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

Condensed notes for the final weeks, aligned to the OxfordAQA International AS and A-level Mathematics specification (9660, Version 5.2), section P2.10 (first part). This content is part of International A-level (A2) Unit P2 and is not examined at International AS. For full explanations, read the study guide, then test yourself with the practice questions.

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Column vectors are written across the line here: (3, −4, 12) is a column of 3, −4, 12. Questions use columns, but i, j, k is allowed.

Key results

Idea	Result
Vector in 3D	(x, y, z) = x i + y j + z k
Magnitude	√(x² + y² + z²)
Add, subtract	Component by component
Scalar multiple	Multiply every component
Parallel	a = λb for a number λ (negative λ allowed)
Position vector	OA, same components as the coordinates of A
Vector from A to B	AB = OB − OA (“end minus start”)
Distance between A and B	|AB|
Midpoint of AB	OA + ½AB = (OA + OB)/2
Point dividing AB in ratio m : n	OA + m/(m + n) × AB
Line	r = a + tb
Collinear points	AB = λAC

Magnitude and scalar multiples

Squares remove signs, so a negative component never gives a negative contribution.

|(1, −2, 2)| = √(1 + 4 + 4) = 3
|(3, 4, 12)| = √(9 + 16 + 144) = 13

Doubling a vector doubles its length. Multiplying by −1 reverses it. A unit vector in the direction of a is a divided by |a|.

Geometry: a + b is the diagonal of the parallelogram on a and b, and also the triangle path (a then b). a − b is the vector from the head of b to the head of a.

Reminder. a = (1, 3, −2), b = (4, 0, 1). Then 2a − b = (2 − 4, 6 − 0, −4 − 1) = (−2, 6, −5).

Method: show two vectors are parallel, or find an unknown

1. Compare one pair of components to find λ.
2. Apply λ to every other component. All must agree.
3. If an unknown is in the question, use λ to solve for it.

Method: AB, distance, midpoint, ratio

1. Write AB = OB − OA.
2. Distance is the magnitude of AB.
3. For a point P with AP : PB = m : n, the fraction along AB is m/(m + n), so OP = OA + m/(m + n) × AB.

Reminder. P (1, 2, 3), Q (4, 0, −1). PQ = (3, −2, −4), so |PQ| = √(9 + 4 + 16) = √29.

Method: proving three points are collinear

1. Find two vectors between the points, for example AB and BC.
2. Show that one vector is a number times the other.
3. Say that they share a point, so the three points lie on one line.

Parallel vectors alone are not enough: the shared point is what makes the points collinear.

Vector equation of a line

r = a + tb     a = a point on the line     b = direction     t any real number

• Through two points A and B: use a = OA and b = AB.
• Equivalent forms: a can be any point of the line, b any non-zero multiple of the direction.
• To test whether a point is on the line: find t from one component, then check the rest.
• In 2D, eliminate t to get y = mx + c. The gradient is (y-direction)/(x-direction).

Reminder (2D). r = (4, 1) + t(−3, 2) gives x = 4 − 3t and y = 1 + 2t. So t = (4 − x)/3 and y = 1 + 2(4 − x)/3 = (11 − 2x)/3.

Method: intersection of two lines

Use different letters: s for line 1, t for line 2.

1. Are the directions parallel? If yes, check whether one point of line 1 lies on line 2. If it does, the lines are the same; if not, they are parallel and distinct. Stop.
2. If not parallel, equate components to form equations in s and t.
3. Solve using two of the equations.
4. In 2D: that is the answer. In 3D: test the third equation.
5. Third equation holds: lines intersect. Substitute s into line 1 for the point, and check with t in line 2.
6. Third equation fails: lines are skew.

Worked reminders for lines

Intersection in 3D. Lines r = (2, 0, 1) + s(1, 1, 2) and r = (5, 1, 0) + t(2, 0, −1).

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

The y equation gives s = 1. Then z gives t = −3. Check x: 2 + 1 = 3 and 5 − 6 = −1. These differ, so the lines are skew (the directions are not multiples of each other).

Changing one letter. If the question gives both lines with the same parameter λ, rename one of them μ before you start.

Unknown constant. If a line contains an unknown such as k, solve for s and t from the two equations without k, then use the remaining equation to find k.

Geometry with vectors. In a parallelogram ABCD, AB = DC, so OD = OA + OC − OB. The diagonals AC and BD bisect each other, so the midpoint of AC equals the midpoint of BD.

Must-know distinctions

Situation	Directions	Common point	Dimension
Intersecting	Not parallel	Exactly one	2D or 3D
Parallel, distinct	Parallel	None	2D or 3D
Same line	Parallel	Infinitely many	2D or 3D
Skew	Not parallel	None	3D only

• A position vector OA is fixed to the origin. A general vector AB is only a displacement.
• A direction vector can be any non-zero multiple: (2, −4, 6) and (−1, 2, −3) give parallel lines.
• A vector equation of a line has one free parameter. A parameter has no meaning outside its own line.

Quick self-test

1. Find the magnitude of (6, −2, 3).
2. Find 2p − 3q when p = (1, 4, −2) and q = (−2, 0, 1).
3. For what value of m is (m, −9, 12) parallel to (2, 3, −4)?
4. A is (3, −1, 2) and B is (−1, 5, 4). Find AB and the distance AB.
5. Find the midpoint of the points in question 4.
6. Write a vector equation of the line through (1, 1, 0) and (3, 0, 4).
7. Does (7, −2, 8) lie on the line r = (1, 1, 0) + t(2, −1, 4)?
8. Write r = (0, 5) + t(2, 1) in the form y = mx + c.
9. Lines r = (2, 0) + s(1, 3) and r = (0, 4) + t(2, 1) meet. Find the point.
10. Lines r = (0, 0, 1) + s(1, 1, 0) and r = (1, 0, 0) + t(0, 1, 1): are they intersecting, parallel or skew?

Answers

1. √(36 + 4 + 9) = √49 = 7.
2. 2p = (2, 8, −4), 3q = (−6, 0, 3). So 2p − 3q = (8, 8, −7).
3. λ = −3 from the second component (−9 = λ × 3). The third gives 12 = (−3)(−4) = 12, which agrees. So m = −3 × 2 = −6.
4. AB = (−4, 6, 2). Distance = √(16 + 36 + 4) = √56 = 2√14.
5. ½(3 − 1, −1 + 5, 2 + 4) = (1, 2, 3).
6. Direction (2, −1, 4). r = (1, 1, 0) + t(2, −1, 4).
7. From x: 1 + 2t = 7, t = 3. Then y = 1 − 3 = −2 and z = 12. The point has z = 8, so no.
8. x = 2t, y = 5 + t, so t = x/2 and y = x/2 + 5.
9. x: 2 + s = 2t; y: 3s = 4 + t. From the first, t = 1 + s/2. Then 3s = 4 + 1 + s/2, so 5s/2 = 5, s = 2, t = 2. Point (4, 6). Check with line 2: (4, 4 + 2) = (4, 6).
10. Directions (1, 1, 0) and (0, 1, 1) are not multiples. x: s = 1. y: s = t, so t = 1. z: 1 + 0 = 1 for the first line, and t = 1 for the second. Both are 1, so the third equation holds. They intersect at (1, 1, 1).

Where marks are usually lost

• Writing AB = OA − OB, which reverses the vector and the sign of every component.
• Reusing t for both lines, which forces the two parameters to be equal.
• Solving two equations in 3D and stating an intersection without checking the third.
• Saying “not parallel, so skew” without showing that the lines fail to meet.
• Giving the position vector when the question asks for coordinates, or the reverse.
• Making a sign slip when subtracting a negative component, such as 5 − (−2).
• Leaving a distance as a decimal when the answer is a surd and the question needs an exact value.
• Testing a point on a line by checking only one coordinate.
• Forgetting that a negative scalar multiple is still parallel.

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).