Consider the following two lines in 3D space: \[L_1: r = \begin{pmatrix} 1 \\ -1 \\ 2 \end{pmatrix} + \lambda \begin{pmatrix} 2 \\ 1 \\ -3 \end{pmatrix} \quad \text{and} \quad L_2: r = \begin{pmatrix} 4 \\ 2 \\ -1 \end{pmatrix} + \mu \begin{pmatrix} -1 \\ 2 \\ 1 \end{pmatrix}\] Show algebraically that these two lines are skew (they are neither parallel nor do they intersect).

No $\lambda, \mu$ satisfy all three component equations ($-4.8 \ne -3$) and the directions are not parallel, so the lines are skew

Two lines have vector equations $l_1: \mathbf{r} = (2\mathbf{i} + 3\mathbf{j}) + t(\mathbf{i} - \mathbf{j} + \mathbf{k})$ and $l_2: \mathbf{r} = (-3\mathbf{j} - 2\mathbf{k}) + s(\mathbf{i} + \mathbf{j} + \mathbf{k})$. Prove that the lines intersect and find the position vector of their point of intersection.

M1 for setting up equations for the components: $2 + t = s$, $3 - t = -3 + s$, and $t = -2 + s$. M1 for solving the first two equations simultaneously. From $s = 2+t$, substituting into the second gives $3 - t = -3 + (2+t)$. M1 for solving to find $2t = 4 \implies t = 2$, which means $s = 4$. M1 for checking consistency in the third (z) equation: $t = 2$, and $-2 + s = -2 + 4 = 2$. Since $2=2$, the lines intersect. A1 for substituting $t=2$ or $s=4$ back to find the point of intersection: $4\mat

IAL P4 Vectors: Exam Questions & Answers

P4 Vectors notes

Magnitude, unit vectors and position vectors

$| 2 i - 3 j + 6 k | = \sqrt{4 + 9 + 36} = 7$ , so its unit vector is $\frac{1}{7} (2 i - 3 j + 6 k)$ . With position vectors $a$ and $b$ , $\vec{A B} = b - a$ .

The scalar product and angles

$a \cdot b = a_{1} b_{1} + a_{2} b_{2} + a_{3} b_{3} = | a | | b | \cos θ$ . For $a = i + 2 j + 2 k$ and $b = 2 i - j + 2 k$ : $a \cdot b = 4$ , $| a | = | b | = 3$ , so $\cos θ = \frac{4}{9}$ and $θ = {63.6}^{\circ}$ . Two non-zero vectors are perpendicular exactly when their scalar product is 0.

Vector equations of lines

$r = a + λ b$ passes through the point with position vector $a$ in the direction $b$ . Lines with parallel direction vectors are parallel. Otherwise equate components: if values of $λ$ and $μ$ satisfy all three equations the lines meet; if not, they are skew.

The foot of a perpendicular

To find the point on a line closest to $P$ , write $\vec{P X}$ for a general point $X$ on the line and solve $\vec{P X} \cdot b = 0$ . The shortest distance is then $| \vec{P X} |$ .

Key formulas

Vectors Learn it: $a \cdot b = a_{1} b_{1} + a_{2} b_{2} + a_{3} b_{3} = | a | | b | \cos θ$
Line; perpendicular Learn it: $r = a + λ b; a \cdot b = 0$
Magnitude Learn it: $| a | = \sqrt{a_{1}^{2} + a_{2}^{2} + a_{3}^{2}}$

In the formula booklet = printed in Pearson's IAL formulae booklet, so know where to find it. Learn it = not printed there, so learn it by heart. All P4 formulas on one page (free PDF) →

P4 Vectors worked examples

Each mark scheme is checked line by line before it is published here. M marks are for method, A marks for accurate answers that follow from them, B marks for independent results.

Worked example 1 · 7 marks · hard

Line $L_{1}$ has vector equation $r_{1} = (\begin{matrix} 1 \\ 0 \\ 1 \end{matrix}) + λ (\begin{matrix} 2 \\ 1 \\ 1 \end{matrix})$ and line $L_{2}$ has equation $r_{2} = (\begin{matrix} 5 \\ - 3 \\ 0 \end{matrix}) + μ (\begin{matrix} - 1 \\ 2 \\ 1 \end{matrix})$ .

Show that $L_{1}$ and $L_{2}$ intersect.
Find the exact coordinates of the point of intersection.

Mark scheme

M1 for equating the parametric equations: $1 + 2 λ = 5 - μ$ , $λ = - 3 + 2 μ$ , and $1 + λ = μ$ .
M1 for taking two equations to solve for the parameters, e.g., $λ = μ - 1$ into $2 λ + μ = 4$ .
M1 for $2 (μ - 1) + μ = 4 ⟹ 3 μ - 2 = 4 ⟹ 3 μ = 6$ .
A1 for finding $μ = 2$ and $λ = 1$ .
B1 for verifying with the unused equation: $λ - 2 μ = 1 - 4 = - 3$ , which matches the second component equation, showing the lines intersect.
M1 for substituting $λ = 1$ into $L_{1}$ (or $μ = 2$ into $L_{2}$ ).
A1 for the intersection point $(3, 1, 2)$ .

2 steps

Worked example 2 · 8 marks · hard

Consider the following two lines in 3D space: $L_{1} : r = (\begin{matrix} 1 \\ - 1 \\ 2 \end{matrix}) + λ (\begin{matrix} 2 \\ 1 \\ - 3 \end{matrix}) and L_{2} : r = (\begin{matrix} 4 \\ 2 \\ - 1 \end{matrix}) + μ (\begin{matrix} - 1 \\ 2 \\ 1 \end{matrix})$ Show algebraically that these two lines are skew (they are neither parallel nor do they intersect).

Mark scheme

Set the components equal:
$1 + 2 λ = 4 - μ ⟹ 2 λ + μ = 3$ (Eq 1). M1
$- 1 + λ = 2 + 2 μ ⟹ λ - 2 μ = 3$ (Eq 2). M1
$2 - 3 λ = - 1 + μ ⟹ - 3 λ - μ = - 3$ (Eq 3). M1
Solve Eq 1 and Eq 2: $2 (3 + 2 μ) + μ = 3 ⟹ 6 + 5 μ = 3 ⟹ μ = - 0.6$ . M1
Substitute $μ = - 0.6 ⟹ λ = 1.8$ . A1
Check in Eq 3: $- 3 (1.8) - (- 0.6) = - 5.4 + 0.6 = - 4.8$ . M1
However, the RHS of Eq 3 is $- 3$ . Since $- 4.8 \neq - 3$ , there is no consistent solution. B1
Therefore, the lines do not intersect. Since their direction vectors are not scalar multiples, they are skew. A1

Worked example 3 · 5 marks · hard

Two lines have vector equations $l_{1} : r = (2 i + 3 j) + t (i - j + k)$ and $l_{2} : r = (- 3 j - 2 k) + s (i + j + k)$ . Prove that the lines intersect and find the position vector of their point of intersection.

Mark scheme

M1 for setting up equations for the components: $2 + t = s$ , $3 - t = - 3 + s$ , and $t = - 2 + s$ .
M1 for solving the first two equations simultaneously. From $s = 2 + t$ , substituting into the second gives $3 - t = - 3 + (2 + t)$ .
M1 for solving to find $2 t = 4 ⟹ t = 2$ , which means $s = 4$ .
M1 for checking consistency in the third (z) equation: $t = 2$ , and $- 2 + s = - 2 + 4 = 2$ . Since $2 = 2$ , the lines intersect.
A1 for substituting $t = 2$ or $s = 4$ back to find the point of intersection: $4 i + j + 2 k$ .

5 steps · 1 to fill in

Common P4 vectors mistakes

Using position vectors for the angle between lines. The angle between two lines uses their direction vectors, not the points they pass through.
Checking only two equations. Two lines meet only if one pair of values for λ and μ fits all three component equations. Check the third.
AB the wrong way round. $\vec{A B} = b - a$ , not $a - b$ .
The obtuse angle. If the scalar product is negative, $θ$ is obtuse; for the acute angle between lines, use $180^{\circ} - θ$ .

Every P4 mistake from the examiner reports, with the paper it came from →

P4 Vectors FAQ

How do I show two lines are skew?

Show their direction vectors are not parallel, then show that the equations for the components have no solution: values of λ and μ that fit two components do not fit the third.

How do I find the angle between two lines?

Take their direction vectors b₁ and b₂ and use cos θ = (b₁ · b₂) / (|b₁||b₂|). Give the acute angle unless the question asks otherwise.

Is a calculator allowed in IAL P4?

Yes. A calculator is allowed in every Edexcel IAL maths unit, P4 included. Method marks are for the working you write down, so show it.

How long is the IAL P4 exam?

Each IAL maths unit, P4 included, is one written paper of 1 hour 30 minutes worth 75 marks.

How often does vectors come up in P4?

In our mapping of the 4 Pearson P4 papers from June 2024 to June 2025, vectors was tested in 4 of them (5 questions in all). Any topic can be examined in any series.

Next steps

More P4: Partial Fractions · Parametric Equations · Differential Equations · Proof · The Binomial Expansion · Differentiation · Integration

Edexcel IAL P4 Vectors: exam questions and answers, notes and worked examples

Try it first

P4 Vectors notes

Magnitude, unit vectors and position vectors

The scalar product and angles

Vector equations of lines

The foot of a perpendicular

Key formulas

P4 Vectors worked examples

Worked example 1 · 7 marks · hard

Mark scheme

Worked example 2 · 8 marks · hard

Mark scheme

Worked example 3 · 5 marks · hard

Mark scheme

Common P4 vectors mistakes

P4 Vectors FAQ

How do I show two lines are skew?

How do I find the angle between two lines?

Is a calculator allowed in IAL P4?

How long is the IAL P4 exam?

How often does vectors come up in P4?

Next steps

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