OxfordAQA International AS and A-level Mathematics: Vectors and vector equations of lines (9660)

This guide follows the OxfordAQA International AS and A-level Mathematics specification (9660, Version 5.2) and covers the first part of section P2.10: vectors in two and three dimensions, and vector equations of lines. This content is part of International A-level (A2) Unit P2 (Pure Maths); it is not examined at International AS. The scalar product and perpendicular distance sit in other pages, so they are not taught here.

Related pages: revision notes, practice questions, exam preparation advice, the course hub and the printable checklist. To find out which topics need work first, take a free 10-minute diagnostic.

What you must be able to do

The specification says column vectors will be used in questions, but you may use i, j, k notation if you prefer. Both are shown here.

Specification content	What it means in practice
Vectors in two and three dimensions	Read, write and use vectors as columns or as ai + bj + ck
Magnitude	Find the length of a vector
Addition and scalar multiples	Combine vectors by components and explain the geometry
AB = OB − OA	Find the vector between two points from their position vectors
Parallel vectors	Recognise and use vectors that are scalar multiples
Position vectors, distance between two points	Use OA, find AB and
Lines r = a + tb	Write a line from a point and direction, or from two points
Intersection, parallel and skew lines	Decide whether two lines meet, and where

In this page a column vector is written across the line, so (2, −3, 6) means the column with 2 on top, then −3, then 6. In your own answers, write columns vertically or use i, j, k.

Vectors, components and magnitude

A vector has magnitude (size) and direction. In three dimensions the vector with components x, y, z is x i + y j + z k, where i, j, k are the unit vectors along the three axes. In two dimensions there is no k.

The magnitude is found with Pythagoras in three dimensions:

|(x, y, z)| = √(x² + y² + z²)

Worked example 1. Find the magnitude of p = (2, −3, 6) and of q = 4i − 4j + 7k.

|p| = √(4 + 9 + 36) = √49 = 7
|q| = √(16 + 16 + 49) = √81 = 9

The signs do not matter, because every component is squared. A vector whose magnitude is 1 is a unit vector; dividing any non-zero vector by its magnitude gives a unit vector in the same direction.

Adding vectors and multiplying by scalars

Add and subtract component by component. A scalar multiple multiplies every component.

Geometrically, draw a, then draw b beginning where a finishes (the triangle rule). The sum a + b is the single arrow that goes straight from the first starting point to the last finishing point. Two vectors drawn from the same point make a parallelogram, and a + b is the diagonal. The vector −a is the reverse of a: equal size, pointing back the other way. A scalar multiple ka stretches a by the factor |k|, and reverses it if k is negative.

Worked example 2. For a = (2, −1, 3) and b = (−1, 4, 0), find 3a − 2b.

3a = (6, −3, 9)
2b = (−2, 8, 0)
3a − 2b = (6 − (−2), −3 − 8, 9 − 0) = (8, −11, 9)

Parallel vectors

Two non-zero vectors are parallel when a = λb for some number λ, that is, when each is a stretched (or reversed) copy of the other. A negative λ means they point in opposite directions, but they are still parallel. Compare one pair of components to find λ, then use it on the others.

Worked example 3. The vector (k, 6, −9) is parallel to (2, −4, 6). Find k.

Compare the y-components: 6 = λ × (−4), so λ = −3/2
Check z: −9 = (−3/2) × 6 = −9, which agrees
Then k = (−3/2) × 2 = −3

If the z-check had failed, the vectors could not be parallel for any k.

Position vectors, AB and distance

The position vector of a point A is the vector OA from the origin O to A. It has the same components as the coordinates of A. For any two points, the vector from A to B is

AB = OB − OA

You go from A back to the origin, then out to B. This is why AB = −BA. The distance between A and B is |AB|, found with the magnitude formula.

Worked example 4. Points A (−2, 3, 1) and B (4, −3, 4) are given. Find the vector AB, the length of the segment AB, the midpoint M, and the point T that is one third of the way from A to B.

AB = (4 − (−2), −3 − 3, 4 − 1) = (6, −6, 3)
|AB| = √(36 + 36 + 9) = √81 = 9
OM = OA + ½AB = (−2 + 3, 3 − 3, 1 + 3/2) = (1, 0, 5/2)
OT = OA + ⅓AB = (−2 + 2, 3 − 2, 1 + 1) = (0, 1, 2)

T divides AB in the ratio 1 : 2, so the fraction of the way along is 1/(1 + 2).

Vectors also prove geometric facts. Suppose AC = λAB for some number λ. Then AB and AC point along the same direction and both start at A, so A, B and C all sit on one straight line. This is the usual test for collinear points.

Vector equations of lines

A line is fixed by a point on it and a direction. If a is the position vector of a point on the line and b is a direction vector, then

r = a + t b

where t is a scalar parameter that can take any real value. Choosing a value of t picks out one point on the line. In components, r = (x, y, z), which gives three separate equations in t.

To find the line through two points A and B, use A as the point and AB as the direction (or B as the point, or any multiple of AB as the direction). Different-looking equations can describe the same line.

Worked example 5. Find a vector equation of the line through A (2, −1, 5) and B (4, 1, 6). Test whether C (8, 5, 8) and D (6, 3, 8) lie on it.

AB = (2, 2, 1)
r = (2, −1, 5) + t(2, 2, 1)

For C: x gives 2 + 2t = 8, so t = 3. Then y = −1 + 6 = 5 and z = 5 + 3 = 8, which match C. So C is on the line.

For D: x gives 2 + 2t = 6, so t = 2. Then z = 5 + 2 = 7, but D has z = 8. So D is not on the line.

Always use one coordinate to find t, then check every other coordinate.

Lines in two dimensions

In 2D the same form works with two components. To change to y = mx + c, eliminate t.

Worked example 6. Write r = (3, −1) + t(2, 5) in the form y = mx + c.

x = 3 + 2t, so t = (x − 3)/2
y = −1 + 5t = −1 + 5(x − 3)/2 = (5x − 17)/2

So y = 5x/2 − 17/2. The gradient is the ratio of the direction components, 5/2.

Intersection, parallel and skew lines

Take two lines r = a₁ + s b₁ and r = a₂ + t b₂. Use a different letter for each parameter, because the two lines reach the point at different parameter values.

Step 1. Compare the direction vectors.

• If b₁ is a scalar multiple of b₂, the lines are parallel. They are either the same line or never meet. Test whether a point of one line lies on the other.
• If not, go to step 2.

Step 2. Equate the position vectors, writing one equation per component.

Step 3. Solve two equations for s and t.

Step 4. In 2D, two non-parallel lines always meet, so you have the answer. In 3D, substitute s and t into the third equation.

• If it holds, the lines intersect. Put s into line 1 (or t into line 2) to get the point.
• If it fails, the lines never meet and are not parallel: they are skew. Skew lines exist only in three dimensions.

Worked example 7 (2D). Find where r = (1, 4) + s(2, −1) meets r = (−1, −4) + t(1, 1).

x: 1 + 2s = −1 + t
y: 4 − s = −4 + t

From the second, t = 8 − s. Substitute into the first: 1 + 2s = −1 + 8 − s, so 3s = 6, s = 2 and t = 6. The point is (1 + 4, 4 − 2) = (5, 2). Check with line 2: (−1 + 6, −4 + 6) = (5, 2).

Worked example 8 (3D, intersecting). Line l₁: r = (1, 2, −1) + s(2, 1, 3). Line l₂: r = (7, −1, 2) + t(−2, 5, 3).

x: 1 + 2s = 7 − 2t
y: 2 + s = −1 + 5t
z: −1 + 3s = 2 + 3t

The z equation gives s = 1 + t. Put that into x: 1 + 2 + 2t = 7 − 2t, so t = 1 and s = 2. Check in y: 2 + 2 = 4 and −1 + 5 = 4. It holds, so the lines intersect at (1 + 4, 2 + 2, −1 + 6) = (5, 4, 5).

Worked example 9 (3D, skew). Use l₁ from above and l₃: r = (0, 3, 1) + u(1, 2, −1). The directions (2, 1, 3) and (1, 2, −1) are not multiples, so the lines are not parallel.

x: 1 + 2s = u
y: 2 + s = 3 + 2u

Substituting u = 1 + 2s into the second equation: 2 + s = 3 + 2 + 4s, so s = −1 and u = −1. Third component: l₁ gives −1 + 3(−1) = −4, but l₃ gives 1 − (−1) = 2. These differ, so the lines do not meet. They are skew.

Worked example 10 (parallel). Line A: r = (1, 0, 2) + s(2, −4, 6). Line B: r = (3, 1, 0) + t(−1, 2, −3). The direction of A is −2 times the direction of B, so the lines are parallel. Does (3, 1, 0) lie on A? The x-value gives 1 + 2s = 3, s = 1, but then y = −4, not 1. It does not, so the lines are parallel and distinct.

Common errors

• Using the same letter for both parameters. The two lines usually reach the common point at different values, so use s and t.
• Finding AB as OA − OB. The vector from A to B is OB − OA, “end minus start”.
• Stopping after two equations in 3D. The third equation is the whole test for intersection against skew.
• Calling two lines skew when the directions are parallel. Parallel lines are never skew.
• Writing the direction as the coordinates of a point on the line. The direction vector is the difference of two points, not either point.
• Treating negative scalars as “not parallel”. (−2, 4) and (1, −2) are parallel.
• Dropping a sign when subtracting a negative component.

Examiner insight

Show the equations you set up, the values of s and t, and the check in the unused component. A stated conclusion (“lines intersect” or “lines are skew”) with the comparison written beside it is easier to credit than a bare statement. Give exact values, and give column vectors for position vectors unless you are asked for coordinates.

Next steps

Work through the practice questions, then use the revision notes to recall the methods. Tick off the topic on the checklist and see the course hub.

Official syllabus

OxfordAQA International AS and A-level Mathematics (9660) specification, Version 5.2 (International AS exams from May/June 2018, A-level from May/June 2019), published by OxfordAQA. This page covers section P2.10, first part (Unit P2, International A-level only).