3D Vectors in 9709 P3: Lines, Intersections and Angles

Vectors is topic 3.7 of Pure Mathematics 3 in the Cambridge International AS & A Level Mathematics 9709 syllabus (2026–2027). Many students find it the most visual topic on Paper 3, and the questions are among the most predictable. The skill is not complicated algebra. It is turning a geometric statement ("the lines meet", "the lines are perpendicular", "the nearest point") into the right vector equation.

What the syllabus requires (and what it doesn't)

Candidates should be able to:

• use standard vector notations, in two and three dimensions
• add and subtract vectors, multiply by a scalar, and interpret these geometrically (e.g. the midpoint of AB has position vector ½(OA + OB))
• calculate magnitudes, and use unit, displacement and position vectors
• understand the line equation r = a + tb and find it from given information
• decide whether two lines are parallel, intersecting or skew, and find the point of intersection
• use the scalar product, for example to find the angle between two lines or the foot of the perpendicular from a point to a line, including in 3D objects such as cuboids and tetrahedra

The syllabus also lists what is not required: the vector product, the shortest distance between two skew lines, the equation of the common perpendicular, and the general form of the ratio theorem. Equations of planes do not appear in the 2026–2027 content, so you don't need to revise them for this syllabus.

The visual method

Before writing any algebra, sketch the situation, even roughly:

1. Mark the known points and draw each line through its point in its direction.
2. Write down what the question is really asking: a common point, an angle between directions, or a perpendicular.
3. Translate it into vectors:
   • common point → equate components
   • angle → scalar product of direction vectors
   • perpendicular → scalar product = 0

Worked examples

All examples use the same two lines, so you can see how one setting generates a full question.

Example 1: equation of a line

Question. Find a vector equation of the line L₁ through A(1, 2, −1) and B(3, −1, 2).

Solution. AB = (2, −3, 3), so L₁: r = (1, 2, −1) + t(2, −3, 3). Any point on the line, and any multiple of the direction vector, gives an equally correct equation.

Example 2: do the lines intersect?

Question. L₂ has equation r = (1, −3, 6) + s(1, 1, −2). Show that L₁ and L₂ intersect and find the point of intersection.

Solution.

• Equate components:
  • x: 1 + 2t = 1 + s
  • y: 2 − 3t = −3 + s
  • z: −1 + 3t = 6 − 2s
• From x: s = 2t. Substituting into y: 2 − 3t = −3 + 2t, so t = 1 and s = 2.
• Check z: −1 + 3 = 2 and 6 − 4 = 2. ✓
• The lines intersect at (3, −1, 2).

The check in the third equation is essential. It is the step that distinguishes intersecting lines from skew lines, and it carries a mark.

Example 3: the angle between the lines

Question. Find the acute angle between L₁ and L₂.

Solution.

• Use the direction vectors b₁ = (2, −3, 3) and b₂ = (1, 1, −2).
• b₁·b₂ = 2 − 3 − 6 = −7
• |b₁| = √22 and |b₂| = √6
• cos θ = |−7|/(√22 √6) = 7/√132 ≈ 0.609, so θ ≈ 52.5°

Use the direction vectors, not the position vectors, and take the modulus of the scalar product to get the acute angle.

Example 4: skew lines

Question. Show that the lines r = t(1, 0, 0) and r = (0, 1, 0) + u(0, 0, 1) are skew.

Solution.

• The directions (1, 0, 0) and (0, 0, 1) are not multiples of each other, so the lines are not parallel.
• Equating components: x gives t = 0, z gives 0 = u, and y gives 0 = 1, which is a contradiction.
• The lines are neither parallel nor intersecting, so they are skew.

Example 5: foot of the perpendicular

Question. Find the foot of the perpendicular from P(6, 1, 2) to L₁, and the perpendicular distance from P to L₁.

Solution.

• A general point on L₁ is Q(1 + 2t, 2 − 3t, −1 + 3t), so PQ = (−5 + 2t, 1 − 3t, −3 + 3t).
• PQ is perpendicular to (2, −3, 3): 2(−5 + 2t) − 3(1 − 3t) + 3(−3 + 3t) = 0, so −22 + 22t = 0 and t = 1.
• The foot is Q(3, −1, 2).
• PQ = (−3, −2, 0), so the distance is √13 ≈ 3.61.

Example 6: magnitude and unit vector

Question. Find the unit vector in the direction of (2, −3, 6).

Solution. The magnitude is √(4 + 9 + 36) = 7, so the unit vector is (1/7)(2, −3, 6).

Working in 3D objects

The syllabus notes that questions may involve cuboids or pyramids. Set up axes at a convenient corner, write the coordinates of each vertex, and the question becomes one of the examples above. Common tasks are the angle between two diagonals, or the point where a line meets an edge.

Mistakes that cost marks

1. Using position vectors instead of direction vectors in the angle formula.
2. Using the same parameter (t) for both lines when testing for intersection.
3. Forgetting to check the third equation.
4. Concluding "skew" without first showing that the lines are not parallel.
5. Giving an obtuse angle when the acute angle was asked for.
6. Rounding too early. Angles in degrees should be given to 1 decimal place unless the question says otherwise.

How to practise

• Build topic sets from the 9709 topical questions bank, then attempt full Paper 3s from the 9709 past papers library.
• Pair vectors with the other big Paper 3 topic in 9709 P3 integration: 10 exam-style questions.
• See where vectors fit in the course in the 9709 Syllabus 2026–2027 guide.

Frequently asked questions

Are planes in 9709 Paper 3? Not in the 2026–2027 syllabus. Topic 3.7 covers lines, intersections, angles and perpendiculars only.

Do I need the vector (cross) product? No. The syllabus states that knowledge of the vector product is not required.

Is the shortest distance between skew lines examined? No. The syllabus says this calculation is not required. You need only show that lines are skew.

How many decimal places for angles? 1 decimal place for angles in degrees, unless the question specifies otherwise.

Sources: Cambridge International AS & A Level Mathematics 9709 syllabus for examination in 2026–2027: topic 3.7 (Vectors) with its notes, and Degrees of accuracy. All examples are original.

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