9231 Further Pure Mathematics — Paper 1
x y z r·n = p n line a × b |a||b|sinθ n̂ ⊥ to both a and b skew lines
9231 Further Pure › Paper 1 › Lesson 1.6

Vectors in
3D Space

Planes in three forms, the vector product, lines and planes in 3D — distances, angles, intersections, skew lines.

🎯 Cambridge 9231 P1 📐 Section 1.6 ⭐ Vector Product Key ⏱ ~110 min
1.6 Lesson

1. Equations of a Plane — Three Forms

A plane in 3D space can be expressed in three equivalent forms. You must be able to use all three and convert between them.

Form 1 — Cartesian
ax + by + cz = d
Normal:(a, b, c) is the normal vector
Use:Easiest for substituting points and finding intersections
Find:Normal n = (a,b,c), point P on plane → d = a·Px+b·Py+c·Pz
Form 2 — Scalar Product (Normal)
r · n = p
n:Normal vector to plane
p:= a · n for any point a on plane
Use:Angles, perpendicular distances, detecting parallel lines
Link:n = (a,b,c), p = d (same as Form 1)
Form 3 — Parametric (Vector)
r = a + λb + μc
a:Position vector of one point on plane
b, c:Two non-parallel direction vectors in the plane
n:= b × c (vector product of the two directions)
Use:Given three points or two directions through a point

Converting Between Forms

Key conversion steps
Form 3 → Form 1Compute n = b × c, then d = n · a. Write ax+by+cz=d.
Form 1 → Form 2Read off n = (a,b,c), p = d. Write r · (a,b,c) = d.
Form 2 → Form 1Expand r · n = p with r = (x,y,z).
Form 1 → Form 3Find any point a on plane, find two independent direction vectors b, c perpendicular to n.

2. The Vector Product a × b

The vector product (cross product) of two vectors produces a new vector perpendicular to both. It is the key tool for finding normals to planes and shortest distances between skew lines.

⭐ Two Equivalent Definitions
Geometrica × b = |a||b| sinθ n̂   (θ = angle between a and b, n̂ = unit normal)
|a × b|= area of parallelogram formed by a and b
i j k
a₁a₂a₃
b₁b₂b₃
a × b = (a₂b₃ − a₃b₂)i  −  (a₁b₃ − a₃b₁)j  +  (a₁b₂ − a₂b₁)k
⚠ Anti-Commutative — Order Matters
a × b= −(b × a)reversing order negates the result
a × a= 0cross product with itself = zero vector
i × j = kj × k = i   k × i = j   (cyclic)

Key Applications of the Vector Product

Normal to a Plane n = b × c where b and c are two vectors lying in the plane
Area of Triangle / Parallelogram Area △ = ½|a × b|   Area ▱ = |a × b| where a and b are two edge vectors
Direction of Common Perpendicular n = d₁ × d₂ d₁, d₂ = direction vectors of the two skew lines
Line of Intersection of Two Planes direction = n₁ × n₂ n₁, n₂ = normals to the two planes

3. Lines and Planes — Seven Problem Types

Problem Type 1 — Line relative to a Plane (3 cases)
  1. 1Compute d · n (dot product of line direction d with plane normal n).
  2. 2If d · n ≠ 0: line INTERSECTS plane. Proceed to find the point.
  3. 3If d · n = 0: line is parallel to or lies IN the plane. Test by substituting the line's base point a into the plane equation.
  4. 4If a satisfies the plane equation: line LIES IN the plane. Otherwise: line is PARALLEL.
Problem Type 2 — Point of Intersection of Line and Plane
  1. 1Write the line parametrically: x = a₁+λd₁, y = a₂+λd₂, z = a₃+λd₃.
  2. 2Substitute these into the Cartesian plane equation ax+by+cz=d.
  3. 3Solve for λ.
  4. 4Substitute λ back into the line equation to get the coordinates of the intersection point.
Problem Type 3 — Foot of Perpendicular from Point P to Plane
  1. 1Write the line through P with direction n (the plane's normal): r = p + λn.
  2. 2Substitute into the plane equation to find λ.
  3. 3The foot F = p + λn (the intersection point of this perpendicular line and the plane).
  4. 4Distance from P to plane = |PF| = |λ| · |n|.
⭐ Perpendicular Distance from Point (x₀,y₀,z₀) to Plane ax+by+cz=d
Distance = |ax₀ + by₀ + cz₀ − d| / √(a²+b²+c²)

This is the quickest method for distance from a point to a plane — use it when the plane is in Cartesian form.

Problem Type 4 — Angle Between a Line and a Plane
  1. 1The angle φ between a line (direction d) and a plane (normal n) satisfies:
  2. 2sin φ = |d · n| / (|d| |n|) — note: use sin not cos, and take absolute value.
  3. 3This is because φ is the complement of the angle between d and n.
  4. 4Always check: if the answer gives an obtuse angle, use 180° minus it to get the acute angle.
Problem Type 5 — Angle Between Two Planes
  1. 1The angle between two planes equals the angle between their normals n₁ and n₂.
  2. 2cos θ = |n₁ · n₂| / (|n₁| |n₂|) — take absolute value to always get acute angle.
  3. 3If the dot product is zero, the planes are perpendicular.
Problem Type 6 — Line of Intersection of Two Planes
  1. 1Direction of line = n₁ × n₂ (cross product of the two normals).
  2. 2Find a point on the line: set one coordinate (e.g. z = 0) and solve the two plane equations simultaneously for x and y.
  3. 3Write the line equation: r = (found point) + t(n₁ × n₂).
Problem Type 7 — Shortest Distance Between Skew Lines
  1. 1Two skew lines: r = a + λd₁ and r = b + μd₂.
  2. 2Find common perpendicular direction: n = d₁ × d₂.
  3. 3Shortest distance = |(a − b) · n̂| where n̂ = n/|n| is the unit common perpendicular.
  4. 4Equivalently: distance = |(a − b) · (d₁ × d₂)| / |d₁ × d₂|.
  5. 5To find the equation of the common perpendicular: it passes through one line, direction n. Find the point on each line where the perpendicular meets it.
💡 Key Memory Rule — sin vs cos
Line ↔ Line anglecos θ = |d₁ · d₂| / (|d₁||d₂|)
Plane ↔ Plane anglecos θ = |n₁ · n₂| / (|n₁||n₂|)
Line ↔ Plane anglesin φ = |d · n| / (|d||n|)sin because line is in plane when d⊥n
Example 1 Vector Product and Plane Equation ★★☆ Challenging
Three points A(1, 2, −1), B(3, 0, 2), C(−1, 1, 3) lie on a plane Π. Find: (a) the vector product AB × AC, (b) the equation of Π in all three forms, (c) the perpendicular distance from the origin to Π.
1
Direction vectors in the plane
AB= B − A = (3−1, 0−2, 2−(−1)) = (2, −2, 3)
AC= C − A = (−1−1, 1−2, 3−(−1)) = (−2, −1, 4)
2
Part (a) — Vector product AB × AC
i component= (−2)(4) − (3)(−1) = −8 + 3 = −5
j component= −[(2)(4) − (3)(−2)] = −[8 + 6] = −14
k component= (2)(−1) − (−2)(−2) = −2 − 4 = −6
AB × AC= (−5, −14, −6) or equivalently n = (5, 14, 6)
3
Part (b) — Three forms using n = (5, 14, 6) and point A(1,2,−1)
d = n · A= 5(1) + 14(2) + 6(−1) = 5 + 28 − 6 = 27
Form 15x + 14y + 6z = 27
Form 2r · (5, 14, 6) = 27
Form 3r = (1,2,−1) + λ(2,−2,3) + μ(−2,−1,4)
4
Part (c) — Distance from origin to Π
Distance= |5(0)+14(0)+6(0)−27| / √(25+196+36)
= 27 / √257 = 27/√257 ≈ 1.684
Example 2 Line–Plane Intersection and Foot of Perpendicular ★★☆ Challenging
The plane Π has equation 2x − y + 3z = 10. The line L has equation r = (1, 2, 1) + λ(1, −1, 2). (a) Find the point of intersection of L and Π. (b) Find the acute angle between L and Π. (c) Find the foot of the perpendicular from the point P(4, 0, 5) to Π.
1
Part (a) — Parametric substitution
Point on L(1+λ, 2−λ, 1+2λ)
Substitute2(1+λ) − (2−λ) + 3(1+2λ) = 10
2 + 2λ − 2 + λ + 3 + 6λ = 10
3 + 9λ = 10 → λ = 7/9
Point= (1+7/9, 2−7/9, 1+14/9) = (16/9, 11/9, 23/9)
2
Part (b) — Angle between line and plane
d= (1, −1, 2), n = (2, −1, 3)
d · n= 2 + 1 + 6 = 9
|d|= √(1+1+4) = √6   |n| = √(4+1+9) = √14
sin φ= 9 / (√6 · √14) = 9/√84 = 9/(2√21)
φ= arcsin(9/√84) ≈ 79.7°
3
Part (c) — Foot of perpendicular from P(4, 0, 5)
Line through Pr = (4, 0, 5) + t(2, −1, 3)   (direction = n)
Point= (4+2t, −t, 5+3t)
Sub into plane:2(4+2t) − (−t) + 3(5+3t) = 10
8+4t + t + 15+9t = 10 → 14t = −13 → t = −13/14
Foot F= (4 − 26/14, 13/14, 5 − 39/14) = (30/14, 13/14, 31/14)
Example 3 Angle Between Planes and Line of Intersection ★★☆ Challenging
Two planes: Π₁: x + 2y − z = 3 and Π₂: 2x − y + 3z = 1. (a) Find the acute angle between Π₁ and Π₂. (b) Find the equation of the line of intersection of Π₁ and Π₂.
1
Part (a) — Angle between planes
n₁ = (1,2,−1)n₂ = (2,−1,3)
n₁ · n₂= 2 − 2 − 3 = −3
|n₁|= √6   |n₂| = √14
cos θ= |−3| / (√6 · √14) = 3/√84 = 3/(2√21)
θ= arccos(3/√84) ≈ 70.9°
2
Part (b) — Direction of line = n₁ × n₂
i= (2)(3) − (−1)(−1) = 6 − 1 = 5
j= −[(1)(3) − (−1)(2)] = −[3+2] = −5
k= (1)(−1) − (2)(2) = −1 − 4 = −5
Direction= (5, −5, −5) = 5(1, −1, −1)
3
Find a point on both planes — set z = 0
Π₁: z=0x + 2y = 3
Π₂: z=02x − y = 1
Solve:x + 2y = 3, 2x − y = 1 → 5x = 5 → x = 1, y = 1
Point= (1, 1, 0)
Line of intersectionr = (1, 1, 0) + t(1, −1, −1)
Example 4 Skew Lines — Shortest Distance and Common Perpendicular ★★★ A* Level
Line L₁: r = (2, 1, 0) + λ(1, 2, 1) and Line L₂: r = (1, 0, 3) + μ(2, 1, −1). (a) Show the lines are skew. (b) Find the shortest distance between them. (c) Find the equation of the common perpendicular.
1
Part (a) — Show lines are skew

Two lines are skew if they are not parallel and do not intersect.

Parallel?d₁ = (1,2,1), d₂ = (2,1,−1): not scalar multiples → not parallel ✓

Test for intersection: set equal componentwise.

x: 2+λ = 1+2μy: 1+2λ = μ    z: λ = 3−μ
From y: μ = 1+2λSubstitute into z: λ = 3−(1+2λ) = 2−2λ → 3λ = 2 → λ = 2/3, μ = 7/3
Check x:2+2/3 = 8/3   vs   1+14/3 = 17/3   → 8/3 ≠ 17/3

Inconsistent → no intersection. Lines are skew. ✓

2
Part (b) — Shortest distance
n = d₁ × d₂= (1,2,1) × (2,1,−1)
i= (2)(−1) − (1)(1) = −3
j= −[(1)(−1) − (1)(2)] = −(−3) = 3
k= (1)(1) − (2)(2) = 1 − 4 = −3
n= (−3, 3, −3) = −3(1, −1, 1)   so use n = (1,−1,1)
a − b= (2,1,0) − (1,0,3) = (1,1,−3)
(a−b)·n= 1−1−3 = −3
|n|= √3
Distance= |−3|/√3 = 3/√3 = √3
3
Part (c) — Common perpendicular direction and equation

The common perpendicular has direction n = (1, −1, 1). We need to find where it meets each line.

A point on L₁ is (2+λ, 1+2λ, λ). The common perpendicular from this point has direction (1,−1,1). For it to reach L₂, it must satisfy the system — but for the equation we can state:

Common ⊥passes through a point on L₁, direction (1,−1,1)
Full method:Find point on L₁ whose perpendicular (direction n) reaches L₂

The complete common perpendicular calculation requires solving λ and μ simultaneously — this is typically set as a full A* question with several marks allocated.

Interactive Vector Calculator

Compute vector products, plane equations, distances, and angles instantly. Enter your vectors and choose the operation.

⚙️ 3D Vector Operations

Plane: ax+by+cz=d  |  Point: (x₀,y₀,z₀)

L₁: r = a + λd₁  |  L₂: r = b + μd₂

Practice Questions

Question 1 — Plane Equations
The plane Π passes through A(2, −1, 3) and has normal vector n = (1, 2, −2). (a) Find the equation of Π in Cartesian form. (b) Write the equation in scalar product form r · n = p. (c) Find the perpendicular distance from B(5, 1, 0) to Π. (d) Find whether the point C(4, 3, 1) lies on Π.
For (a): use d = n·A = 1(2)+2(−1)+(−2)(3). For (c): use the distance formula |n·B − d|/|n|. For (d): substitute C into the equation and check if it equals d.
✓ Solution
(a) d= 1(2)+2(−1)+(−2)(3) = 2−2−6 = −6
Plane Πx + 2y − 2z = −6
(b)r · (1, 2, −2) = −6
(c) |n|= √(1+4+4) = 3
n·B= 5+2+0 = 7
Distance= |7−(−6)|/3 = 13/3 ≈ 4.33
(d) n·C= 4+6−2 = 8 ≠ −6 → C does NOT lie on Π
Question 2 — Vector Product
For vectors a = 2i − j + 3k and b = i + 4j − k, find: (a) a × b, (b) |a × b|, (c) the area of the triangle with sides OA and OB where A and B have position vectors a and b, (d) a unit vector perpendicular to both a and b.
Use the component formula. For (c): area of triangle = ½|a×b|. For (d): divide a×b by its magnitude.
✓ Solution
(a) i= (−1)(−1)−(3)(4) = 1−12 = −11
j= −[(2)(−1)−(3)(1)] = −(−2−3) = 5
k= (2)(4)−(−1)(1) = 8+1 = 9
a×b= (−11, 5, 9)
(b) |a×b|= √(121+25+81) = √227
(c) Area △= ½√227 ≈ 7.54
(d) Unit ⊥= (1/√227)(−11, 5, 9)
Question 3 — Line and Plane
The line L: r = (3, 1, −2) + λ(2, 1, 3) and the plane Π: 3x − y + 2z = 8. (a) Find the point of intersection of L and Π. (b) Find the acute angle between L and Π. (c) Find the foot of perpendicular from P(1, 2, 1) to Π, and hence the distance from P to Π.
(a) Substitute parametric form of L into plane. (b) sinφ = |d·n|/(|d||n|). (c) Line through P with direction n=(3,−1,2), find where it hits the plane.
✓ Solution
(a) Point(3+2λ, 1+λ, −2+3λ)
Sub:3(3+2λ)−(1+λ)+2(−2+3λ)=8 → 9+6λ−1−λ−4+6λ=8 → 4+11λ=8 → λ=4/11
Point= (41/11, 15/11, −10/11)
(b) d·n= 6−1+6=11, |d|=√14, |n|=√14
sinφ= 11/14 → φ = arcsin(11/14) ≈ 51.8°
(c) Liner=(1,2,1)+t(3,−1,2): (1+3t,2−t,1+2t)
Sub:3(1+3t)−(2−t)+2(1+2t)=8 → 3+9t−2+t+2+4t=8 → 3+14t=8 → t=5/14
Foot F= (29/14, 23/14, 24/14)
Distance= |5/14|·|n| = (5/14)·√14 = 5/√14 ≈ 1.34
Question 4 — Skew Lines A*
L₁: r = (1, 0, 2) + s(1, 1, 0) and L₂: r = (2, 1, 3) + t(0, 1, 2). (a) Prove the lines are skew. (b) Find the shortest distance between L₁ and L₂. (c) Find the coordinates of the points on L₁ and L₂ that are closest to each other.
(a) Check not parallel, then test intersection — show system is inconsistent. (b) n = d₁×d₂, distance = |(a−b)·n|/|n|. (c) The shortest segment PQ where P on L₁ and Q on L₂ must satisfy PQ ⊥ d₁ and PQ ⊥ d₂ simultaneously. Set P=(1+s, s, 2) and Q=(2, 1+t, 3+2t), PQ = Q−P, then PQ·d₁=0 and PQ·d₂=0 gives two equations in s and t.
✓ Solution
(a) d₁=(1,1,0), d₂=(0,1,2)Not scalar multiples → not parallel ✓
Equate:x: 1+s=2→s=1; y: s=1+t→t=0; z: 2=3+0=3 → 2≠3 ✗ → skew ✓
(b) n=d₁×d₂i: 1·2−0·1=2, j: −(1·2−0·0)=−2, k: 1·1−1·0=1 → n=(2,−2,1)
a−b=(1,0,2)−(2,1,3)=(−1,−1,−1)
(a−b)·n= −2+2−1 = −1
|n|=√9=3Distance = 1/3 ≈ 0.333
(c) PQ=(1−s,1+t,1+2t)PQ·d₁=0: (1−s)+(1+t)=0 → 2−s+t=0 ①
PQ·d₂=0:(1+t)+(2+4t)=0 → 3+5t=0 → t=−3/5, s=2+t=7/5
P on L₁= (1+7/5, 7/5, 2) = (12/5, 7/5, 2)
Q on L₂= (2, 1−3/5, 3−6/5) = (2, 2/5, 9/5)

Formula Reference Sheet

Complete reference for Vectors in 3D — Cambridge 9231 P1, Section 1.6.

Three Forms of a Plane
Cartesian: ax+by+cz=d
Normal: r·n = p
Parametric: r=a+λb+μc
n=(a,b,c), p=d=n·a
Normal form: n = b×c
Vector Product a×b
= |a||b|sinθ n̂
i: a₂b₃−a₃b₂
j: −(a₁b₃−a₃b₁)
k: a₁b₂−a₂b₁
a×b = −(b×a)
Distances
Pt to plane: |n·p−d|/|n|
Skew lines: |(a−b)·(d₁×d₂)|/|d₁×d₂|
Foot of ⊥: r=p+tn, find t, sub back
Angles
Line∥line: cosθ=|d₁·d₂|/(|d₁||d₂|)
Plane∥plane: cosθ=|n₁·n₂|/(|n₁||n₂|)
Line–plane: sinφ=|d·n|/(|d||n|)
Line–Plane Relationships
d·n≠0: line intersects
d·n=0, a not on Π: parallel
d·n=0, a on Π: line in plane
Intersection: sub parametric into Π
Skew Lines and Intersections
Line of Π₁∩Π₂: direction=n₁×n₂
Point: set z=0, solve system
Skew: not parallel, no intersection
Common ⊥ direction: d₁×d₂
📋 Cambridge Exam Strategy — Vectors
  • Show the j-component negation explicitly when computing cross products — the minus sign on j is the most common source of errors.
  • For line–plane angle, always use sin not cos. A line lying in a plane makes 0° with the plane but 90° with its normal — that is why sin appears.
  • When finding the line of intersection of two planes, the direction is n₁×n₂ and a point is found by setting z=0 (or whichever coordinate gives a clean system).
  • To show lines are skew: first confirm not parallel (directions not proportional), then show no intersection by solving the parametric system and checking it is inconsistent in one equation.
  • For skew distance, use the formula |(a−b)·(d₁×d₂)|/|d₁×d₂| directly — it is faster than finding the common perpendicular first.
  • Always convert between plane forms fluently — Cambridge may give Form 3 and ask a question that requires Form 1.
← Lesson 1.5: Polar Coordinates 9231 P1 · Lesson 1.6 of 7