Why is "Using the same parameter $\lambda$ for both lines" a common mistake in Problem solving using equations of lines?

Why it happens: Copying the from the first line into the second without renaming. How to avoid it: Always relabel the second line's parameter as (or t). Two independent lines need two independent parameters. Source: 9231 Paper 1 Examiner Reports — recurring comment

Why is "Not checking the third equation after solving for $\lambda$ and $\mu$" a common mistake in Problem solving using equations of lines?

Why it happens: Assuming that solving two equations proves intersection; forgetting the system is over-determined (3 equations, 2 unknowns). How to avoid it: Substitute the values into the unused third equation. Consistent → intersect; contradiction → skew. Source: 9231 Paper 1 mark schemes

Why is "Using position vectors instead of directions in the angle formula" a common mistake in Problem solving using equations of lines?

Why it happens: Confusing the line's point a with its direction d. How to avoid it: The angle between lines depends only on d_1 and d_2 — the points are irrelevant. Source: 9231 Paper 1 Examiner Reports

Why is "Calling parallel lines 'skew'" a common mistake in Problem solving using equations of lines?

Why it happens: Jumping straight to the intersection equations, getting a contradiction, and concluding skew. How to avoid it: Test parallel FIRST. Parallel lines (directions proportional) are never skew, even though they don't meet. Source: 9231 Paper 1 Examiner Reports

Why is "Giving an obtuse angle between two lines" a common mistake in Problem solving using equations of lines?

Why it happens: Not taking the modulus of the scalar product when it is negative. How to avoid it: Use = |d_1·d_2||d_1||d_2| — the angle between lines is conventionally acute. Source: 9231 Paper 1 — angle questions

Why is "Showing lines intersect but not stating the point of intersection" a common mistake in Problem solving using equations of lines?

Why it happens: Stopping at the consistency check; overlooking the 'find the point' part of the question. How to avoid it: Once consistency is confirmed, substitute (or ) back into a line equation and state the coordinates. Source: 9231 Paper 1 Examiner Reports

Cambridge International A Levels Further Mathematics (9231)

Problem solving using equations of lines

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Detailed Study Notes

Detailed notes on Vectors for Cambridge International A Levels Further Mathematics, covering key concepts, explanations, examples, and exam-focused revision points.

Problem Solving Using Equations of Lines — Cambridge International AS & A Level Further Mathematics 9231 Paper 1 (2026-2027 syllabus)

A line in 3D is written $\mathbf{r} = \mathbf{a} + \lambda\mathbf{d}$ — a point plus a direction. Master finding intersections, the angle between two lines, and classifying a pair of lines as intersecting, parallel or skew.

At a glance

Vector line equation: $\mathbf{r} = \mathbf{a} + \lambda\mathbf{d}$ — $\mathbf{a}$ is a point on the line, $\mathbf{d}$ is its direction.
Intersection: equate the two lines, solve TWO component equations for $\lambda$ and $\mu$ , then CHECK the third.
Angle between lines: use $\cos\theta = \dfrac{\mathbf{d}_1\cdot\mathbf{d}_2}{|\mathbf{d}_1||\mathbf{d}_2|}$ on the DIRECTION vectors only.
Parallel $\Leftrightarrow$ directions are scalar multiples. Skew $\Leftrightarrow$ not parallel AND no intersection (3D only).
Different parameters: use $\lambda$ for line 1 and $\mu$ for line 2 — never the same letter for both.

What you’ll learn

Mapped to the Cambridge International A Level 9231 syllabus (2026-2027).

1.6d — Use the equation of a line in the form $\mathbf{r} = \mathbf{a} + \lambda\mathbf{d}$ to solve problems involving points, directions and intersections.
1.6e — Determine whether two lines are parallel, intersect, or are skew, and find the point of intersection where it exists.
1.6f — Find the angle between two lines using the scalar product of their direction vectors.

The vector equation of a line — point plus direction

r = a + λd: start at a point a, then walk any distance along a direction d.

The key idea. A straight line is completely described by one point on it and its direction. In vector form:

$\mathbf{r} = \mathbf{a} + \lambda\mathbf{d}$

$\mathbf{a}$ is the position vector of a known point on the line.
$\mathbf{d}$ is the direction vector — any vector pointing along the line.
$\lambda$ is a scalar parameter: as $\lambda$ varies over all real numbers, $\mathbf{r}$ traces out every point of the line.

Building the equation from two points. If a line passes through $A$ and $B$ , take $\mathbf{a}$ as the position vector of $A$ and direction $\mathbf{d} = \overrightarrow{AB} = \mathbf{b} - \mathbf{a}$ : $\mathbf{r} = \mathbf{a} + \lambda(\mathbf{b} - \mathbf{a})$

Direction is only fixed up to a scalar. $\mathbf{d} = (2,4,-2)$ and $\mathbf{d} = (1,2,-1)$ describe the same direction. Always reduce to simplest form, and remember two lines can share a direction (parallel) even when written with different multiples.

Starting at point A, every point r on the line is reached by adding some multiple λ of the direction d.

Cambridge tip. Read each component of $\mathbf{r} = \mathbf{a} + \lambda\mathbf{d}$ as a separate parametric equation: $x = a_1 + \lambda d_1$ , $y = a_2 + \lambda d_2$ , $z = a_3 + \lambda d_3$ . This is exactly what you compare when finding intersections.

$\mathbf{r} = \mathbf{a} + \lambda\mathbf{d}$ = point $\mathbf{a}$ plus parameter $\lambda$ times direction $\mathbf{d}$ .
Through two points: $\mathbf{d} = \mathbf{b} - \mathbf{a}$ .
Direction is fixed only up to a scalar multiple — simplify it.

See the full worked example for problem solving using equations of lines →

The angle between two lines

Use the scalar product of the DIRECTION vectors — the points are irrelevant.

The angle between two lines depends only on their directions, not on where they sit in space. So you work with the direction vectors $\mathbf{d}_1$ and $\mathbf{d}_2$ :

$\cos\theta = \frac{\mathbf{d}_1\cdot\mathbf{d}_2}{|\mathbf{d}_1|\,|\mathbf{d}_2|}$

Method.

Pick out the two direction vectors (ignore the position vectors entirely).
Compute the scalar product $\mathbf{d}_1\cdot\mathbf{d}_2$ and the two magnitudes.
Divide, then take $\cos^{-1}$ .

The acute-angle convention. Lines (unlike vectors) have no fixed direction, so the "angle between two lines" is taken to be acute (between $0°$ and $90°$ ). If the scalar product comes out negative, take the modulus of the cosine: $\cos\theta = \frac{|\mathbf{d}_1\cdot\mathbf{d}_2|}{|\mathbf{d}_1|\,|\mathbf{d}_2|}$

Perpendicular lines have $\mathbf{d}_1\cdot\mathbf{d}_2 = 0$ ; parallel lines have $\theta = 0$ .

The angle θ between the lines is found from the scalar product of d₁ and d₂; take it acute.

Cambridge tip. A frequent error is using the position vectors $\mathbf{a}_1, \mathbf{a}_2$ instead of the directions. The angle between lines never involves the points — only the $\mathbf{d}$ 's.

$\cos\theta = \dfrac{\mathbf{d}_1\cdot\mathbf{d}_2}{|\mathbf{d}_1||\mathbf{d}_2|}$ — directions only.
Take the modulus for the acute angle convention.
Perpendicular $\Leftrightarrow \mathbf{d}_1\cdot\mathbf{d}_2 = 0$ .

See the full worked example for problem solving using equations of lines →

Do two lines intersect? Solve and check

Equate the lines, solve two components for λ and μ, then test the third.

To test whether two lines $\mathbf{r}_1 = \mathbf{a}_1 + \lambda\mathbf{d}_1$ and $\mathbf{r}_2 = \mathbf{a}_2 + \mu\mathbf{d}_2$ meet, set them equal: $\mathbf{a}_1 + \lambda\mathbf{d}_1 = \mathbf{a}_2 + \mu\mathbf{d}_2$

This is three scalar equations (one per component) in two unknowns $\lambda, \mu$ .

The procedure (memorise this).

Use different parameters — $\lambda$ for line 1, $\mu$ for line 2. Using the same letter is a fatal error.
Solve any two of the three component equations simultaneously for $\lambda$ and $\mu$ .
Substitute into the THIRD equation.
- If it is satisfied → the lines intersect.
- If it is not satisfied (a contradiction) → the lines do not intersect.
Find the point by putting $\lambda$ (or $\mu$ ) back into the relevant line equation.

Why the third check is essential. In 3D, two lines almost always fail to meet. Two equations can always be solved for two unknowns, but that solution only describes a genuine crossing if it is consistent with the third component. Skipping the check is the single biggest error in this topic.

When the two values of λ and μ from any two components also satisfy the third, the lines genuinely cross.

Cambridge tip. Lay your three equations out neatly labelled (1), (2), (3). State which two you solved and which one you checked — examiners reward a clearly signposted method even if a number slips.

Equate the lines → three equations in $\lambda$ and $\mu$ .
Solve two for $\lambda, \mu$ ; substitute into the third to check consistency.
Consistent → intersect (find the point); inconsistent → no intersection.

See the full worked example for problem solving using equations of lines →

Classifying a pair of lines — parallel, intersecting or skew

First test parallel (directions multiples?); if not, test intersection; if neither, skew.

Given two lines in 3D, there are exactly three possibilities. Use this decision flow:

Step 1 — Are the directions parallel? Check whether $\mathbf{d}_2 = k\mathbf{d}_1$ for some scalar $k$ .

Yes, and the lines coincide (a point of one lies on the other) → the lines are the same line.
Yes, but they are distinct → the lines are parallel (never meet).

Step 2 — If NOT parallel, do they intersect? Run the solve-two-check-third procedure.

Consistent → the lines intersect at a point.
Inconsistent → the lines are skew.

What "skew" means. Skew lines are lines that are neither parallel nor intersecting. They exist only in 3D (in 2D, two non-parallel lines always meet). Think of two flight paths at different altitudes crossing over a city — they cross in plan view but never actually touch.

Summary table.

Directions parallel + share a point → same line.
Directions parallel + no shared point → parallel.
Directions not parallel + consistent → intersecting.
Directions not parallel + inconsistent → skew.

Skew lines: not parallel, yet they never intersect — one passes in front of the other with a gap between them.

Cambridge tip. Always test parallel first. If you jump straight to solving simultaneous equations on parallel lines you get a contradiction and may wrongly call them skew — parallel lines are NOT skew.

Three outcomes: same line, parallel, intersecting, or skew.
Test parallel first ( $\mathbf{d}_2 = k\mathbf{d}_1$ ?); if not, test intersection.
Skew = not parallel AND no intersection — a purely 3D situation.

See the full worked example for problem solving using equations of lines →

Quick recap

$\mathbf{r} = \mathbf{a} + \lambda\mathbf{d}$ : $\mathbf{a}$ a point, $\mathbf{d}$ a direction, $\lambda$ a free parameter.
Angle between lines: $\cos\theta = \frac{|\mathbf{d}_1\cdot\mathbf{d}_2|}{|\mathbf{d}_1||\mathbf{d}_2|}$ — directions only, take acute.
Intersection: equate lines, solve two components for $\lambda, \mu$ , CHECK the third.
Use different parameters $\lambda$ and $\mu$ for the two lines.
Parallel $\Leftrightarrow$ directions are scalar multiples; skew $\Leftrightarrow$ not parallel and no intersection.
Always test parallel before testing intersection.

Memorise this

Verbatim phrases and definitions Cambridge mark schemes credit.

Line equation $\mathbf{r} = \mathbf{a} + \lambda\mathbf{d}$ .
Angle between lines $\cos\theta = \frac{|\mathbf{d}_1\cdot\mathbf{d}_2|}{|\mathbf{d}_1||\mathbf{d}_2|}$ .
Intersection procedure: solve two components, check the third.
Skew = neither parallel nor intersecting (3D only).
Different parameters $\lambda, \mu$ for the two lines.

How it’s examined

Topic 1.6 line problems are a Paper 1 staple (typically 6-10 marks). A standard question gives two lines and asks you to (a) determine whether they intersect / are parallel / are skew, (b) find the point of intersection if it exists, and (c) find the acute angle between them. Marks are lost through (i) using the same parameter for both lines, (ii) omitting the third-component consistency check, and (iii) using position vectors instead of directions in the angle formula. Clear labelling of equations and an explicit consistency check secure the method marks.

Sources: Cambridge International AS & A Level Further Mathematics 9231 syllabus (2026-2027); 9231 Examiner Reports 2022-2024 (Paper 1); 9231/11 & 9231/13 Further Pure Mathematics 1 — past papers and mark schemes. Last reviewed 2026-06-23.

Take this whole topic with you

Step-by-step worked examples — Problem solving using equations of lines

Step-by-step solutions to past-paper-style questions on problem solving using equations of lines, written exactly the way a tutor would explain them at the board.

Question type:

1Vector equation of a line through two points
Getting startedDirect calculation• line-equation, direction
Question
Find a vector equation of the line through $A(1,2,3)$ and $B(4,0,5)$ .
Step-by-step solution
1. Step 1
  Direction vector $\mathbf{d} = \overrightarrow{AB} = \mathbf{b} - \mathbf{a}$ .
  $\mathbf{d} = (4-1,\,0-2,\,5-3) = (3,\,-2,\,2)$
2. Step 2
  Use $A$ as the point. Write $\mathbf{r} = \mathbf{a} + \lambda\mathbf{d}$ .
  $\mathbf{r} = \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix} + \lambda\begin{pmatrix} 3 \\ -2 \\ 2 \end{pmatrix}$
Answer
$\mathbf{r} = (1,2,3) + \lambda(3,-2,2)$ .
Examiner tip
Using $B$ as the point, or $\overrightarrow{BA}$ as the direction, gives an equally valid (different-looking) equation. The direction may also be scaled.
2Is a point on the line?
Getting startedIdentify & classify• line-equation, point
Question
Does the point $P(7,-2,9)$ lie on the line $\mathbf{r} = (1,2,3) + \lambda(3,-2,2)$ ?
Step-by-step solution
1. Step 1
  Equate components with $P$ and find $\lambda$ from the $x$ -equation.
  $7 = 1 + 3\lambda \;\Rightarrow\; \lambda = 2$
2. Step 2
  Check the same $\lambda$ works for $y$ and $z$ .
  $y: 2 - 2(2) = -2\;\checkmark \qquad z: 3 + 2(2) = 7 \ne 9$
3. Step 3
  The $z$ -equation fails for $\lambda = 2$ , so $P$ is not on the line.
Answer
No — at $\lambda = 2$ the $z$ -component gives $7$ , not $9$ , so $P$ does not lie on the line.
Examiner tip
A point lies on a line only if a SINGLE value of $\lambda$ satisfies all three component equations. One failure rules it out.
3Angle between two lines
Building confidenceMulti-step problem• angle, scalar-product
Question
Find the acute angle between the lines with directions $\mathbf{d}_1 = (2,1,2)$ and $\mathbf{d}_2 = (1,-2,2)$ , to the nearest $0.1°$ .
Step-by-step solution
1. Step 1
  Scalar product of the directions.
  $\mathbf{d}_1\cdot\mathbf{d}_2 = (2)(1)+(1)(-2)+(2)(2) = 2-2+4 = 4$
2. Step 2
  Magnitudes.
  $|\mathbf{d}_1| = \sqrt{4+1+4} = 3, \qquad |\mathbf{d}_2| = \sqrt{1+4+4} = 3$
3. Step 3
  Apply the cosine formula (take modulus for the acute angle).
  $\cos\theta = \frac{|4|}{3\times 3} = \frac{4}{9}$
4. Step 4
  Inverse cosine.
  $\theta = \cos^{-1}\!\left(\tfrac{4}{9}\right) = 63.6°$
Answer
The acute angle between the lines is $63.6°$ (1 d.p.).
Examiner tip
Both magnitudes are $3$ here — convenient, but always compute them. Use only the DIRECTION vectors, never the points on the lines.
4Finding the point of intersection
Building confidenceMulti-step problem• intersection, simultaneous
Question
Show that the lines $\ell_1: \mathbf{r} = (1,0,1) + \lambda(2,1,1)$ and $\ell_2: \mathbf{r} = (5,3,0) + \mu(1,1,-1)$ intersect, and find the point of intersection.
Step-by-step solution
1. Step 1
  Equate the lines component by component (different parameters $\lambda, \mu$ ).
  $(1): 1+2\lambda = 5+\mu, \quad (2): \lambda = 3+\mu, \quad (3): 1+\lambda = -\mu$
2. Step 2
  Solve (2) and (3) simultaneously. From (2) $\lambda - \mu = 3$ ; from (3) $\lambda + \mu = -1$ .
  $\text{Adding: } 2\lambda = 2 \Rightarrow \lambda = 1, \;\; \mu = -2$
3. Step 3
  Check the remaining equation (1).
  $1 + 2(1) = 3 \;\text{ and }\; 5 + (-2) = 3 \;\checkmark$
4. Step 4
  Consistent — so they intersect. Substitute $\lambda = 1$ into $\ell_1$ .
  $\mathbf{r} = (1,0,1) + 1\cdot(2,1,1) = (3,\,1,\,2)$
Answer
The lines intersect at the point $(3,\,1,\,2)$ .
Examiner tip
Always verify the point in BOTH lines: $\ell_2$ at $\mu=-2$ gives $(5,3,0)+(-2)(1,1,-1)=(3,1,2)$ . The third-equation check is what proves intersection.
5Classifying lines as skew
StretchShow that / prove• skew, intersection
Question
Show that the lines $\ell_1: \mathbf{r} = (1,0,1) + \lambda(2,1,1)$ and $\ell_2: \mathbf{r} = (5,3,0) + \mu(1,1,2)$ are skew.
Step-by-step solution
1. Step 1
  Test parallel first. Is $(1,1,2)$ a multiple of $(2,1,1)$ ? $\frac{1}{2} \ne \frac{1}{1}$ — no. So not parallel.
2. Step 2
  Equate the lines to test for intersection.
  $(1): 1+2\lambda = 5+\mu, \quad (2): \lambda = 3+\mu, \quad (3): 1+\lambda = 2\mu$
3. Step 3
  Solve (1) and (2). From (2) $\mu = \lambda - 3$ ; sub into (1): $1+2\lambda = 5 + \lambda - 3$ .
  $1 + 2\lambda = 2 + \lambda \Rightarrow \lambda = 1, \;\; \mu = -2$
4. Step 4
  Check (3).
  $1 + 1 = 2 \quad\text{but}\quad 2(-2) = -4 \Rightarrow 2 \ne -4$
5. Step 5
  (3) is contradicted, so the lines do not intersect. Not parallel + no intersection $\Rightarrow$ skew.
Answer
The lines are not parallel and the consistency check fails ( $2 \ne -4$ ), so they are skew.
Examiner tip
A 'show that ... skew' answer must do BOTH: rule out parallel AND show the intersection equations are inconsistent. Stating only one is incomplete.
6Showing two lines are perpendicular
StretchShow that / prove• perpendicular, angle
Question
The lines $\ell_1$ and $\ell_2$ have directions $\mathbf{d}_1 = (3,-1,2)$ and $\mathbf{d}_2 = (1,1,-1)$ . Show that $\ell_1$ and $\ell_2$ are perpendicular.
Step-by-step solution
1. Step 1
  Perpendicular lines have $\mathbf{d}_1\cdot\mathbf{d}_2 = 0$ — compute the scalar product.
  $\mathbf{d}_1\cdot\mathbf{d}_2 = (3)(1) + (-1)(1) + (2)(-1) = 3 - 1 - 2 = 0$
2. Step 2
  The scalar product is $0$ , so the directions — and hence the lines — are perpendicular.
Answer
$\mathbf{d}_1\cdot\mathbf{d}_2 = 0$ , so the lines are perpendicular.
Examiner tip
Perpendicularity of LINES is a statement about their DIRECTIONS only — the lines need not even intersect. Show the full $3-1-2$ working before concluding '= 0'.

Model Answers — Problem solving using equations of lines

High-scoring sample answers for problem solving using equations of lines on the Cambridge IGCSE paper, with examiner-style notes mapping each response to the mark scheme and assessment objectives.

Question 1
3 marks
Find a vector equation of the line passing through $A(2,-1,4)$ and $B(5,1,2)$ .
Model answer
Direction vector: $\mathbf{d} = \overrightarrow{AB} = \mathbf{b} - \mathbf{a} = (5-2,\;1-(-1),\;2-4) = (3,\,2,\,-2)$

Using $A$ as the point on the line: $\mathbf{r} = \begin{pmatrix} 2 \\ -1 \\ 4 \end{pmatrix} + \lambda\begin{pmatrix} 3 \\ 2 \\ -2 \end{pmatrix}$
Why this scores
M1 for the direction $\overrightarrow{AB}$ , A1 for correct components $(3,2,-2)$ , A1 for a complete equation $\mathbf{r} = \mathbf{a} + \lambda\mathbf{d}$ . Any valid point/direction combination is accepted.
Question 2
4 marks
Find, to the nearest $0.1°$ , the acute angle between the lines with direction vectors $(1,2,2)$ and $(2,-1,0)$ .
Model answer
Scalar product: $\mathbf{d}_1\cdot\mathbf{d}_2 = (1)(2) + (2)(-1) + (2)(0) = 2 - 2 + 0 = 0$

Since the scalar product is $0$ , the lines are perpendicular, so: $\cos\theta = \frac{|0|}{|\mathbf{d}_1||\mathbf{d}_2|} = 0 \;\Rightarrow\; \theta = 90.0°$
Why this scores
M1 for the scalar product, A1 for $0$ , M1 for recognising $\cos\theta = 0$ , A1 for $90.0°$ . Computing the magnitudes is unnecessary once the dot product is zero — but state why.
Question 3
8 marks
The lines $\ell_1: \mathbf{r} = (2,1,3) + \lambda(1,2,1)$ and $\ell_2: \mathbf{r} = (5,4,3) + \mu(2,1,-1)$ are given. (a) Show that the lines intersect and find the point of intersection. (b) Find the acute angle between them, to the nearest $0.1°$ .
Model answer
(a) Equate the lines, using parameters $\lambda$ and $\mu$ : $(1):\; 2 + \lambda = 5 + 2\mu, \qquad (2):\; 1 + 2\lambda = 4 + \mu, \qquad (3):\; 3 + \lambda = 3 - \mu$

From (1): $\lambda - 2\mu = 3$ . From (2): $2\lambda - \mu = 3$ .

Solve: from the first, $\lambda = 3 + 2\mu$ ; substitute into the second: $2(3 + 2\mu) - \mu = 3 \;\Rightarrow\; 6 + 3\mu = 3 \;\Rightarrow\; \mu = -1,\; \lambda = 1$

Check (3): $3 + 1 = 4$ and $3 - (-1) = 4$ , so $4 = 4\;\checkmark$ . The system is consistent, so the lines intersect.

Point of intersection — substitute $\lambda = 1$ into $\ell_1$ : $\mathbf{r} = (2,1,3) + 1\cdot(1,2,1) = (3,\,3,\,4)$ (Verify in $\ell_2$ with $\mu = -1$ : $(5,4,3) + (-1)(2,1,-1) = (3,3,4)\;\checkmark$ .)

(b) Angle between the directions: $\mathbf{d}_1\cdot\mathbf{d}_2 = (1)(2)+(2)(1)+(1)(-1) = 2+2-1 = 3$ $|\mathbf{d}_1| = \sqrt{1+4+1} = \sqrt{6}, \quad |\mathbf{d}_2| = \sqrt{4+1+1} = \sqrt{6}$ $\cos\theta = \frac{|3|}{\sqrt6\cdot\sqrt6} = \frac{3}{6} = \frac12 \;\Rightarrow\; \theta = 60.0°$
Why this scores
Part (a): M1 to set up and equate, M1 to solve two equations, A1 for $\lambda=1,\mu=-1$ , A1 for testing the third equation, showing consistency and stating the point $(3,3,4)$ . Part (b): M1 scalar product, M1 magnitudes, A1 $\cos\theta=\tfrac12$ , A1 $60.0°$ . The third-equation check is the proof of intersection.
Question 4
5 marks
Two lines are $\ell_1: \mathbf{r} = (1,2,0) + \lambda(2,-1,3)$ and $\ell_2: \mathbf{r} = (4,0,1) + \mu(-4,2,-6)$ . Determine whether the lines are parallel, and state whether they are the same line.
Model answer
Test parallel. Is $\mathbf{d}_2$ a scalar multiple of $\mathbf{d}_1$ ? $\mathbf{d}_2 = (-4,\,2,\,-6) = -2\,(2,\,-1,\,3) = -2\,\mathbf{d}_1$ Yes — so the lines are parallel.

Are they the same line? Check whether a point of $\ell_2$ , say $(4,0,1)$ , lies on $\ell_1$ . From the $x$ -component: $4 = 1 + 2\lambda \;\Rightarrow\; \lambda = \tfrac{3}{2}$ Check $y$ : $2 - \tfrac32 = \tfrac12 \ne 0$ .

The point $(4,0,1)$ does not lie on $\ell_1$ , so the lines are distinct parallel lines, not the same line.
Why this scores
M1 for testing direction proportionality, A1 for $\mathbf{d}_2 = -2\mathbf{d}_1$ (parallel); M1 for substituting a point of one line into the other, A1 for the $y$ -mismatch, A1 for concluding 'distinct parallel'. A common error is calling parallel lines skew — parallel lines are never skew.
Question 5
7 marks
The line $\ell: \mathbf{r} = (1,0,2) + \lambda(2,1,2)$ and the point $P(4,3,4)$ are given. Find the foot $F$ of the perpendicular from $P$ to $\ell$ , and hence the shortest distance from $P$ to $\ell$ .
Model answer
General point on $\ell$ : $F = (1+2\lambda,\; \lambda,\; 2+2\lambda)$ .

Vector $\overrightarrow{PF}$ : $\overrightarrow{PF} = F - P = (1+2\lambda - 4,\; \lambda - 3,\; 2+2\lambda - 4) = (2\lambda - 3,\; \lambda - 3,\; 2\lambda - 2)$

$\overrightarrow{PF}$ is perpendicular to the direction $\mathbf{d} = (2,1,2)$ , so $\overrightarrow{PF}\cdot\mathbf{d} = 0$ : $2(2\lambda - 3) + 1(\lambda - 3) + 2(2\lambda - 2) = 0$ $4\lambda - 6 + \lambda - 3 + 4\lambda - 4 = 0 \;\Rightarrow\; 9\lambda - 13 = 0 \;\Rightarrow\; \lambda = \tfrac{13}{9}$

Foot of perpendicular: $F = \left(1 + \tfrac{26}{9},\; \tfrac{13}{9},\; 2 + \tfrac{26}{9}\right) = \left(\tfrac{35}{9},\; \tfrac{13}{9},\; \tfrac{44}{9}\right)$

Shortest distance $= |\overrightarrow{PF}|$ . With $\lambda = \tfrac{13}{9}$ : $\overrightarrow{PF} = \left(\tfrac{26}{9} - 3,\; \tfrac{13}{9} - 3,\; \tfrac{26}{9} - 2\right) = \left(-\tfrac{1}{9},\; -\tfrac{14}{9},\; \tfrac{8}{9}\right)$ $|\overrightarrow{PF}| = \tfrac{1}{9}\sqrt{(-1)^2 + (-14)^2 + 8^2} = \tfrac{1}{9}\sqrt{1 + 196 + 64} = \tfrac{1}{9}\sqrt{261} = \tfrac{\sqrt{261}}{9} \approx 1.80$
Why this scores
M1 for the general point, M1 for $\overrightarrow{PF}$ , M1 for setting $\overrightarrow{PF}\cdot\mathbf{d}=0$ , A1 for $\lambda = \tfrac{13}{9}$ , A1 for $F$ ; M1 magnitude, A1 for $\tfrac{\sqrt{261}}{9} \approx 1.80$ . The perpendicular condition $\overrightarrow{PF}\cdot\mathbf{d}=0$ is the engine of the whole question.
Question 6
8 marks
The lines $\ell_1: \mathbf{r} = (1,1,1) + \lambda(1,2,-1)$ and $\ell_2: \mathbf{r} = (2,0,k) + \mu(1,-1,1)$ intersect. Find the value of $k$ and the point of intersection.
Model answer
Equate the lines (parameters $\lambda, \mu$ ): $(1):\; 1 + \lambda = 2 + \mu, \qquad (2):\; 1 + 2\lambda = -\mu, \qquad (3):\; 1 - \lambda = k + \mu$

Solve (1) and (2) for $\lambda, \mu$ . From (1): $\lambda - \mu = 1$ . From (2): $2\lambda + \mu = -1$ .

Adding: $3\lambda = 0 \Rightarrow \lambda = 0$ , and then $\mu = \lambda - 1 = -1$ .

Use (3) to find $k$ (the lines intersect, so (3) must be consistent): $1 - 0 = k + (-1) \;\Rightarrow\; k = 2$

Point of intersection — substitute $\lambda = 0$ into $\ell_1$ : $\mathbf{r} = (1,1,1) + 0\cdot(1,2,-1) = (1,\,1,\,1)$

Verify in $\ell_2$ with $\mu = -1$ and $k = 2$ : $(2,0,2) + (-1)(1,-1,1) = (1,1,1)\;\checkmark$
Why this scores
M1 to equate and form three equations, M1 to solve the two not containing $k$ , A1 for $\lambda = 0, \mu = -1$ , M1 A1 for $k = 2$ from the third equation, M1 A1 for the point $(1,1,1)$ , plus the verification. Here the third equation FIXES the unknown rather than being a pass/fail check.

Key Formulae — Problem solving using equations of lines

The formulae you need to memorise for problem solving using equations of lines on the Cambridge IGCSE paper, with every variable defined in plain English and a note on when to use it.

Vector equation of a line
$\mathbf{r} = \mathbf{a} + \lambda\mathbf{d}$
$\mathbf{a}$
position vector of a known point on the line
$\mathbf{d}$
direction vector of the line
$\lambda$
scalar parameter (any real number)
When to use
The standard form for any line in 3D. Read each component as a parametric equation when comparing lines.
Direction from two points
$\mathbf{d} = \overrightarrow{AB} = \mathbf{b} - \mathbf{a}$
$\mathbf{a}, \mathbf{b}$
position vectors of points $A$ and $B$ on the line
When to use
Build a line through two known points. Any scalar multiple of $\mathbf{d}$ is an equally valid direction.
Angle between two lines
$\cos\theta = \frac{|\mathbf{d}_1\cdot\mathbf{d}_2|}{|\mathbf{d}_1|\,|\mathbf{d}_2|}$
$\mathbf{d}_1, \mathbf{d}_2$
direction vectors of the two lines
$\theta$
the acute angle between the lines
When to use
Uses the DIRECTIONS only. Take the modulus of the scalar product to guarantee an acute angle.
Perpendicularity of lines
$\mathbf{d}_1\cdot\mathbf{d}_2 = 0 \;\Leftrightarrow\; \ell_1 \perp \ell_2$
$\mathbf{d}_1\cdot\mathbf{d}_2$
scalar product of the two direction vectors
When to use
Two lines are perpendicular exactly when their direction vectors have zero scalar product (whether or not they meet).
Intersection consistency
$\mathbf{a}_1 + \lambda\mathbf{d}_1 = \mathbf{a}_2 + \mu\mathbf{d}_2$
$\lambda, \mu$
the (distinct) parameters of the two lines
When to use
Equate the lines to get three equations. Solve two for $\lambda, \mu$ ; the lines intersect iff the third is then satisfied.

Key Definitions and Keywords — Problem solving using equations of lines

Definitions to memorise and the exact keywords mark schemes credit for problem solving using equations of lines answers — sharpened from recent examiner reports for the 2026 Cambridge IGCSE sitting.

Direction vector
Examiner keyword
A vector $\mathbf{d}$ pointing along a line. Fixed only up to a non-zero scalar multiple — $(2,4,-2)$ and $(1,2,-1)$ give the same direction.
Parameter
Examiner keyword
The scalar $\lambda$ (or $\mu$ ) in $\mathbf{r} = \mathbf{a} + \lambda\mathbf{d}$ that selects a point on the line. Use different letters for different lines.
Skew lines
Examiner keyword
Two lines that are neither parallel nor intersecting. They exist only in three dimensions and never meet.
Parallel lines
Examiner keyword
Lines whose direction vectors are scalar multiples of each other. If they also share a point they coincide (the same line); otherwise they never meet.
Point of intersection
Examiner keyword
The single point lying on both lines, found by substituting the consistent parameter value back into either line equation.
Foot of the perpendicular
The point $F$ on a line closest to a given point $P$ ; characterised by $\overrightarrow{PF}\cdot\mathbf{d} = 0$ , where $\mathbf{d}$ is the line's direction.

Common Mistakes and Misconceptions — Problem solving using equations of lines

The traps other students keep falling into on problem solving using equations of lines questions — taken from recent Cambridge IGCSE examiner reports and mark schemes — and how to avoid them.

✕Using the same parameter $\lambda$ for both lines
9231 Paper 1 Examiner Reports — recurring comment
Why it happens
Copying the $\lambda$ from the first line into the second without renaming.
How to avoid it
Always relabel the second line's parameter as $\mu$ (or $t$ ). Two independent lines need two independent parameters.
✕Not checking the third equation after solving for $\lambda$ and $\mu$
9231 Paper 1 mark schemes
Why it happens
Assuming that solving two equations proves intersection; forgetting the system is over-determined (3 equations, 2 unknowns).
How to avoid it
Substitute the values into the unused third equation. Consistent → intersect; contradiction → skew.
✕Using position vectors instead of directions in the angle formula
9231 Paper 1 Examiner Reports
Why it happens
Confusing the line's point $\mathbf{a}$ with its direction $\mathbf{d}$ .
How to avoid it
The angle between lines depends only on $\mathbf{d}_1$ and $\mathbf{d}_2$ — the points are irrelevant.
✕Calling parallel lines 'skew'
9231 Paper 1 Examiner Reports
Why it happens
Jumping straight to the intersection equations, getting a contradiction, and concluding skew.
How to avoid it
Test parallel FIRST. Parallel lines (directions proportional) are never skew, even though they don't meet.
✕Giving an obtuse angle between two lines
9231 Paper 1 — angle questions
Why it happens
Not taking the modulus of the scalar product when it is negative.
How to avoid it
Use $\cos\theta = \frac{|\mathbf{d}_1\cdot\mathbf{d}_2|}{|\mathbf{d}_1||\mathbf{d}_2|}$ — the angle between lines is conventionally acute.
✕Showing lines intersect but not stating the point of intersection
9231 Paper 1 Examiner Reports
Why it happens
Stopping at the consistency check; overlooking the 'find the point' part of the question.
How to avoid it
Once consistency is confirmed, substitute $\lambda$ (or $\mu$ ) back into a line equation and state the coordinates.

Problem solving using equations of lines

Detailed Study Notes

At a glance

What you’ll learn

Quick recap

Memorise this

How it’s examined

Take this whole topic with you

1Vector equation of a line through two points

2Is a point on the line?

3Angle between two lines

4Finding the point of intersection

5Classifying lines as skew

6Showing two lines are perpendicular

Vector equation of a line

Direction from two points

Angle between two lines

Perpendicularity of lines

Intersection consistency

Direction vector

Parameter

Skew lines

Parallel lines

Point of intersection

Foot of the perpendicular

✕Using the same parameter λ\lambdaλ for both lines

✕Not checking the third equation after solving for λ\lambdaλ and μ\muμ

✕Using position vectors instead of directions in the angle formula

✕Calling parallel lines 'skew'

✕Giving an obtuse angle between two lines

✕Showing lines intersect but not stating the point of intersection

✕Using the same parameter $\lambda$ for both lines

✕Not checking the third equation after solving for $\lambda$ and $\mu$