Double-check sign rules: subtracting a negative flips to plus; multiplying a negative by a negative is positive.

⚠ Common mistakes

Sign errors on subtraction: 4 − (−3) = 7, not 1. The double-negative is the most common slip.
Forgetting a zero component: writing n = 3j − 2k as (3, −2)^T instead of (0, 3, −2)^T.
Scaling only some components: 2(i − j + 3k) = 2i − 2j + 6k, not 2i − j + 3k.
Combining different base vectors: 2i + 3j ≠ 5(i + j) — i and j are independent and cannot be merged.
Rearranging an equation incorrectly: from 2x + a = b, the next step is 2x = b − a, not x = b − a.

Next up — Position & Displacement Vectors. The position vector of a point is just the vector from the origin to that point; the displacement from A to B is the difference of their position vectors: AB with an arrow = b − a. Adding and subtracting fluently — the skill you just practiced — is exactly what those questions test.

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