Vector Equations of Lines

This lesson covers the vector equation of a straight line, parallel lines, intersecting lines, and skew lines in 3D — as required by the Edexcel A-Level Mathematics specification (9MA0).

---

The Vector Equation of a Line

A straight line in 3D can be described by giving:

• A point on the line (with position vector a)
• A direction vector d (parallel to the line)

The vector equation of the line is:

r = a + td

where:

• r is the position vector of a general point on the line
• a is the position vector of a known point on the line
• d is the direction vector of the line
• t is a scalar parameter (t can take any real value)

As t varies, r traces out every point on the line.

Example

Write the vector equation of the line through the point (2, 3, -1) with direction vector (1, -2, 4).

r = (2, 3, -1) + t(1, -2, 4)

In parametric form: x = 2 + t, y = 3 - 2t, z = -1 + 4t

---

Finding the Equation from Two Points

If the line passes through points A and B, then the direction vector is AB = b - a, and the equation is:

r = a + t(b - a)

Example: Find the vector equation of the line through A(1, 0, 3) and B(4, 2, -1).

Direction: AB = (3, 2, -4)

Equation: r = (1, 0, 3) + t(3, 2, -4)

---

Parallel Lines

Two lines are parallel if their direction vectors are scalar multiples of each other.

Line 1: r = a₁ + td₁ Line 2: r = a₂ + sd₂

The lines are parallel if d₁ = kd₂ for some scalar k.

Example: Line 1: r = (1, 2, 3) + t(2, -1, 4) Line 2: r = (0, 1, -1) + s(4, -2, 8)

Since (4, -2, 8) = 2(2, -1, 4), the direction vectors are parallel, so the lines are parallel.

---

Intersection of Two Lines

To find where two lines intersect, set their position vectors equal and solve for the parameters.

Line 1: r = a₁ + td₁ Line 2: r = a₂ + sd₂

At intersection: a₁ + td₁ = a₂ + sd₂

This gives a system of simultaneous equations (one for each component). If a consistent solution for t and s exists, the lines intersect.

Worked Example

Find the point of intersection of: Line 1: r = (1, 2, 3) + t(2, 1, -1) Line 2: r = (3, 0, 1) + s(1, -1, 2)

Setting equal: 1 + 2t = 3 + s ... (i) 2 + t = -s ... (ii) 3 - t = 1 + 2s ... (iii)

From (ii): s = -(2 + t)

Substitute into (i): 1 + 2t = 3 - (2 + t) = 1 - t 3t = 0 → t = 0, so s = -2

Check in (iii): 3 - 0 = 1 + 2(-2) → 3 = -3. This is FALSE.

Since the solution does not satisfy all three equations, the lines do not intersect.

---

Skew Lines

In 3D, two lines that are not parallel and do not intersect are called skew lines.

Skew lines exist in different planes and never meet, even though they are not parallel. (This cannot happen in 2D — in 2D, non-parallel lines always intersect.)

How to Identify Skew Lines

1. Check if the direction vectors are parallel. If yes, the lines are parallel (not skew).
2. If not parallel, try to find an intersection point. If no consistent solution exists, the lines are skew.

From the worked example above, the lines are not parallel (since (2, 1, -1) ≠ k(1, -1, 2) for any k) and do not intersect. Therefore, they are skew.

---

Angle Between Two Lines

The acute angle between two lines with direction vectors d₁ and d₂ is found using the scalar product:

cos θ = |d₁ · d₂| / (|d₁| × |d₂|)

The modulus sign ensures we get the acute angle.

Example

Find the acute angle between: Line 1 with direction (1, 2, -1) Line 2 with direction (3, -1, 2)

d₁ · d₂ = 3 - 2 - 2 = -1 |d₁| = √(1 + 4 + 1) = √6 |d₂| = √(9 + 1 + 4) = √14

cos θ = |-1| / (√6 × √14) = 1/√84 = 1/(2√21)

θ = arccos(1/(2√21)) ≈ 83.7°

---

A Point on a Line

To check if a point lies on a line, substitute the point's coordinates into the parametric equations and check that a single, consistent value of t is obtained.

Example: Does the point (5, 4, 1) lie on the line r = (1, 2, 3) + t(2, 1, -1)?

From x: 5 = 1 + 2t → t = 2 From y: 4 = 2 + t → t = 2 From z: 1 = 3 - t → t = 2

All components give t = 2, so the point lies on the line.

---

Shortest Distance from a Point to a Line

To find the shortest distance from a point P to a line r = a + td:

1. Let Q be the foot of the perpendicular from P to the line.
2. Q has position vector a + td for some value of t.
3. PQ is perpendicular to d, so PQ · d = 0.
4. Solve for t, then find |PQ|.

Example

Find the shortest distance from P(3, 1, 2) to the line r = (1, 0, -1) + t(1, 1, 1).

Q = (1 + t, t, -1 + t)

PQ = Q - P = (t - 2, t - 1, t - 3)

PQ · d = 0: (t - 2)(1) + (t - 1)(1) + (t - 3)(1) = 0 t - 2 + t - 1 + t - 3 = 0 3t - 6 = 0 → t = 2

Q = (3, 2, 1) PQ = (0, 1, -1) |PQ| = √(0 + 1 + 1) = √2

---

Common Mistakes

1. Using the same parameter for both lines — Use different letters (t and s) for different lines.
2. Not checking all three equations — Finding t and s from two equations is not enough; you must verify the solution in the third equation.
3. Forgetting the modulus for acute angle — Take |d₁ · d₂| to ensure the angle is acute.
4. Confusing parallel with coincident — Parallel lines may not be the same line. Check if a point from one line lies on the other.

---

Summary

• Vector equation of a line: r = a + td.
• Parallel lines: direction vectors are scalar multiples.
• Intersecting lines: set equal and solve simultaneously — check all three equations.
• Skew lines: not parallel AND no intersection (3D only).
• Angle between lines: cos θ = |d₁ · d₂| / (|d₁||d₂|).
• To check if a point lies on a line, substitute and verify all components give the same t.

A-Level Deep Dive: Vector Equations of Lines

Spec mapping

Edexcel 9MA0-02 specification section 12 — Vectors, Year 2 sub-strand 12.5 covers vectors to solve problems in pure mathematics and in context, including forces, velocity and geometry. Understand the vector and parametric form $\mathbf{r} = \mathbf{a} + \lambda \mathbf{d}$r=a+λd for a line in two and three dimensions (refer to the official specification document for exact wording). This builds on Year 1 section 11 (2D vectors) and is examined in Paper 2 alongside scalar product, 3D coordinate work, and synoptically in Mechanics (Paper 3) for displacement and velocity. The specification expects fluent translation between Cartesian $(x, y, z)$(x,y,z) form, parametric form, and the vector equation form. The Edexcel formula booklet does not list the line equation $\mathbf{r} = \mathbf{a} + \lambda \mathbf{d}$r=a+λd — it must be quoted from memory.

---

Worked example with full mark scheme

Question (8 marks):

Two lines $\ell_1$ℓ1​ and $\ell_2$ℓ2​ have vector equations:

$\ell_1: \mathbf{r} = \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix} + \lambda \begin{pmatrix} 2 \\ -1 \\ 1 \end{pmatrix}, \quad \ell_2: \mathbf{r} = \begin{pmatrix} 4 \\ 0 \\ 5 \end{pmatrix} + \mu \begin{pmatrix} 1 \\ 1 \\ -1 \end{pmatrix}$ℓ1​:r=​123​​+λ​2−11​​,ℓ2​:r=​405​​+μ​11−1​​

(a) Show that $\ell_1$ℓ1​ and $\ell_2$ℓ2​ intersect, and find the point of intersection. (5)

(b) Find the acute angle between the lines, giving your answer to one decimal place. (3)

Solution with mark scheme:

(a) Step 1 — equate components using different parameters.

For intersection, the position vectors must be equal for some $\lambda$λ and $\mu$μ:

$\begin{pmatrix} 1 + 2\lambda \\ 2 - \lambda \\ 3 + \lambda \end{pmatrix} = \begin{pmatrix} 4 + \mu \\ \mu \\ 5 - \mu \end{pmatrix}$​1+2λ2−λ3+λ​​=​4+μμ5−μ​​

M1 — equating component-wise with distinct parameters. Using $\lambda$λ for both lines is the single most common error and earns zero on this step.

Step 2 — solve a 2-by-2 system.

From the $x$x-equation: $1 + 2\lambda = 4 + \mu$1+2λ=4+μ, so $\mu = 2\lambda - 3$μ=2λ−3. From the $y$y-equation: $2 - \lambda = \mu$2−λ=μ, so $\mu = 2 - \lambda$μ=2−λ.

Equating: $2\lambda - 3 = 2 - \lambda \implies 3\lambda = 5 \implies \lambda = \dfrac{5}{3}$2λ−3=2−λ⟹3λ=5⟹λ=35​.

Hence $\mu = 2 - \dfrac{5}{3} = \dfrac{1}{3}$μ=2−35​=31​.

M1 — solving any two equations simultaneously for $\lambda$λ and $\mu$μ. A1 — correct values $\lambda = \tfrac{5}{3}$λ=35​, $\mu = \tfrac{1}{3}$μ=31​.

Step 3 — verify the third equation.

Substitute into the $z$z-equation: LHS $= 3 + \tfrac{5}{3} = \tfrac{14}{3}$=3+35​=314​; RHS $= 5 - \tfrac{1}{3} = \tfrac{14}{3}$=5−31​=314​. Consistent, so the lines do intersect.

M1 — checking the third component (this is what shows intersection rather than skewness).

Step 4 — compute the point.

Substituting $\lambda = \tfrac{5}{3}$λ=35​ into $\ell_1$ℓ1​:

$\mathbf{r} = \begin{pmatrix} 1 + \tfrac{10}{3} \\ 2 - \tfrac{5}{3} \\ 3 + \tfrac{5}{3} \end{pmatrix} = \begin{pmatrix} \tfrac{13}{3} \\ \tfrac{1}{3} \\ \tfrac{14}{3} \end{pmatrix}$r=​1+310​2−35​3+35​​​=​313​31​314​​​

A1 — point of intersection $\left(\tfrac{13}{3}, \tfrac{1}{3}, \tfrac{14}{3}\right)$(313​,31​,314​).

(b) Step 1 — apply the scalar-product angle formula.

The angle between lines is determined by their direction vectors:

$\cos\theta = \dfrac{\mathbf{d}_1 \cdot \mathbf{d}_2}{|\mathbf{d}_1||\mathbf{d}_2|}$cosθ=∣d1​∣∣d2​∣d1​⋅d2​​

$\mathbf{d}_1 \cdot \mathbf{d}_2 = (2)(1) + (-1)(1) + (1)(-1) = 2 - 1 - 1 = 0$d1​⋅d2​=(2)(1)+(−1)(1)+(1)(−1)=2−1−1=0.

M1 — correct dot product. A1 — value $0$0, so $\cos\theta = 0$cosθ=0.

Step 2 — interpret.

$\theta = 90°$θ=90°, i.e. the lines are perpendicular.

A1 — $\theta = 90.0°$θ=90.0°.

Total: 8 marks (M4 A4).

---

Specimen question modelled on the Edexcel 9MA0 Paper 2 format

Question (6 marks): The line $\ell$ℓ passes through the points $A(1, 0, 2)$A(1,0,2) and $B(4, 3, -1)$B(4,3,−1).

(a) Find a vector equation for $\ell$ℓ. (2)

(b) Determine whether the point $P(7, 6, -4)$P(7,6,−4) lies on $\ell$ℓ. (2)

(c) The line $m$m has equation $\mathbf{r} = (2\mathbf{i} + \mathbf{j} + \mathbf{k}) + t(\mathbf{i} - \mathbf{j} + 2\mathbf{k})$r=(2i+j+k)+t(i−j+2k). Determine whether $\ell$ℓ and $m$m are parallel, intersecting, or skew. (2)

Mark scheme decomposition by AO:

(a)

• M1 (AO1.1a) — direction vector $\overrightarrow{AB} = \mathbf{b} - \mathbf{a} = (3, 3, -3)$AB=b−a=(3,3,−3) or any non-zero scalar multiple.
• A1 (AO1.1b) — vector equation $\mathbf{r} = (1, 0, 2) + \lambda(3, 3, -3)$r=(1,0,2)+λ(3,3,−3) or equivalent.

(b)

• M1 (AO2.1) — equating $P$P to the parametric form: $1 + 3\lambda = 7$1+3λ=7 gives $\lambda = 2$λ=2; verify $y$y and $z$z.
• A1 (AO2.4) — at $\lambda = 2$λ=2, $y = 6$y=6 ✓, $z = 2 - 6 = -4$z=2−6=−4 ✓, so $P$P lies on $\ell$ℓ.

(c)

• M1 (AO3.1a) — directions $(3,3,-3)$(3,3,−3) and $(1,-1,2)$(1,−1,2) are not scalar multiples (parallel rejected); set up component equations and obtain a contradiction in the third component.
• A1 (AO3.2a) — conclude skew.

Total: 6 marks split AO1 = 3, AO2 = 2, AO3 = 1. Edexcel uses three-part vector questions to escalate AO demand: routine setup (AO1), point-on-line verification (AO2), and the parallel/intersecting/skew classification (AO3).

---

Synoptic links

Connects to:

1. Section 10 — Scalar product: the angle between two lines is computed via $\cos\theta = \dfrac{\mathbf{d}_1 \cdot \mathbf{d}_2}{|\mathbf{d}_1||\mathbf{d}_2|}$cosθ=∣d1​∣∣d2​∣d1​⋅d2​​ applied to direction vectors. Perpendicularity of lines reduces to $\mathbf{d}_1 \cdot \mathbf{d}_2 = 0$d1​⋅d2​=0, mirroring the gradient-product rule $m_1 m_2 = -1$m1​m2​=−1 from coordinate geometry.
2. Section 3 — Coordinate geometry in the $(x,y)$(x,y)-plane: the 2D Cartesian form $y = mx + c$y=mx+c is the same line as $\mathbf{r} = (0, c) + \lambda(1, m)$r=(0,c)+λ(1,m). Eliminating $\lambda$λ from $x = \lambda$x=λ and $y = c + m\lambda$y=c+mλ recovers Cartesian form, exposing parametric form as a generalisation.
3. 3D vectors (section 11.5): vector equations of lines extend trivially into $\mathbb{R}^3$R3 where Cartesian form is no longer a single equation but a system, making the parametric form essential rather than optional.
4. Mechanics (9MA0-03) — motion in a straight line: a particle's position $\mathbf{r}(t) = \mathbf{r}_0 + t\mathbf{v}$r(t)=r0​+tv is literally a vector line equation, with parameter $t$t as time and direction $\mathbf{v}$v as velocity. Closest-approach problems between two particles are exactly skew-line geometry questions.
5. Computer graphics and ray-tracing: rendering a 3D scene casts a ray $\mathbf{r} = \mathbf{c} + \lambda \mathbf{d}$r=c+λd from the camera through each pixel and tests intersection with surfaces — a direct industrial application of A-Level line geometry.

---

Mark-scheme literacy

Vector-line questions split AO marks roughly as: