MathematicsCore20 min read

Vectors in a Plane

Quantities that carry a direction, and the algebra that handles them

This topic appears in:

10th Class Mathematics · ch.62nd Year Mathematics · ch.7O Level Mathematics (4024) · ch.7
01

Some quantities need a direction to make sense

A scalar is fully described by a number and a unit: a mass of 5 kg, a temperature of 30°C, a length of 8 m. A vector also needs a direction: a displacement of 8 m north-east, a velocity of 60 km/h due south.

A vector is drawn as an arrow: its length represents the magnitude and its direction represents the direction. Two arrows of the same length pointing the same way represent the same vector, wherever they are drawn on the page. That is why a vector can be slid around freely without changing.

v = ⟨a, b⟩ = a i + b j|v| = √(a² + b²)unit vector v̂ = v / |v|i and j are the unit vectors along the x and y axes
a
the horizontal componentpositive to the right
b
the vertical componentpositive upward
|v|
the magnitude, or lengthalways ≥ 0
02

Adding vectors: components, or the triangle rule

There are two ways to add, and they always agree. Geometrically, place the tail of the second arrow at the head of the first; the sum is the arrow from the start of the first to the end of the second. This is the triangle rule, and its twin, the parallelogram rule, is the same statement drawn differently.

Algebraically, just add the components: ⟨a, b⟩ + ⟨c, d⟩ = ⟨a + c, b + d⟩. This is far quicker and is what the exam expects unless a diagram is demanded.

Subtraction follows from the fact that −v is v reversed: u − v = u + (−v). And multiplying by a scalar k stretches the vector by a factor |k|, reversing it if k is negative.

Notice the resultant closes the triangle. Reverse one vector and the resultant becomes the difference — subtraction is nothing but addition of the reversed vector.

03

Position vectors and the vector joining two points

The position vector of a point A is the vector from the origin to A, written OA or a. If A is the point (3, 5) then a = ⟨3, 5⟩ — the position vector and the coordinates carry the same numbers.

The vector joining two points is the difference of their position vectors, and the order matters.

AB = b − a(head minus tail)|AB| = √[(x₂ − x₁)² + (y₂ − y₁)²]midpoint of AB = (a + b) / 2AB = −BA, so getting the order the wrong way round reverses the answer
Worked example

A is (2, −1) and B is (7, 11). Find AB, its magnitude, and a unit vector in its direction.

  1. AB = b − a = ⟨7, 11⟩ − ⟨2, −1⟩ = ⟨5, 12⟩.Head minus tail. Writing a − b would give the vector from B to A instead.
  2. |AB| = √(25 + 144) = √169 = 13.The magnitude is the distance formula, which is Pythagoras again.
  3. Unit vector = ⟨5, 12⟩ / 13 = ⟨5/13, 12/13⟩.Dividing by the magnitude leaves the direction unchanged and the length equal to 1.
  4. Check: (5/13)² + (12/13)² = (25 + 144)/169 = 1 ✓Any unit vector must satisfy this, so it is a free check.

AB = ⟨5, 12⟩, |AB| = 13, unit vector ⟨5/13, 12/13⟩

04

Parallel vectors, and using vectors to prove geometry

Two vectors are parallel when one is a scalar multiple of the other: u = k v. If k is positive they point the same way; if negative, opposite ways. This one fact does most of the geometric work in the chapter.

To prove a quadrilateral is a parallelogram, show one pair of opposite sides is equal and parallel — as vectors, show AB = DC. To show three points are collinear, show that AB is a scalar multiple of BC; since the two share the point B, the three must lie on one line.

Equal vectors say more than parallel ones

Writing AB = DC asserts two things at once: the sides are the same length and they point the same way. That single equation is enough to prove ABCD is a parallelogram, which is why the vector proof is so much shorter than the coordinate one.

Before you leave this chapter

  1. Scalar = magnitude only. Vector = magnitude and direction.
  2. Add and subtract componentwise; the triangle rule is the same operation drawn.
  3. AB = b − a, head minus tail. AB = −BA.
  4. |v| = √(a² + b²), and the unit vector is v divided by its own magnitude.
  5. u parallel to v means u = kv. Equal vectors are parallel and the same length.
06

Components from magnitude and direction, and back again

Vectors reach you in two forms. Sometimes you are given the components ⟨a, b⟩; sometimes you are given a magnitude and a bearing or angle, as in "a force of 40 N at 30° above the horizontal". Converting between the two is pure right-angled trigonometry.

Going from magnitude and angle to components, drop a perpendicular and read off the two sides. Going back, use Pythagoras for the length and the tangent ratio for the angle.

a = |v| cos θb = |v| sin θ|v| = √(a² + b²)θ = tan⁻¹(b / a)θ is measured from the positive x-axis; check the quadrant before accepting the calculator value
Worked example

A vector has magnitude 20 and makes an angle of 60° with the positive x-axis. Find its components.

  1. a = 20 cos 60° = 20 × 0.5 = 10.The horizontal component uses cosine, because the adjacent side lies along the x-axis.
  2. b = 20 sin 60° = 20 × 0.866 = 17.3.The vertical component uses sine.
  3. So v = ⟨10, 17.3⟩.
  4. Check: √(100 + 300) = √400 = 20 ✓Recomputing the magnitude from the components confirms both numbers at once.

v = ⟨10, 17.3⟩

07

Proving a geometrical result with vectors

The reason vectors are worth learning is that a geometrical proof written with them is often three lines long. The method is always the same: express every vector in the diagram in terms of two chosen ones, then compare.

Worked example

In triangle OAB, M is the midpoint of OA and N is the midpoint of OB. Show that MN is parallel to AB and half its length.

  1. Let OA = a and OB = b. Then OM = ½a and ON = ½b.Choose two vectors from one point and write everything else in terms of them.
  2. MN = ON − OM = ½b − ½a = ½(b − a).Head minus tail, exactly as for position vectors.
  3. AB = b − a.
  4. So MN = ½ AB, a scalar multiple, which proves MN is parallel to AB and half as long. ∎One equation delivers both conclusions at once — the direction from "scalar multiple" and the length from the factor ½.

MN = ½AB, so MN ∥ AB and |MN| = ½|AB|

Practice questions

6 questions · 20 marks · full working on every one

Try each one on paper first, then open the working. The marks are shown where they are actually awarded, because that is where they are actually lost.

Short questions

3 · 6 marks

Two marks each, in the style of the short-question section of the paper. Answer in two or three lines.

SQ1[2 marks]
Distinguish between a scalar and a vector, giving one example of each.
Model answer

A scalar has magnitude only — for example a mass of 5 kg. A vector has both magnitude and direction — for example a displacement of 5 m due east.

Examiner tip. Give examples from different quantities. Offering "distance and displacement" is fine but the pair must be clearly labelled which is which.

SQ2[2 marks]
If u = ⟨3, −4⟩, find |u| and a unit vector in the direction of u.
Model answer

|u| = √(9 + 16) = 5, and the unit vector is ⟨3/5, −4/5⟩.

Examiner tip. Divide both components by the magnitude, not just one. Checking that the squares add to 1 confirms the answer instantly.

SQ3[2 marks]
Given AB = ⟨4, −2⟩, write down BA and explain the relationship.
Model answer

BA = ⟨−4, 2⟩. It is the same length but in the exactly opposite direction, so BA = −AB.

Examiner tip. State the relationship in symbols. Just writing the components scores one of the two marks.

Solved numericals

2 · 8 marks

Full working, one step per line, with the marks shown where they are awarded.

N1[4 marks]
P is (1, 3), Q is (5, 6) and R is (9, 9). Show that P, Q and R are collinear.
Full working
  1. PQ = ⟨5 − 1, 6 − 3⟩ = ⟨4, 3⟩head minus tail[1]
  2. QR = ⟨9 − 5, 9 − 6⟩ = ⟨4, 3⟩[1]
  3. PQ = QR, so PQ is a scalar multiple of QR with k = 1, hence they are parallel[1]
  4. They share the point Q, so P, Q and R lie on a single straight line and are collinearthe shared point is essential — parallel alone is not enough[1]

PQ = QR = ⟨4, 3⟩ and they share Q, so the three points are collinear (with Q the midpoint of PR).

Examiner tip. The final mark is always for the sentence about the common point. Two parallel vectors that do not meet prove nothing about collinearity.

N2[4 marks]
Given a = ⟨2, 5⟩ and b = ⟨−3, 1⟩, find 2a − 3b and its magnitude.
Full working
  1. 2a = ⟨4, 10⟩both components multiplied[1]
  2. 3b = ⟨−9, 3⟩[1]
  3. 2a − 3b = ⟨4 − (−9), 10 − 3⟩ = ⟨13, 7⟩subtracting a negative is where marks are lost[1]
  4. |2a − 3b| = √(169 + 49) = √218 ≈ 14.8exact surd or 3 s.f. both accepted[1]

⟨13, 7⟩, magnitude √218 ≈ 14.8

Examiner tip. Scale each vector first and write the result down before subtracting. Trying to do both operations in one line is where the sign of the −9 goes wrong.

Long questions

1 · 6 marks

Theory and numerical together, as they appear in the long-question section.

LQ1[6 marks]
A quadrilateral has vertices A(1, 1), B(5, 2), C(6, 6) and D(2, 5).
  1. Find the vectors AB and DC.
  2. Show that ABCD is a parallelogram.
  3. Find the position vector of the point where the diagonals meet.
Mark scheme
  1. AB = ⟨5 − 1, 2 − 1⟩ = ⟨4, 1⟩[1]
  2. DC = ⟨6 − 2, 6 − 5⟩ = ⟨4, 1⟩take the vertices in the order that makes DC correspond to AB[1]
  3. AB = DC, so AB and DC are equal in length and parallel[1]
  4. One pair of opposite sides equal and parallel is sufficient for a parallelogramthe reason must be stated, not just the equality[1]
  5. The diagonals of a parallelogram bisect each other, so they meet at the midpoint of ACor the midpoint of BD, which gives the same point[1]
  6. Midpoint = (a + c)/2 = (⟨1,1⟩ + ⟨6,6⟩)/2 = ⟨3.5, 3.5⟩checking with BD: (⟨5,2⟩ + ⟨2,5⟩)/2 = ⟨3.5, 3.5⟩ ✓[1]

(a) AB = DC = ⟨4, 1⟩ (b) equal and parallel, so a parallelogram (c) ⟨3.5, 3.5⟩

Examiner tip. Computing both midpoints in part (c) is a genuine proof that the diagonals bisect each other, and costs one extra line for a guaranteed check.