Vectors in Three Dimensions

This lesson extends vectors to three dimensions using the i, j, k notation — as required by the Edexcel A-Level Mathematics specification (9MA0). You need to perform vector operations in 3D, calculate magnitudes, and find unit vectors in three dimensions.

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Extending to 3D

In three dimensions, we add a z-component to our vectors. A 3D vector is written as:

a = (a₁, a₂, a₃) or equivalently a₁i + a₂j + a₃k

where:

• i = (1, 0, 0) — unit vector along the x-axis
• j = (0, 1, 0) — unit vector along the y-axis
• k = (0, 0, 1) — unit vector along the z-axis

Example: v = (3, -1, 5) = 3i - j + 5k

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Magnitude in 3D

The magnitude of a = (a₁, a₂, a₃) is:

|a| = √(a₁² + a₂² + a₃²)

This is the 3D extension of Pythagoras' theorem.

Example 1: |a| where a = (2, 3, 6) |a| = √(4 + 9 + 36) = √49 = 7

Example 2: |b| where b = (1, -2, 2) |b| = √(1 + 4 + 4) = √9 = 3

Example 3: |c| where c = (4, 0, -3) |c| = √(16 + 0 + 9) = √25 = 5

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Operations in 3D

All the operations from 2D extend naturally to 3D by including the third component.

Addition

(a₁, a₂, a₃) + (b₁, b₂, b₃) = (a₁ + b₁, a₂ + b₂, a₃ + b₃)

Example: (1, 3, -2) + (4, -1, 5) = (5, 2, 3)

Subtraction

(a₁, a₂, a₃) - (b₁, b₂, b₃) = (a₁ - b₁, a₂ - b₂, a₃ - b₃)

Example: (5, 2, 3) - (1, 3, -2) = (4, -1, 5)

Scalar Multiplication

k(a₁, a₂, a₃) = (ka₁, ka₂, ka₃)

Example: 2(3, -1, 4) = (6, -2, 8)

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Unit Vectors in 3D

The unit vector in the direction of a is:

â = a/|a|

Example: Find the unit vector in the direction of a = (2, -1, 2).

|a| = √(4 + 1 + 4) = √9 = 3

â = (2/3, -1/3, 2/3)

Check: |(2/3, -1/3, 2/3)| = √(4/9 + 1/9 + 4/9) = √(9/9) = 1 ✓

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Distance Between Two Points in 3D

The distance between points A(x₁, y₁, z₁) and B(x₂, y₂, z₂) is:

d = |AB| = √[(x₂ - x₁)² + (y₂ - y₁)² + (z₂ - z₁)²]

Example: Find the distance between A(1, 2, 3) and B(4, 6, 3).

d = √[(3)² + (4)² + (0)²] = √(9 + 16) = √25 = 5

Example: Find the distance between P(0, 0, 0) and Q(1, 2, 2).

d = √(1 + 4 + 4) = √9 = 3

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Parallel Vectors in 3D

Two 3D vectors are parallel if one is a scalar multiple of the other:

b = ka

This means b₁/a₁ = b₂/a₂ = b₃/a₃ = k (assuming no component is zero).

Example: Are a = (2, -3, 1) and b = (6, -9, 3) parallel?

6/2 = 3, (-9)/(-3) = 3, 3/1 = 3. All ratios are equal, so b = 3a and the vectors are parallel.

Example: Are p = (1, 4, 2) and q = (3, 12, 7) parallel?

3/1 = 3, 12/4 = 3, 7/2 = 3.5. The ratios are not all equal, so the vectors are not parallel.

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Zero Vector and Negative Vector

The Zero Vector

The zero vector is 0 = (0, 0, 0). It has zero magnitude and no defined direction.

For any vector a: a + 0 = a and a - a = 0.

The Negative Vector

The negative of a = (a₁, a₂, a₃) is -a = (-a₁, -a₂, -a₃).

It has the same magnitude as a but points in the opposite direction.

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Working with Forces in 3D

In mechanics, forces in three dimensions are handled as 3D vectors.

Example: Three forces act on a particle: F₁ = (2, 3, -1) N, F₂ = (-1, 0, 4) N, F₃ = (3, -5, -1) N

Find the resultant force and determine whether the particle is in equilibrium.

Resultant = F₁ + F₂ + F₃ = (2 - 1 + 3, 3 + 0 - 5, -1 + 4 - 1) = (4, -2, 2) N

Since the resultant is not the zero vector, the particle is not in equilibrium.

Magnitude of resultant = √(16 + 4 + 4) = √24 = 2√6 N

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Worked Example: Finding a Vector with Given Magnitude

Find a vector of magnitude 10 in the direction of a = (1, 2, 2).

Step 1: Find the unit vector. |a| = √(1 + 4 + 4) = 3 â = (1/3, 2/3, 2/3)

Step 2: Multiply by 10. 10â = (10/3, 20/3, 20/3)

Check: magnitude = (10/3)√(1 + 4 + 4) = (10/3)(3) = 10 ✓

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Comparing 2D and 3D Vector Operations

Operation	2D	3D
Magnitude	√(x² + y²)	√(x² + y² + z²)
Addition	(a₁ + b₁, a₂ + b₂)	(a₁ + b₁, a₂ + b₂, a₃ + b₃)
Scalar mult.	k(a₁, a₂)	k(a₁, a₂, a₃)
Unit vector	a/	a
Unit basis	i, j	i, j, k

All the algebraic rules are the same — just extended with an extra component.

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Summary

• 3D vectors are written as (a₁, a₂, a₃) or a₁i + a₂j + a₃k.
• Magnitude: |a| = √(a₁² + a₂² + a₃²).
• All 2D operations (addition, subtraction, scalar multiplication) extend naturally to 3D.
• The distance between two 3D points uses the 3D distance formula.
• Two vectors are parallel if one is a scalar multiple of the other (all component ratios are equal).
• Unit vectors in 3D: â = a/|a|.

A-Level Deep Dive: Vectors in Three Dimensions

Spec mapping

Edexcel 9MA0 specification, Paper 2 — Pure Mathematics, section 12 — Vectors covers vectors in two dimensions and in three dimensions; calculate the magnitude and direction of a vector and convert between component form and magnitude/direction form; add vectors diagrammatically and perform the algebraic operations of vector addition and multiplication by scalars; understand and use position vectors; calculate the distance between two points represented by position vectors (refer to the official specification document for exact wording). Three-dimensional vectors are a Year 2 extension of the Year 1 two-dimensional content. The unit vectors $\mathbf{i}$i, $\mathbf{j}$j, $\mathbf{k}$k along the $x$x, $y$y, $z$z axes form an orthonormal basis. The magnitude of $\mathbf{v} = x\mathbf{i} + y\mathbf{j} + z\mathbf{k}$v=xi+yj+zk is $|\mathbf{v}| = \sqrt{x^2 + y^2 + z^2}$∣v∣=x2+y2+z2​, and the distance between points $A(x_1, y_1, z_1)$A(x1​,y1​,z1​) and $B(x_2, y_2, z_2)$B(x2​,y2​,z2​) is $\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}$(x2​−x1​)2+(y2​−y1​)2+(z2​−z1​)2​. Vectors interlock with section 9 (mechanics — forces, motion in 3D for Paper 3) and the dot product (Pure Year 2). The Edexcel formula booklet does not include the 3D distance formula explicitly — it is treated as an extension of Pythagoras and must be reconstructed.

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Worked example with full mark scheme

Question (8 marks): Points $A(1, -2, 3)$A(1,−2,3), $B(4, 0, -1)$B(4,0,−1) and $C(2, 5, 2)$C(2,5,2) are given.

(a) Find the distance $AB$AB. (2)

(b) Find the angle $\angle BAC$∠BAC to the nearest degree using the dot product. (6)

Solution with mark scheme:

(a) Step 1 — form the displacement vector.

$\overrightarrow{AB} = \mathbf{b} - \mathbf{a} = (4 - 1)\mathbf{i} + (0 - (-2))\mathbf{j} + (-1 - 3)\mathbf{k} = 3\mathbf{i} + 2\mathbf{j} - 4\mathbf{k}$AB=b−a=(4−1)i+(0−(−2))j+(−1−3)k=3i+2j−4k

M1 — correct subtraction of position vectors component-by-component. A common slip is to add rather than subtract, or to mishandle the double negative on the $y$y-component ($0 - (-2) = 2$0−(−2)=2, not $-2$−2).

Step 2 — apply the magnitude formula.

$|\overrightarrow{AB}| = \sqrt{3^2 + 2^2 + (-4)^2} = \sqrt{9 + 4 + 16} = \sqrt{29}$∣AB∣=32+22+(−4)2​=9+4+16​=29​

A1 — exact surd form $\sqrt{29}$29​. Writing $5.39$5.39 without the surd loses this mark unless the question explicitly asks for a decimal.

(b) Step 1 — form both displacement vectors from $A$A.

$\overrightarrow{AB} = 3\mathbf{i} + 2\mathbf{j} - 4\mathbf{k}$AB=3i+2j−4k (from part (a)).

$\overrightarrow{AC} = \mathbf{c} - \mathbf{a} = (2 - 1)\mathbf{i} + (5 - (-2))\mathbf{j} + (2 - 3)\mathbf{k} = \mathbf{i} + 7\mathbf{j} - \mathbf{k}$AC=c−a=(2−1)i+(5−(−2))j+(2−3)k=i+7j−k

M1 — both vectors expressed from the common vertex $A$A. The angle $\angle BAC$∠BAC is the angle at $A$A, so both vectors must originate there.

Step 2 — compute the dot product.

$\overrightarrow{AB} \cdot \overrightarrow{AC} = (3)(1) + (2)(7) + (-4)(-1) = 3 + 14 + 4 = 21$AB⋅AC=(3)(1)+(2)(7)+(−4)(−1)=3+14+4=21

M1 — correct component-wise dot product, including sign care on the $\mathbf{k}$k term.

Step 3 — compute the magnitudes.

$|\overrightarrow{AB}| = \sqrt{29}$∣AB∣=29​ (from part (a)). $|\overrightarrow{AC}| = \sqrt{1^2 + 7^2 + (-1)^2} = \sqrt{51}$∣AC∣=12+72+(−1)2​=51​.

M1 — correct magnitudes.

Step 4 — apply the cosine formula.

$\cos\theta = \dfrac{\overrightarrow{AB} \cdot \overrightarrow{AC}}{|\overrightarrow{AB}|\,|\overrightarrow{AC}|} = \dfrac{21}{\sqrt{29}\sqrt{51}} = \dfrac{21}{\sqrt{1479}}$cosθ=∣AB∣∣AC∣AB⋅AC​=29​51​21​=1479​21​

A1 — correct exact expression for $\cos\theta$cosθ.

Step 5 — evaluate.

$\sqrt{1479} \approx 38.46$1479​≈38.46, so $\cos\theta \approx 0.5460$cosθ≈0.5460, giving $\theta \approx 56.9°\approx 57°$θ≈56.9°≈57°.

A1 — final answer $57°$57° to the nearest degree.

Total: 8 marks (M4 A4).

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Specimen question modelled on the Edexcel 9MA0 Paper 2 format

Question (6 marks): Relative to a fixed origin $O$O, point $P$P has position vector $2\mathbf{i} - \mathbf{j} + 4\mathbf{k}$2i−j+4k and point $Q$Q has position vector $-\mathbf{i} + 3\mathbf{j} + 2\mathbf{k}$−i+3j+2k.

(a) Find the position vector of the midpoint $M$M of $PQ$PQ. (2)

(b) Show that $|OM|^2 = \tfrac{41}{4}$∣OM∣2=441​. (2)

(c) Find a unit vector parallel to $\overrightarrow{OM}$OM. (2)

Mark scheme decomposition by AO:

(a)

• M1 (AO1.1a) — using the midpoint formula $\mathbf{m} = \tfrac{1}{2}(\mathbf{p} + \mathbf{q})$m=21​(p+q).
• A1 (AO1.1b) — $\mathbf{m} = \tfrac{1}{2}\mathbf{i} + \mathbf{j} + 3\mathbf{k}$m=21​i+j+3k.

(b)

• M1 (AO1.1b) — applying the magnitude-squared formula $|OM|^2 = (1/2)^2 + 1^2 + 3^2$∣OM∣2=(1/2)2+12+32.
• A1 (AO2.1) — evaluating to $\tfrac{1}{4} + 1 + 9 = \tfrac{41}{4}$41​+1+9=441​, matching the printed target.

(c)

• M1 (AO1.1b) — dividing the vector by its magnitude $\sqrt{41}/2$41​/2.
• A1 (AO1.1b) — $\hat{\mathbf{m}} = \dfrac{1}{\sqrt{41}}(\mathbf{i} + 2\mathbf{j} + 6\mathbf{k})$m^=41​1​(i+2j+6k).

Total: 6 marks split AO1 = 5, AO2 = 1. Year 2 vectors questions are AO1-heavy; AO2 marks are reserved for connecting magnitude, dot product, and unit-vector reasoning into multi-step arguments.

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Synoptic links

Connects to:

1. Section 12 — 2D vectors (Year 1): every 3D operation (addition, scalar multiplication, magnitude, position vectors) reduces to the 2D case when $z = 0$z=0. The 3D distance formula $\sqrt{(\Delta x)^2 + (\Delta y)^2 + (\Delta z)^2}$(Δx)2+(Δy)2+(Δz)2​ is just Pythagoras applied twice — once in the $xy$xy-plane, once vertically.

2. Section 12 — Dot product: $\mathbf{a} \cdot \mathbf{b} = a_1 b_1 + a_2 b_2 + a_3 b_3 = |\mathbf{a}||\mathbf{b}|\cos\theta$a⋅b=a1​b1​+a2​b2​+a3​b3​=∣a∣∣b∣cosθ. Perpendicular vectors satisfy $\mathbf{a} \cdot \mathbf{b} = 0$a⋅b=0. This is the only tool A-Level provides for finding angles in 3D, since the cross product is Further Maths.

3. Section 12 — Vector lines (Year 2): lines in 3D are written $\mathbf{r} = \mathbf{a} + t\mathbf{d}$r=a+td where $\mathbf{a}$a is a position vector and $\mathbf{d}$d a direction vector. Distance between lines, intersection points, and angles between lines all reduce to magnitude / dot-product calculations on 3D vectors.

4. Paper 3 Mechanics — 3D forces and motion: force vectors in 3D ($\mathbf{F} = F_x \mathbf{i} + F_y \mathbf{j} + F_z \mathbf{k}$F=Fx​i+Fy​j+Fz​k) are added componentwise; equilibrium requires the resultant to be $\mathbf{0}$0 in each component. The same magnitude/direction logic applies.

5. A-Level Physics: electric and magnetic field vectors, angular velocity, torque, and angular momentum are intrinsically 3D. The $\mathbf{i}, \mathbf{j}, \mathbf{k}$i,j,k notation used in physics is identical to the maths convention.

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Mark-scheme literacy

3D vector questions on 9MA0-02 split AO marks as follows:

AO	Typical share	Earned by
AO1 (knowledge / procedure)	65–75%	Computing magnitudes, dot products, midpoints, unit vectors via standard formulas
AO2 (reasoning / interpretation)	20–30%	Recognising perpendicularity from $\mathbf{a} \cdot \mathbf{b} = 0$a⋅b=0, choosing the right vectors from a common vertex when finding angles, justifying parallelism via scalar multiples
AO3 (problem-solving)	0–10%	Multi-step geometric reasoning — e.g. finding an unknown coordinate so that a triangle is right-angled