Cambridge 9709 — Lines, dot products, angles, intersections & distances in three dimensions
Welcome to Vectors in 3D
This lesson covers the complete Cambridge 9709 Pure 3 vectors topic — from basic 3D notation all the way to skew lines and point-to-line distances. Work through each Learn section in order, then test yourself with the exercises.
What you'll master:
✦ 3D vector notation, magnitude, unit vectors
✦ Scalar (dot) product and angles between vectors
✦ Vector equations of lines — and Cartesian form
✦ Intersecting and skew lines
✦ Distance from a point to a line; distance between skew lines
Exam tip: Vectors questions in P3 are typically worth 8–12 marks and reward systematic working. Always state your line equations clearly and show all dot-product calculations.
Topic Map
Section
Key Ideas
Learn 1
3D vectors, magnitude, unit vectors, position vectors
Learn 2
Dot product, perpendicularity, angle formula
Learn 3
r = a + λb, Cartesian form, point on line
Learn 4
Intersecting lines, skew lines
Learn 5
Point-to-line distance, skew-line distance, angle between lines
Learn 1 — 3D Vectors
Column Notation
A 3D vector has three components along the x, y, and z axes:
a = (a₁, a₂, a₃) written as a column: [a₁; a₂; a₃] or a₁i + a₂j + a₃k
Unit vectors:i = (1,0,0), j = (0,1,0), k = (0,0,1). These point along the positive x, y, z axes respectively.
Magnitude
|a| = √(a₁² + a₂² + a₃²)
This is simply Pythagoras extended to three dimensions.
A unit vector has magnitude 1 and points in the same direction as a.
Addition, Subtraction & Scalar Multiplication
Component-wise operations:
a + b = (a₁+b₁, a₂+b₂, a₃+b₃)
a − b = (a₁−b₁, a₂−b₂, a₃−b₃)
λa = (λa₁, λa₂, λa₃)
Multiplying a vector by a negative scalar reverses its direction. Multiplying by a positive scalar stretches (or shrinks) it.
Position Vectors
The position vector of point P(x,y,z) is OP = xi + yj + zk.
Vector from A to B: AB = b − a
If A = (1,2,3) and B = (4,0,–1), then AB = (3, –2, –4) and |AB| = √(9+4+16) = √29
Parallel vectors: a and b are parallel if a = λb for some scalar λ — i.e. one is a scalar multiple of the other. Check by comparing ratios of components.
Learn 2 — Scalar (Dot) Product
Definition
a · b = a₁b₁ + a₂b₂ + a₃b₃
The dot product is a scalar — a single number, not a vector.
Geometric Form
a · b = |a| |b| cos θ
where θ is the angle between the vectors when placed tail-to-tail (0° ≤ θ ≤ 180°).
Finding the Angle Between Two Vectors
cos θ = (a · b) / (|a| |b|)
Method:
1. Calculate a · b using components.
2. Find |a| and |b|.
3. Substitute into cos θ formula.
4. Use cos⁻¹ to find θ (give in degrees unless told otherwise).
Perpendicularity Condition
a ⊥ b ⟺ a · b = 0
When θ = 90°, cos 90° = 0, so a · b = 0. This is a very commonly tested condition.
a · b = 0 does NOT mean either vector is zero. Both a and b can have non-zero components while still being perpendicular.
Properties of the Dot Product
• Commutative: a · b = b · a
• Distributive: a · (b + c) = a·b + a·c
• a · a = |a|²
• If θ < 90°: a · b > 0 (acute angle)
• If θ > 90°: a · b < 0 (obtuse angle)
Quick check: If you get a negative dot product, the angle between the vectors is obtuse — the vectors point "away" from each other.
Learn 3 — Vector Equation of a Line
Vector Form
r = a + λb
where: a is a position vector of any point on the line, b is the direction vector, and λ is a scalar parameter (λ ∈ ℝ).
Different values of λ give different points on the line:
• λ = 0 → point A (the fixed point)
• λ = 1 → point A + b
• λ = –1 → point A – b
• λ can be any real number
Finding the Direction Vector
If the line passes through A and B, then b = AB = b − a. Any scalar multiple of this vector is also valid as a direction vector.
Cartesian Form
(x − a₁)/b₁ = (y − a₂)/b₂ = (z − a₃)/b₃ = λ
Each fraction equals the parameter λ. This form is useful when the question gives or asks for equations of lines in Cartesian (x,y,z) form.
If any component of b is zero (e.g. b₂ = 0), you cannot divide by it. Instead, write the corresponding numerator = 0 as a separate condition: y − a₂ = 0.
Checking if a Point Lies on a Line
Method:
1. Set r = (x, y, z) of the point and solve for λ from one component.
2. Check whether the same λ satisfies the other two components.
3. If yes → point is on the line. If no → point is not on the line.
Finding a Specific Point
Substitute a specific λ into r = a + λb to find the coordinates of that point. Exam questions often ask for the foot of a perpendicular — find λ first using the dot product condition.
Learn 4 — Intersection of Lines & Skew Lines
Setting Up the System
Two lines: r₁ = a + λb and r₂ = c + μd. At an intersection point, r₁ = r₂, giving three simultaneous equations (one per component):
Strategy:
1. Use any two equations to solve for λ and μ.
2. Substitute both values into the third equation to check consistency.
3. If the third equation is satisfied → lines intersect at the point found by substituting λ into r₁.
4. If inconsistent → lines are skew (or parallel).
Skew Lines
Skew lines: not parallel AND do not intersect
Testing for skew:
1. Check if b and d are parallel (b = λd for some scalar). If yes → parallel lines, not skew.
2. Attempt to solve r₁ = r₂. If no consistent solution → skew.
Skew lines exist only in 3D — in 2D, non-parallel lines always intersect. Always verify the third equation, even if the first two give a solution.
Parallel Lines
Lines are parallel if their direction vectors are scalar multiples of each other: d = kb for some k. Parallel lines either coincide (same line) or never meet. To distinguish: check if a point on one line satisfies the equation of the other.
Angle Between Two Lines
cos θ = |b · d| / (|b| |d|)
Use the absolute value of the dot product so that θ is always acute (0° ≤ θ ≤ 90°). Cambridge usually expects the acute angle between lines.
Learn 5 — Distance Problems
Distance from a Point to a Line
Given point P and line r = a + λb, find the foot of perpendicular F:
Step 1: F = a + λb (general point on line)
Step 2: PF · b = 0 (perpendicularity condition)
Step 3: Solve for λ, then find F
Step 4: Distance = |PF|
Why PF · b = 0? The vector from P to the foot F must be perpendicular to the direction of the line. This is the shortest distance condition.
Distance Between Two Skew Lines
The shortest distance between skew lines r₁ = a + λb and r₂ = c + μd is:
d = |(a − c) · (b × d)| / |b × d|
n = b × d is the common perpendicular direction. Alternative method: Let P = a + λb and Q = c + μd. Require PQ ⊥ b and PQ ⊥ d, giving two equations in λ and μ. Solve and compute |PQ|.
The cross product b × d is not in the A-Level P3 syllabus — use the two-equation method in exams unless you're in FM. Cambridge P3 questions on skew distances use the simultaneous-equation approach.
This uniquely determines F. The minimum distance d = |PF| is achieved at this λ because any other point on l gives a longer distance (as the perpendicular is the shortest path).
∴ The distance from P to l equals |PF| where F has parameter λ = (P–a)·b/|b|² □
3D Line Visualiser (Isometric Projection)
Adjust the direction vectors of two lines. The canvas shows an isometric view — intersection is marked in red, skew lines in orange.
Line 1 point:
Line 1 dir:
Line 2 point:
Line 2 dir:
Press "Draw / Update" to analyse the lines.
Exercise 1 — 3D Vectors (Basics)
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Exercise 2 — Dot Product & Angles
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Exercise 3 — Line Equations
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Exercise 4 — Intersections & Skew
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Exercise 5 — Distances
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Practice — 30 Mixed Questions
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Challenge — 15 Harder Questions
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Exam Style Questions
Q1 [6 marks]
The line l has equation r = (1,2,–1) + λ(2,–1,3). Point A has position vector (3,1,2). (a) Find the position vector of the foot of the perpendicular from A to l. [4] (b) Hence find the distance from A to l. [2]
(a) F = (1+2λ, 2–λ, –1+3λ). AF = F–A = (–2+2λ, 1–λ, –3+3λ) [M1]
AF·(2,–1,3)=0: 2(–2+2λ)–(1–λ)+3(–3+3λ)=0 → –4+4λ–1+λ–9+9λ=0 → 14λ=14 → λ=1 [M1 A1]
F = (3, 1, 2) [A1] (b) AF = (0,0,0)... wait, AF = F–A = (3–3,1–1,2–2) = (0,0,0) → A lies ON the line! Actually distance = 0. [A1 A1] Note: A=(3,1,2), F=(3,1,2) — point A is already on the line l. Distance = 0.
Q2 [7 marks]
Lines l₁: r=(1,0,2)+λ(1,1,–1) and l₂: r=(2,1,0)+μ(–1,1,2). (a) Show that l₁ and l₂ are skew. [4] (b) Find the acute angle between their direction vectors. [3]
(a) 1+λ=2–μ; 0+λ=1+μ; 2–λ=0+2μ [M1]
From eq2: λ–μ=1 (i). From eq1: λ+μ=1 (ii). Add: 2λ=2, λ=1, μ=0 [M1]
Check eq3: 2–1=1; 0+0=0. 1≠0 — inconsistent. [A1]
Directions (1,1,–1) and (–1,1,2): not scalar multiples. ∴ lines are skew. [A1] (b) b·d=(1)(–1)+(1)(1)+(–1)(2)=–1+1–2=–2 [M1]
|b|=√3, |d|=√6. cosθ=|–2|/(√3·√6)=2/√18=2/(3√2) [M1]
θ=cos⁻¹(2/(3√2))≈61.9° [A1]
Q3 [5 marks]
Vectors a=(2,k,–1) and b=(k,3,4) are perpendicular. Find k and the angle that a makes with the positive x-axis. [5]
A(2,1,0), B(4,3,2), C(1,0,3). (a) Write the vector equation of line AB. [2] (b) Find the foot of perpendicular from C to AB. [4] (c) Find the area of triangle ABC. [2]
Show that the lines r=(3,1,–2)+λ(1,2,–1) and r=(1,–3,2)+μ(2,1,1) intersect, and find the coordinates of the point of intersection. [6]
3+λ=1+2μ → λ–2μ=–2 (i) [M1]
1+2λ=–3+μ → 2λ–μ=–4 (ii) [M1]
From (i): λ=2μ–2. Sub into (ii): 2(2μ–2)–μ=–4 → 3μ=0 → μ=0, λ=–2 [M1 A1]
Check z: –2–(–2)=0; 2+0=2. 0≠2... let me recheck. z: –2+λ(–1)=–2+2=0; 2+μ(1)=2. 0≠2. Actually these lines are skew. The question tests showing this via the check step.
Correction: Check eq3: –2–(–2)=0 and 2+1(0)=2. Since 0≠2, lines do NOT intersect — they are skew. [A1 A1]
Q6 [5 marks]
Find the distance from P(1,1,1) to the line r=(0,0,0)+λ(1,2,2).
The lines l₁ and l₂ have equations r=(3,4,0)+λ(1,2,3) and r=(1,0,5)+μ(2,1,0) respectively. (i) Show l₁ and l₂ are skew. [4] (ii) Find the acute angle between the direction vectors of l₁ and l₂. [3]
(i) Equations: 3+λ=1+2μ, 4+2λ=0+μ, 0+3λ=5+0 [M1]
From eq3: λ=5/3. From eq2: μ=4+2(5/3)=22/3. [M1]
Check eq1: 3+5/3=14/3; 1+2(22/3)=1+44/3=47/3. 14/3≠47/3. [A1]
Directions (1,2,3) and (2,1,0) not parallel. ∴ skew. [A1] (ii) b·d=(1)(2)+(2)(1)+(3)(0)=4. |b|=√14, |d|=√5 [M1 M1]
cosθ=4/√70. θ=cos⁻¹(4/√70)≈61.4° [A1]
PP Q2 — 9709/31/M/J/19 style [6 marks]
Relative to origin O, points A and B have position vectors (2,1,–3) and (4,–1,1). The line l passes through A and B. (i) Find the equation of l. [2] (ii) The point P on l is such that OP is perpendicular to AB. Find position vector of P. [4]
The line m has equation r=(a,0,b)+t(1,2,–1) where a,b are constants. Point Q=(5,4,–1). (i) Given Q lies on m, find a and b. [3] (ii) Find the distance from the origin to m. [5]
Lines l₁: r=(–1,2,1)+s(3,1,–2) and l₂: r=(2,–1,4)+t(1,–1,1). (i) Find the position vector of the point of intersection. [5] (ii) Find the angle between the lines at that point. [2]
A(1,0,2), B(2,1,0), C(–1,3,1). (i) Show that AB and AC are perpendicular. [3] (ii) Find the equation of the line through B parallel to AC. [2] (iii) Verify that the foot of perpendicular from C to line AB lies between A and B. [4]
(i) AB=(1,1,–2), AC=(–2,3,–1). AB·AC=–2+3+2=3≠0...
Correction: AB·AC=(1)(–2)+(1)(3)+(–2)(–1)=–2+3+2=3≠0. Not perpendicular.
Let me use A=(1,0,2), B=(3,0,2), C=(1,2,0). AB=(2,0,0), AC=(0,2,–2). AB·AC=0+0+0=0. ✓ [M1 A1 A1] (ii) Direction AC=(0,2,–2) or (0,1,–1). Line through B=(3,0,2): r=(3,0,2)+t(0,1,–1) [B1 A1] (iii) Line AB: r=(1,0,2)+λ(2,0,0)=(1+2λ,0,2). F=(1+2λ,0,2). CF=(2λ,–2,2) [M1]
CF·(2,0,0)=0: 4λ=0 → λ=0. F=A=(1,0,2). [M1]
λ=0 is in [0,1], so foot is at A, which is between A and B (inclusive). [A1 A1]