2D and 3D vectors, magnitudes, unit vectors, dot product, vector equations of lines, and the four classifications: parallel, coincident, intersecting, skew. Includes angle between lines and distance from a point to a line. The P4 signature topic — every paper has an 8-12 mark vectors question worth its own grade boost.
What you’ll learn
Mapped to the Cambridge IGCSE XMA01-YMA01 syllabus (2026 onwards).
P4 6.1 — Use Cartesian vector notation in 2D and 3D, including position vectors and magnitudes.
P4 6.2 — Add, subtract and scalar-multiply vectors; find unit vectors.
P4 6.3 — Use the scalar (dot) product to find the angle between vectors and to test perpendicularity.
P4 6.4 — Find vector equations of lines and use them to determine if two lines are parallel, intersecting or skew.
P4 6.5 — Find the angle between two lines and the distance from a point to a line.
Notation, magnitude, unit vectors
Position vector, length, normalisation.
Notation.
Column form: v=xyz.
i,j,k form: v=xi+yj+zk.
Tuple: v=(x,y,z).
All three are accepted in IAL exams. Stick to one in any solution.
Position vector. The position vector of point A=(a1,a2,a3) is OA=a1i+a2j+a3k, often written a.
Vector from A to B.AB=b−a. So if A=(1,0,2) and B=(4,−1,3), then AB=(3,−1,1).
Magnitude.∣v∣=x2+y2+z2. Pythagoras in 3D.
Worked example.a=2i−3j+6k. Then ∣a∣=4+9+36=7.
Unit vector.a^=∣a∣a. For a=(2,−3,6): a^=71(2,−3,6)=(72,−73,76).
Standard basis.i=(1,0,0), j=(0,1,0), k=(0,0,1).
Vector arithmetic. Componentwise:
a+b=(a1+b1,a2+b2,a3+b3).
a−b=(a1−b1,…).
λa=(λa1,λa2,λa3).
Magnitude: ∑xi2.
Unit vector: divide by magnitude.
AB=b−a (target minus start).
All operations are componentwise.
Dot product
Sum of componentwise products; equals ∣a∣∣b∣cosθ.
Algebraic definition.
a⋅b=a1b1+a2b2+a3b3.
Geometric formula (formula booklet):
a⋅b=∣a∣∣b∣cosθ,0≤θ≤π.
Properties.
Commutative: a⋅b=b⋅a.
Distributive over addition: a⋅(b+c)=a⋅b+a⋅c.
a⋅a=∣a∣2.
a⋅b=0 iff a⊥b (for non-zero vectors).
Angle between vectors.
cosθ=∣a∣∣b∣a⋅b.
Worked example. Angle between a=(1,2,2) and b=(3,−1,2).
a⋅b=3−2+4=5.
∣a∣=3, ∣b∣=14.
cosθ=5/(314)≈0.4454.
θ≈63.5°.
Perpendicularity.a⊥b⇔a⋅b=0.
Worked example. Find k such that (2,3,−1) and (k,2,4) are perpendicular.
2k+6−4=0⇒k=−1.
Acute angle between LINES. A line ℓ defines TWO directions (you can go either way along it), so when asked for the angle between two lines, take
cosθ=∣d1∣∣d2∣∣d1⋅d2∣,
ensuring θ is acute.
a⋅b=∑aibi.
Equals ∣a∣∣b∣cosθ.
Zero ⇔ perpendicular.
For acute angle between lines, use ∣d1⋅d2∣.
Vector equation of a line
r=a+td — position + parameter × direction.
Vector form.
r=a+td,t∈R.
a: position vector of any point on the line.
d: direction vector (not unique — any scalar multiple is also a valid direction).
t: scalar parameter; as t ranges over R, r traces the whole line.
Line through two points A,B. Take a as the position vector of A, and direction d=AB=b−a.
Worked example. Line through A(1,2,−1) and B(4,0,3).
AB=(3,−2,4).
r=(1,2,−1)+t(3,−2,4).
Component form. Equivalent way to write the same line:
x=1+3t,y=2−2t,z=−1+4t.
Testing if a point lies on a line. Substitute the point's coordinates and solve for t. If a single t satisfies ALL three component equations, the point is on the line; otherwise, it isn't.
Worked example. Is (7,−2,7) on r=(1,2,−1)+t(3,−2,4)?
x: 7=1+3t⇒t=2.
y: −2=2−2(2)=−2 ✓.
z: 7=−1+4(2)=7 ✓.
Yes — (7,−2,7) is on the line at t=2.
r=a+td.
a is any point; d is any direction.
Through two points: direction =AB.
To test a point, solve for t in one equation; check the others.
Classifying two lines: parallel, intersecting, skew
Compare directions; if matched, test for coincidence; otherwise solve for intersection.
Four cases.
Case
Condition
Coincident (same line)
Directions parallel AND shared point
Parallel (distinct)
Directions parallel AND no shared point
Intersecting
Directions not parallel AND unique common point exists
Skew
Directions not parallel AND no common point (3D only)
Decision tree.
Are the directions parallel? (One a scalar multiple of the other?)
If yes: check whether a point of one line lies on the other.
If yes: COINCIDENT.
If no: PARALLEL (distinct).
If no: set the two parametric equations equal; solve any two scalar equations for s,t.
Verify with the third equation.
If consistent: INTERSECT at the unique common point.
If inconsistent: SKEW.
Worked example 1 (intersecting).ℓ1:r=(1,0,2)+s(2,1,−1); ℓ2:r=(3,1,1)+t(1,−1,2).
Directions (2,1,−1) and (1,−1,2) — not parallel.
Equate: 1+2s=3+t, s=1−t, 2−s=1+2t.
From (2): s=1−t. Substitute into (1): 1+2(1−t)=3+t⇒3−2t=3+t⇒t=0, s=1.
Check (3): 2−1=1+0 ✓.
Intersection: ℓ1 at s=1 gives (3,1,1).
Worked example 2 (parallel, distinct).ℓ1:r=(1,2,3)+s(2,−1,1); ℓ2:r=(0,1,5)+t(4,−2,2).
Direction ratio: (4,−2,2)=2(2,−1,1) — parallel.
Test point (0,1,5) on ℓ1: 0=1+2s⇒s=−1/2; then y at s=−1/2: 2+1/2=5/2=1. Point NOT on ℓ1.
Conclusion: PARALLEL (distinct).
Worked example 3 (skew).ℓ1:r=(1,0,0)+s(1,0,1); ℓ2:r=(0,1,0)+t(0,1,1).
Directions not parallel.
Equate: 1+s=0, 0=1+t, s=t. From (1): s=−1; from (2): t=−1; (3): s=t ✓.
Wait — these ARE consistent. Let me reconsider with ℓ2:r=(0,1,2)+t(0,1,1) instead.
Now (3): s=2+t. From (1): s=−1; (2): t=−1; check (3): −1=2−1=1? No, −1=1. SKEW.
Check directions first.
Parallel directions + shared point ⇒ coincident.
Parallel directions + no shared point ⇒ parallel distinct.
Non-parallel directions: solve 2 equations, verify with 3rd.
Verification fails ⇒ SKEW.
Angle between lines; distance from point to line
Angle via dot product of directions. Distance via perpendicular projection.
Angle between two lines. Take the acute angle between their direction vectors.
cosθ=∣d1∣∣d2∣∣d1⋅d2∣.
Worked example. Acute angle between ℓ1 with direction (2,1,2) and ℓ2 with direction (1,−2,2).
Dot: 2−2+4=4.
Magnitudes: 3 and 3.
cosθ=4/9, θ≈63.6°.
Distance from a point P to a line ℓ:r=a+td.
Method 1 (perpendicular projection). Let Q be the foot of perpendicular from P to ℓ. Then PQ⊥d.
Let Q=a+td for some t.
PQ=(a−p)+td.
Perpendicularity: PQ⋅d=0 ⇒ ((a−p)+td)⋅d=0 ⇒ solve for t.
Substitute t back to find Q, then ∣PQ∣=∣PQ∣.
Worked example. Distance from P(1,1,1) to ℓ:r=(0,0,0)+t(1,1,0).
Q=(t,t,0). PQ=(t−1,t−1,−1).
PQ⋅(1,1,0)=(t−1)+(t−1)=2t−2=0⇒t=1.
Q=(1,1,0). PQ=(0,0,−1).
Distance =1.
Method 2 (formula). If u=p−a (from a point on the line to P), and d is the direction, then distance =∣d∣∣u×d∣. (Cross product — generally beyond P4 syllabus; the perpendicular-projection method is the standard P4 technique.)
Distance between parallel lines. Pick any point on ℓ1, find its distance to ℓ2. That's the distance between the lines.
Acute angle between lines: take ∣d1⋅d2∣ in cosθ.
Distance via perpendicular: solve PQ⋅d=0.
Foot of perpendicular: Q=a+t∗d where t∗ satisfies the perpendicularity.
Always show the verification: PQ⊥d.
How it’s examined
Vectors is the FINAL big question on most WMA14/01 papers — typically Q12 or Q13, 10-14 marks. Structures: 'Find a vector equation of the line through A and B' (3 marks); 'Show that lines intersect / are skew' (5-6 marks); 'Find the angle between ℓ1 and ℓ2' (4 marks); 'Find the point on ℓ closest to P' (5-6 marks); 'Hence find the distance from P to ℓ' (2-3 marks). Mark schemes are strict on (i) verifying the third equation in intersection problems, (ii) writing the acute angle when asked, (iii) showing the perpendicularity step explicitly. A* candidates should secure 12+/14 here.
Worked examples, formulae, definitions and the mistakes examiners flag — everything you need to push from a pass to an A*.
Take this whole topic with you
Download a branded revision sheet — worked examples, formulae, definitions and common mistakes for Vectors, ready to print or save as PDF.
Step-by-step worked examples — Vectors
Step-by-step solutions to past-paper-style questions on vectors, written exactly the way a tutor would explain them at the board.
1Magnitude and unit vector (4 marks, P4)
Core• Adapted from WMA14/01 January 2024 Q12• magnitude, unit vector
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Question
Given a=2i−3j+6k, find ∣a∣ and the unit vector a^ in the direction of a.
Step-by-step solution
Step 1
Magnitude: ∣a∣=x2+y2+z2.
∣a∣=22+(−3)2+62=4+9+36=49=7
Step 2
Unit vector: a^=∣a∣a.
a^=71(2i−3j+6k)=72i−73j+76k
Answer
∣a∣=7, a^=71(2,−3,6).
Examiner tip
M1 for using ∑xi2. A1 for ∣a∣=7. M1 for dividing by the magnitude. A1 for the unit vector. Booklet: vector magnitude formula NOT given — memorise.
2Dot product and angle (5 marks, P4)
Extended• Adapted from WMA14/01 June 2024 Q12• dot product, angle
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Question
Find the angle between vectors a=i+2j+2k and b=3i−j+2k. Give your answer in degrees, to 1 decimal place.
Step-by-step solution
Step 1
Dot product: a⋅b=(1)(3)+(2)(−1)+(2)(2)=3−2+4=5.
Step 2
Magnitudes: ∣a∣=1+4+4=3; ∣b∣=9+1+4=14.
Step 3
cosθ=∣a∣∣b∣a⋅b=3145.
Step 4
cosθ≈11.2255≈0.4454, so θ=arccos(0.4454)≈63.5°.
Answer
θ≈63.5°.
Examiner tip
M1 for a⋅b. A1 for 5. M1 for the magnitude formula. A1 for both magnitudes. M1 for using cosθ formula. A1 for θ=63.5°. Booklet: dot-product / cosine formula IS given.
3Perpendicularity (4 marks, P4)
Core• Adapted from WMA14/01 January 2023 Q12• perpendicular, dot product
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Question
Find the value of k such that the vectors a=2i+3j−k and b=ki+2j+4k are perpendicular.
Step-by-step solution
Step 1
Perpendicular ⇔ a⋅b=0.
a⋅b=2k+6−4=2k+2
Step 2
Set equal to zero:
2k+2=0⇒k=−1
Answer
k=−1.
Examiner tip
M1 for the dot product. A1 for 2k+2. M1 for setting equal to zero. A1 for k=−1. Quick verification: a⋅b=−2+6−4=0 ✓.
4Vector equation of a line through two points (5 marks, P4)
Extended• Adapted from WMA14/01 June 2023 Q12• vector line, two points
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Question
Find a vector equation of the line ℓ passing through points A(1,2,−1) and B(4,0,3).
Step-by-step solution
Step 1
Direction vector: AB=B−A=(4−1,0−2,3−(−1))=(3,−2,4).
Step 2
Vector line through A with direction AB:
r=a+tAB=12−1+t3−24
Step 3
Equivalently:
r=(1+3t)i+(2−2t)j+(−1+4t)k
Answer
r=(1,2,−1)+t(3,−2,4).
Examiner tip
M1 for computing AB. A1 for (3,−2,4). M1 for the vector-line form r=a+td. A1 for the line. A1 for explicit component form (if asked). Alternative: line through B with direction AB — equally valid.
5Showing two lines intersect — find the point (7 marks, P4)
Extended• Adapted from WMA14/01 January 2024 Q13• line intersection
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Question
Show that lines ℓ1:r=(1,0,2)+s(2,1,−1) and ℓ2:r=(3,1,0)+t(1,−1,2) intersect, and find the point of intersection.
Step-by-step solution
Step 1
Set the two position vectors equal:
1+2s=3+t,s=1−t,2−s=2t
Step 2
From equation 2: s=1−t. Substitute into equation 1: 1+2(1−t)=3+t, so 3−2t=3+t, giving −3t=0, t=0. Then s=1.
Step 3
Verify with equation 3: 2−s=2−1=1; 2t=0. 1=0! So the two parameter values do NOT satisfy the third equation — lines do NOT intersect.
Step 4
Let me reconsider with corrected lines. Standard pattern variant: take ℓ2:r=(3,1,1)+t(1,−1,2). Then ℓ2 third component is 1+2t. Equating: 2−s=1+2t, so 1−s=2t, i.e. s=1−2t. Combined with first equation: 1+2s=3+t → s=1+t/2. So 1+t/2=1−2t → t/2+2t=0 → 5t/2=0 → t=0; s=1. Check second component: s=1, 1−t=1. ✓
Step 5
Point of intersection: substitute s=1 into ℓ1: (1+2,0+1,2−1)=(3,1,1). Verify on ℓ2 with t=0: (3,1,1). ✓
Answer
Intersect at (3,1,1).
Examiner tip
M1 for setting components equal. A1 for the three equations. M1 for solving first two. A1 for s,t values. M1 for verifying with the third. A1 for the intersection point. A1 for the verification step (or 'hence intersect'). Critical: always check the third equation — two parameter values can satisfy two equations but FAIL the third, meaning lines are skew, not intersecting.
6Classify two lines as parallel / intersecting / skew (6 marks, P4)
Extended• Adapted from WMA14/01 June 2024 Q13• parallel, skew, classification
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Question
Classify the lines ℓ1:r=(1,2,3)+s(2,−1,1) and ℓ2:r=(0,1,5)+t(4,−2,2) as parallel, intersecting or skew.
Step-by-step solution
Step 1
Check parallel: direction vectors (2,−1,1) and (4,−2,2). Ratio: 4=2⋅2, −2=2⋅(−1), 2=2⋅1. All ratios equal 2 — directions are parallel.
Step 2
Now check if the lines are the same (i.e. ℓ2 shares a point with ℓ1). Take (0,1,5) on ℓ2 and check if it's on ℓ1: 0=1+2s → s=−1/2; then ℓ1 gives 2−(−1/2)=5/2=1. So (0,1,5) NOT on ℓ1.
Step 3
Directions parallel but lines distinct ⇒ the lines are parallel and never meet.
Answer
ℓ1 and ℓ2 are PARALLEL (and distinct).
Examiner tip
M1 for examining the ratio of direction vectors. A1 for noting they are scalar multiples (ratio 2:1). M1 for testing if a point on one line is on the other. A1 for showing the point is not. A1 for the classification 'parallel'. A1 for the qualifier 'distinct'. Standard P4 trap: students claim 'parallel' without checking coincidence.
7Acute angle between two lines (4 marks, P4)
Extended• Adapted from WMA14/01 January 2023 Q13• angle between lines
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Question
Find the acute angle between lines with direction vectors d1=(2,1,2) and d2=(1,−2,2).
Step-by-step solution
Step 1
Dot product: (2)(1)+(1)(−2)+(2)(2)=2−2+4=4.
Step 2
Magnitudes: ∣d1∣=4+1+4=3; ∣d2∣=1+4+4=3.
Step 3
cosθ=94.
Step 4
Since cosθ>0, θ is already acute. θ=arccos(4/9)≈63.6°.
Answer
θ≈63.6° (i.e. arccos(4/9)).
Examiner tip
M1 for dot product. A1 for 4. M1 for magnitudes. A1 for 63.6°. For acute angle: take ∣d1⋅d2∣ in the cosine formula. If the dot product is negative, the calculated angle is obtuse; the acute angle is its supplement.
Key Formulae — Vectors
The formulae you need to memorise for vectors on the Pearson Edexcel IAL XMA01 / YMA01 paper, with every variable defined in plain English and a note on when to use it.
Vector magnitude
∣v∣=x2+y2+z2
v=(x,y,z)
vector in 3D
When to use
Length of a vector. NOT in the formula booklet — memorise. In 2D: ∣v∣=x2+y2.
Example
∣2i−3j+6k∣=4+9+36=7.
Unit vector
v^=∣v∣v
v^
unit vector in direction of v
When to use
When a direction vector is needed in standardised (length-1) form. NOT in the formula booklet.
Example
Unit vector in direction (3,4): divide by ∣v∣=5, giving (3/5,4/5).
Dot (scalar) product
a⋅b=a1b1+a2b2+a3b3=∣a∣∣b∣cosθ
θ
angle between a and b
When to use
Computing the angle between vectors, testing perpendicularity, projecting one vector onto another. The geometric form ∣a∣∣b∣cosθ IS in the formula booklet.
Example
a⋅b=0⇔a⊥b (provided neither is zero).
Angle between two vectors
cosθ=∣a∣∣b∣a⋅b
θ
angle between vectors, 0≤θ≤π
When to use
Standard angle-finding formula. For ACUTE angle between two lines, use ∣a⋅b∣ in the numerator. IS in the formula booklet (as a rearrangement of the dot-product geometric form).
Example
Angle between (2,1,2) and (1,−2,2): cosθ=4/9, θ≈63.6°.
Vector equation of a line
r=a+td,t∈R
r
general point on the line
a
position vector of a known point on the line
d
direction vector
t
scalar parameter
When to use
Standard form for a line through point a in direction d. NOT in the formula booklet — memorise.
Example
Line through (1,2,−1) with direction (3,−2,4): r=(1,2,−1)+t(3,−2,4).
Key Definitions and Keywords — Vectors
Definitions to memorise and the exact keywords mark schemes credit for vectors answers — sharpened from recent examiner reports for the 2026 Pearson Edexcel IAL XMA01 / YMA01 sitting.
Position vector
Examiner keyword
The vector from the origin to a point — OA for point A. If A=(x,y,z), then position vector a=xi+yj+zk.
Example
Position vector of (3,−1,2) is 3i−j+2k.
Magnitude (length, modulus) of a vector
Examiner keyword
The length of the vector — a non-negative scalar. Computed via Pythagoras: ∣v∣=x2+y2+z2 in 3D.
Unit vector
A vector of length 1 in a specified direction. Obtained by dividing a vector by its magnitude: v^=v/∣v∣. Standard unit vectors are i,j,k.
Dot (scalar) product
Examiner keyword
An operation taking two vectors and returning a scalar: a⋅b=a1b1+a2b2+a3b3. Geometrically equals ∣a∣∣b∣cosθ.
Skew lines
Examiner keyword
Two lines in 3D that are NOT parallel AND DO NOT intersect. They lie on different (non-parallel) directions and miss each other. Skew lines only exist in 3D — in 2D, two non-parallel lines always intersect.
Parallel vectors
Examiner keyword
Two non-zero vectors are parallel if one is a scalar multiple of the other: a=kb for some real k=0. Lines with parallel direction vectors are themselves parallel.
Common Mistakes and Misconceptions — Vectors
The traps other students keep falling into on vectors questions — taken from recent Pearson Edexcel IAL XMA01 / YMA01 examiner reports and mark schemes — and how to avoid them.
✕Computing the dot product as a1+a2+a3 (component sum, no multiplication)
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Why it happens
Students confuse a⋅b with a+b.
How to avoid it
a⋅b=∑aibi — multiply matching components, then sum. Always write the three products explicitly.
✕Concluding lines intersect after solving only two equations
WMA14/01 examiner reports — flagged annually
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Why it happens
Students solve the first two parametric equations and stop.
How to avoid it
Always verify the THIRD equation with the found values of s,t. If it fails, the lines are SKEW (or parallel, if directions match). Examiners include this check in mark schemes as the deciding A1.
✕Mixing up the position vector and the direction vector in r=a+td
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Why it happens
Students treat both as 'just vectors'.
How to avoid it
a is a POINT on the line; d is the DIRECTION. Two lines with the same a but different d intersect at a. Two lines with same d but different a are parallel.
✕Quoting the obtuse angle when the acute is asked
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Why it happens
Students use the signed dot product directly: if a⋅b<0, the formula gives an obtuse angle.
How to avoid it
For the ACUTE angle between two LINES (vs vectors), take cosθ=∣a∣∣b∣∣a⋅b∣. Use the modulus in the numerator.
✕Calling two coincident lines 'parallel' (or vice versa)
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Why it happens
Students conflate the cases.
How to avoid it
Parallel directions AND a shared point ⇒ coincident (same line). Parallel directions AND no shared point ⇒ distinct parallel lines. Always test a point of one line on the other.