A number on its own can tell you how much — a temperature, a price, a length. But some quantities also need a direction: a push of 10 newtons means nothing until you say which way, and "walk 3 km" is incomplete without a heading. A vector is the mathematical object that carries both at once — a size and a direction — and this topic is about handling them cleanly.
This first page is the foundation, and it is also the part the CSCA tests most directly: the dated papers almost always include one vector question, and it is usually straightforward component arithmetic — combining two vectors like 3a − 2b. So we will build that skill until it is automatic. We start with what a vector actually is, move to writing it in components, then add, subtract, and scale them — always tying the algebra to a picture so the symbols always mean something you can see.
You can learn anything. Let's do this!
Vectors and scalars
An ordinary number — one with size but no direction — is called a scalar. Mass, time, temperature, and plain lengths are scalars. A vector has both a size (its magnitude) and a direction. We picture it as an arrow: the length of the arrow is the magnitude, and where it points is the direction.
A vector does not care where it sits on the page — only how long it is and which way it points. Slide an arrow somewhere else without rotating or resizing it, and it is the same vector. We write vectors with a small arrow, like a, or as the arrow between two points, like AB (read "the vector from A to B").
A vector is defined by length and direction only, not position. These two arrows are the same vector a.
Vector vs scalar. A scalar has size only (a plain number). A vector has size and direction, drawn as an arrow. Two arrows of equal length pointing the same way are the same vector, wherever they are drawn.
Try it! · Check-in question
Which of the following is a vector quantity rather than a scalar?
A displacement carries both a size (5 km) and a direction (north-east), so it is a vector. The answer is B. Temperature (A), mass (C), and a count of students (D) each have a size but no direction, so they are scalars.
Component form
Drawing arrows is fine for intuition, but to calculate we need numbers. Place the vector on the coordinate plane and describe it by how far it goes across and how far up: those two numbers are its components. A vector that moves 3 right and 2 up is written ⟨3, 2⟩ — the first number is the horizontal change, the second the vertical change.
The most common task is finding the vector between two points. To go from A to B you ask "how far across, how far up" — which is just B's coordinates minus A's. The rule is tip minus tail: the vector AB points from A (the tail) to B (the tip), so you subtract the tail from the tip.
The vector from A(1, 1) to B(4, 3) is ⟨4 − 1, 3 − 1⟩ = ⟨3, 2⟩ — tip minus tail.
Component form and "tip minus tail". A vector is written ⟨x, y⟩, the horizontal change then the vertical change. For two points, AB = ⟨xB − xA, yB − yA⟩ — the tip's coordinates minus the tail's.
Worked example · vector between two points
Find the vector AB from A(2, −1) to B(5, 3).
Use tip minus tail: subtract A's coordinates from B's.
Horizontal: 5 − 2 = 3.
Vertical: 3 − (−1) = 4.
So AB = ⟨3, 4⟩.
The order matters: AB is B minus A, not A minus B. Reversing it would give BA = ⟨−3, −4⟩, the same arrow pointing the opposite way.
Watch the subtraction order.AB means "from A to B," so it is B − A (tip minus tail). A frequent slip is computing A − B, which flips every sign and points the arrow backwards. Whenever you see AB, the second letter is the tip — subtract the first from the second.
Try it! · Check-in question
Given A(−1, 4) and B(3, 1), the vector BA is ( ).
BA goes from B (the tail) to A (the tip). Tip minus tail: A − B = ⟨−1 − 3, 4 − 1⟩ = ⟨−4, 3⟩. The answer is D. Option A computes B − A, which is AB, the arrow pointing the other way. Option B adds the coordinates instead of subtracting: ⟨−1 + 3, 4 + 1⟩ = ⟨2, 5⟩. Option C subtracts in the right order for x but in the wrong order for y: 1 − 4 = −3.
Worked example · a vector to a midpoint
The midpoint of segment AB is M, where A(1, 0) and B(3, 4). Find the vector AM.
First find M, the midpoint, by averaging the coordinates: M = ⟨(1 + 3)/2, (0 + 4)/2⟩ = (2, 2).
Now take tip minus tail for AM: ⟨2 − 1, 2 − 0⟩.
So AM = ⟨1, 2⟩.
A neat check: AM should be exactly half of AB. Here AB = ⟨3 − 1, 4 − 0⟩ = ⟨2, 4⟩, and half of that is ⟨1, 2⟩ ✓ — the midpoint is halfway along, so its vector from A is half the full vector.
Try it! · Check-in question
M is the midpoint of AB, with A(2, 1) and B(6, 5). The vector BM is ( ).
M = ((2 + 6)/2, (1 + 5)/2) = (4, 3). BM starts at B and ends at M, so take M − B: ⟨4 − 6, 3 − 5⟩ = ⟨−2, −2⟩. The answer is A. Equivalently, BM is half of BA = ⟨−4, −4⟩. Option B computes B − M, which is MB, the reversed arrow. Option C is the full BA; it forgets to halve. Option D gives the coordinates of the point M, not the vector from B.
Adding and subtracting vectors
Vectors add component by component: line up the horizontals and add them, line up the verticals and add them. ⟨a, b⟩ + ⟨c, d⟩ = ⟨a + c, b + d⟩. There is nothing more to it algebraically — and the picture explains why.
Geometrically, adding vectors is tip to tail: draw the first arrow, then start the second where the first ended; the sum is the arrow from the very start to the very end. Walk 3-right-2-up, then 1-right-3-up, and you have ended 4-right-5-up — which is exactly ⟨3, 2⟩ + ⟨1, 3⟩ = ⟨4, 5⟩.
Tip to tail: the second arrow starts where the first ends, and the sum runs from the original start to the final tip. The components simply add.
Subtraction works the same way, component by component: ⟨a, b⟩ − ⟨c, d⟩ = ⟨a − c, b − d⟩. (And notice "tip minus tail" from earlier is the same rule written another way: AB = OB − OA, where OA and OB are the vectors from the origin O to A and B.)
Worked example · adding and subtracting
Given a = ⟨5, −2⟩ and b = ⟨1, 4⟩, find a + b and a − b.
For the sum, add matching components: ⟨5 + 1, −2 + 4⟩ = ⟨6, 2⟩.
For the difference, subtract matching components: ⟨5 − 1, −2 − 4⟩ = ⟨4, −6⟩.
Keep the horizontals with the horizontals and the verticals with the verticals — the two slots never mix.
Try it! · Check-in question
If a = ⟨3, −1⟩ and a + b = ⟨1, 6⟩, then b = ( ).
Take a away from the sum: b = (a + b) − a = ⟨1 − 3, 6 − (−1)⟩ = ⟨−2, 7⟩. The answer is C. Check: ⟨3, −1⟩ + ⟨−2, 7⟩ = ⟨1, 6⟩ ✓. Option A adds a instead of taking it away: ⟨1 + 3, 6 + (−1)⟩ = ⟨4, 5⟩. Option B subtracts in the wrong order, a − (a + b), which gives −b. Option D treats 6 − (−1) as 6 − 1 = 5; subtracting −1 adds 1.
Scalar multiplication
Multiplying a vector by a plain number — a scalar — stretches or shrinks it, and flips it if the number is negative. You multiply each component by the scalar: k·⟨x, y⟩ = ⟨kx, ky⟩. Doubling a vector makes it twice as long in the same direction; multiplying by −1 keeps the length but reverses the arrow.
Scalar multiplication. k·⟨x, y⟩ = ⟨kx, ky⟩ — multiply both components by k. The direction stays the same for k > 0 and reverses for k < 0; the length is multiplied by |k|.
Worked example · scaling a vector
Given a = ⟨3, −2⟩, find 4a and −2a.
For 4a, multiply each component by 4: ⟨4·3, 4·(−2)⟩ = ⟨12, −8⟩.
For −2a, multiply each component by −2: ⟨−2·3, −2·(−2)⟩ = ⟨−6, 4⟩.
Notice −2a points the opposite way to a (every sign flipped) and is twice as long. The negative scalar both reverses and stretches.
Try it! · Check-in question
Which vector points in the opposite direction to a = ⟨−1, 3⟩ and is twice as long?
Opposite direction needs a negative scalar. Twice as long needs size 2. So multiply by −2: −2⟨−1, 3⟩ = ⟨2, −6⟩. The answer is C. Option A multiplies by +2: twice as long, but in the same direction. Option B multiplies by −1: the opposite direction, but the same length. Option D multiplies by −1/2: the opposite direction, but half as long.
Combining it all: linear combinations
Here is the question the CSCA actually asks. It hands you two vectors and asks for a combination like 3b − a or a − 2b. This is just the three operations above, run together: scale each vector first, then add or subtract component by component. Take it one slot at a time and it is hard to go wrong.
Computing m·a + n·b. (1) Scale: multiply a by m and b by n, each component. (2) Combine: add (or subtract) the results component by component. Do the horizontal slot and the vertical slot separately so signs never tangle.
Worked example · a linear combination (March 2026, Q44)
Given a = ⟨−2, 3⟩ and b = ⟨2, 1⟩, find a − 2b.
Scale b by 2 first: 2b = ⟨4, 2⟩.
Now subtract, component by component. Horizontal: −2 − 4 = −6.
Vertical: 3 − 2 = 1.
So a − 2b = ⟨−6, 1⟩.
The reliable habit: deal with the scalar multiple before the subtraction, and keep the two components on separate lines. Rushing both at once is where the sign on the −6 gets dropped.
Try it! · Check-in question
Given a = ⟨1, −2⟩ and b = ⟨3, 1⟩, find 2a + b.
Scale first: 2a = ⟨2, −4⟩. Then add b: ⟨2 + 3, −4 + 1⟩ = ⟨5, −3⟩. The answer is A. Option B forgets to scale a and finds a + b = ⟨4, −1⟩. Option C doubles b instead of a: a + 2b = ⟨1 + 6, −2 + 2⟩ = ⟨7, 0⟩. Option D doubles only the first component of a: ⟨2, −2⟩ + ⟨3, 1⟩ = ⟨5, −1⟩.
Try it! · Check-in question
Given a = ⟨4, −1⟩ and b = ⟨2, 3⟩, the vector c with a + 2c = b is ( ).
Solve it like an ordinary equation. Subtract a from both sides: 2c = b − a = ⟨2 − 4, 3 − (−1)⟩ = ⟨−2, 4⟩. Divide by 2: c = ⟨−1, 2⟩. The answer is D. Check: a + 2c = ⟨4 − 2, −1 + 4⟩ = ⟨2, 3⟩ = b ✓. Option A finds 2c but forgets to divide by 2. Option B adds a instead of subtracting it: (a + b)/2 = ⟨6, 2⟩/2 = ⟨3, 1⟩. Option C subtracts in the wrong order: (a − b)/2 = ⟨2, −4⟩/2 = ⟨1, −2⟩.
Magnitude: the length of a vector
The magnitude of a vector is its length — how long the arrow is — written |a|. Since the components ⟨x, y⟩ are the horizontal and vertical legs of a right triangle, and the vector itself is the hypotenuse, the length comes straight from Pythagoras: square the components, add, square-root.
The components are the legs of a right triangle; the vector is the hypotenuse. For ⟨3, 4⟩ the length is √(3² + 4²) = √25 = 5.
Magnitude. For a = ⟨x, y⟩, the length is |a| = √(x² + y²). It is the distance formula applied to a vector — Pythagoras on its two components. A magnitude is never negative.
Worked example · magnitude of a vector
Find the magnitude of a = ⟨−6, 8⟩.
Square each component: (−6)² = 36 and 8² = 64.
Add them: 36 + 64 = 100.
Square-root: |a| = √100 = 10.
The negative sign on the −6 vanishes when squared. A length cannot be negative, and squaring first is what makes sure the answer is positive.
Try it! · Check-in question
The vector ⟨5, k⟩ has magnitude 13. A possible value of k is ( ).
Set the length equal to 13: √(5² + k²) = 13. Square both sides: 25 + k² = 169, so k² = 144 and k = 12 or k = −12. Only 12 is offered. The answer is B. Option A subtracts the lengths: 13 − 5 = 8. Lengths combine by Pythagoras, not by plain subtraction. Option C stops at k² = 144 and forgets the square root. Option D adds the squares instead of subtracting them: 169 + 25 = 194.
Your turn! · Practice questions
Practice · Question 1
Given A(3, −2) and AB = ⟨−4, 7⟩, the coordinates of B are ( ).
Tip minus tail says AB = B − A, so B = A + AB = (3 + (−4), −2 + 7) = (−1, 5). The answer is D. Check: B − A = ⟨−1 − 3, 5 − (−2)⟩ = ⟨−4, 7⟩ ✓. Option A subtracts the vector from A: (3 − (−4), −2 − 7) = (7, −9). Option B gives the vector's components as the point; it forgets to start from A. Option C subtracts A from the vector: (−4 − 3, 7 − (−2)) = (−7, 9).
Practice · Question 2
If a = ⟨6, −1⟩ and a − b = ⟨8, −5⟩, then b = ( ).
Rearrange a − b = ⟨8, −5⟩ to get b alone: b = a − (a − b) = ⟨6 − 8, −1 − (−5)⟩ = ⟨−2, 4⟩. The answer is A. Check: ⟨6, −1⟩ − ⟨−2, 4⟩ = ⟨8, −5⟩ ✓. Option B adds instead: a + (a − b) = ⟨14, −6⟩. Option C subtracts in the wrong order, (a − b) − a, which gives −b. Option D treats −1 − (−5) as −1 − 5 = −6; subtracting −5 adds 5.
Practice · Question 3
If a = ⟨−4, 3⟩, then |−3a| = ( ).
Scale first: −3a = ⟨12, −9⟩. Its length is √(12² + (−9)²) = √(144 + 81) = √225 = 15. The answer is B. A quicker route: |a| = √(16 + 9) = 5, and the length is multiplied by |−3| = 3, giving 15. Option A multiplies the length by −3; a length is never negative. Option C forgets the scalar and gives |a|. Option D adds the sizes of the components, 12 + 9 = 21, instead of using Pythagoras.
Practice · Question 4
Given a = ⟨1, 2⟩ and b = ⟨3, −1⟩, find 2a − 3b.
Scale first: 2a = ⟨2, 4⟩ and 3b = ⟨9, −3⟩. Then subtract: ⟨2 − 9, 4 − (−3)⟩ = ⟨−7, 7⟩. The answer is C. Option A adds 3b instead of subtracting it: ⟨2 + 9, 4 + (−3)⟩ = ⟨11, 1⟩. Option B treats 4 − (−3) as 4 − 3 = 1; subtracting −3 adds 3. Option D forgets to multiply b by 3 and finds 2a − b = ⟨2 − 3, 4 + 1⟩ = ⟨−1, 5⟩.
Practice · Question 5
For the points P(1, −2) and Q(4, 2), the length |PQ| is ( ).
First the vector, tip minus tail: PQ = ⟨4 − 1, 2 − (−2)⟩ = ⟨3, 4⟩. Then its length: √(3² + 4²) = √(9 + 16) = √25 = 5. The answer is A. Option B adds the components, 3 + 4 = 7, without Pythagoras. Option C forgets to square before adding: √(3 + 4) = √7. Option D forgets to subtract P and uses Q's coordinates alone: √(4² + 2²) = √20 = 2√5.
Practice · Question 6
M is the midpoint of AB, with A(−2, 3) and M(1, 1). The vector AB is ( ).
First AM = ⟨1 − (−2), 1 − 3⟩ = ⟨3, −2⟩. M is halfway along AB, so AB is twice AM: AB = ⟨6, −4⟩. The answer is D. Check: B = A + AB = (4, −1), and the midpoint of (−2, 3) and (4, −1) is (1, 1) ✓. Option A stops at AM and forgets to double. Option B gives the coordinates of the point B, not the vector from A. Option C subtracts in the wrong order, A − M, and then doubles; that is BA.
Things to remember
Concept
What to know
Vector vs scalar
A scalar has size only; a vector has size and direction (an arrow). Position doesn't matter — same length and direction means the same vector.
Component form
⟨x, y⟩ = horizontal change, then vertical change.
Tip minus tail
AB = ⟨xB − xA, yB − yA⟩. The second letter is the tip — subtract the first from it. AB and BA are opposites.
Add / subtract
Component by component: ⟨a, b⟩ ± ⟨c, d⟩ = ⟨a ± c, b ± d⟩. Geometrically, addition is tip to tail.
Scalar multiple
k⟨x, y⟩ = ⟨kx, ky⟩. Length × |k|; direction reverses if k < 0.
Linear combination
For m·a + n·b: scale each vector first, then combine. Do each component slot separately.
Magnitude
|⟨x, y⟩| = √(x² + y²) — Pythagoras on the components. Always ≥ 0; squaring removes any minus signs.
