Library › Further Pure Mathematics 3 WFM03 › Vectors

AS Level · Further Pure Mathematics 3 WFM03

Vectors

Revise Vectors for Further Pure Mathematics 3 WFM03 (AS Level) — revision notes and instant AI marking. Free to start.

📖 Revision notes · preview

FP3 · Chapter 5

Vectors

The big idea: once you know how to multiply two vectors to get a third vector perpendicular to both (the vector product), almost every 3-D geometry problem — lines, planes, distances, angles — becomes a case of picking the right combination of dot products and cross products.

Summary Vector Product Volumes Lines Planes Distances Angles Memorise Checklist Exam Tips

Summary

The vector product a × b gives a new vector perpendicular to both a and b, with magnitude ab sin θ.
The vector product has a neat component formula using i, j, k or a determinant.
a × b gives you areas (triangles, parallelograms) almost for free.
The triple scalar product a · (b × c) gives the (signed) volume of a parallelepiped; a sixth of it gives the volume of a tetrahedron.
Lines in 3-D can be written three ways: vector form r = a + λb, Cartesian form (x−l)/α = (y−m)/β = (z−n)/γ, and vector product form r × b = c.
Planes can be written three ways too: scalar product form r·n = d, Cartesian form ax+by+cz=d, and vector form r = a + λb + μc.
There's a whole toolkit of distance formulas: point–to–plane, point–to–line, between parallel planes, between skew lines. They all boil down to projecting one vector onto a unit vector.
Reflections in a plane use the point-to-plane method twice (find the foot of the perpendicular, then double it).
The angle between two lines, or two planes, or a line and a plane, always comes from a dot product of direction/normal vectors — you just have to know which pair to dot, and whether to subtract from 90°.

The Vector (Cross) Product

You already know the scalar product a·b = ab cos θ from earlier work — it takes two vectors and spits out a . The vector product does something different: it takes two vectors and spits out a .

Definition

a × b = ab sin θ n̂

where n̂ is a perpendicular to both a and b, and θ is the angle between a and b (0 ≤ θ ≤ π). The direction of n̂ follows the right-hand corkscrew rule: turn a right-handed screw from a towards b through angle θ, and n̂ points the way the screw moves.

Think of it like this: if a and b are two edges of a flat sheet of card lying on a table, a × b points straight up out of the table (or straight down, depending on which way you rotate from a to b). Its tells you how "spread apart" a and b are (maximum when they're perpendicular, zero when they're parallel).

🧭 Why "corkscrew"? Picture i × j. Turning from i to j is a quarter-turn anticlockwise when viewed from the +z direction — exactly how a normal (right-handed) screw moves forward as you turn it that way. So i × j = k. This gives you the three key results: i × j = k, j × k = i, k × i = j — and reversing any pair flips the sign.

Base vectors

i × j = k, j × k = i, k × i = j
i × k = −j, j × i = −k, k × j = −i

Key properties

Anticommutative: b × a = −(a × b). Swapping the order of a cross product flips the direction of n̂ (the corkscrew turns the other way), so the whole vector negates.
a × a = 0, since θ = 0 so sin θ = 0.
a × b = 0 ⟺ a is parallel to b (or one of them is the zero vector) — because sin θ = 0 only when θ = 0 or π.
Distributive over addition: a × (b + c) = a × b + a × c.
a × b is always perpendicular to both a and b — straight from the definition.

⚠️ Common mistake The vector product is not commutative. Writing b × a = a × b (like you might unconsciously do with numbers) is one of the most common silly slips in this chapter. Always check the order matches what the question asks for.

Component form

In practice you'll almost never use the ab sin θ n̂ definition directly to a cross product — instead you use the component formula, most easily remembered as a determinant:

Component formula

a × b = | i j k |
| a₁ a₂ a₃ |
| b₁ b₂ b₃ | = ( a₂b₃ − a₃b₂ , −(a₁b₃ − a₃b₁) , a₁b₂ − a₂b₁ )

Expand along the top row like any 3×3 determinant, remembering the +−+ sign pattern.

Worked Example — Vector Product

Find a × b where a = (3, 1, −1) and b = (4, 2, −3).

a × b = | i j k ; 3 1 −1 ; 4 2 −3 |
i-component: (1)(−3) − (−1)(2) = −3 + 2 = −1
j-component: −[(3)(−3) − (−1)(4)] = −[−9 + 4] = 5
k-component: (3)(2) − (1)(4) = 6 − 4 = 2
⇒ a × b = (−1, 5, 2)

Practice: Find p × q where p = (2, −1, 0) and q = (1, 3, −2).

Applications of the vector product — areas

Because |a × b| = ab sin θ, and the area of a triangle OAB with sides a and b enclosing angle θ is ½ ab sin θ, we get areas almost for free:

Areas

Area of triangle OAB = ½ |a × b|
Area of parallelogram OADB = |a × b|

A parallelogram is just two copies of the triangle, so it's exactly double.

Worked Example — Area of a Triangle

A is (−1, 2, 1), B is (2, 3, 0), C is (3, 4, −2). Find the area of triangle ABC.

Area = ½ |AB⃗ × AC⃗|
AB⃗ = b − a = (3, 1, −1) AC⃗ = c − a = (4, 2, −3)
AB⃗ × AC⃗ = (−1, 5, 2)
|AB⃗ × AC⃗| = √(1 + 25 + 4) = √30
Area ABC = ½√30

Practice: Find the area of the parallelogram with sides OP and OQ where P = (1, 0, 2) and Q = (0, 3, 1).

Volume of a Parallelepiped & Tetrahedron

Volume of a parallelepiped

A parallelepiped is a "squashed box" — six parallelogram faces, built from three vectors a, b, c meeting at a corner. Its base is the parallelogram formed by b and c, with area bc sin θ. Its height is the component of a perpendicular to that base.

If n̂ is a unit vector perpendicular to the base, the height is h = a·n̂ (with a possible sign flip if the angle is obtuse). Since n̂ = (b × c)/|b × c|, some algebra shows the whole volume collapses into one clean scalar quantity:

Volume of a parallelepiped

V = a · (b × c)

This is called the triple scalar product. Take the modulus if you just want a positive volume — the sign tells you about orientation, not size.

The triple scalar product — component form and notation

Writing it all out in components, the triple scalar product is exactly the determinant with rows a, b, c:

Triple scalar product

a · (b × c) = | a₁ a₂ a₃ |
| b₁ b₂ b₃ |
| c₁ c₂ c₃ |

It's common to write this as the shorthand {a, b, c}. You can show a·(b×c) = (a×b)·c — the dot and cross can swap roles as long as a, b, c stay in the same cyclic order.

🔄 Cyclic symmetry {a, b, c} = {c, a, b} = {b, c, a} (rotate the order — value unchanged)
{a, b, c} = −{a, c, b} = −{c, b, a} = −{b, a, c} (swap any two — value flips sign)

Volume of a tetrahedron

A tetrahedron ABCD is exactly ¹⁄₆ of the parallelepiped built from AB⃗, AC⃗, AD⃗ — because its base triangle is half the area of the parallelepiped's base parallelogram, and the standard "cone" volume formula ⅓ × base × height brings in another factor of ½ (½ × ⅓ = ⅙).

Volume of a tetrahedron

V = ¹⁄₆ |{a, b, c}|

where a, b, c are any three edge vectors from the same vertex.

Worked Example — Volume of a Tetrahedron

A(1, 0, 2), B(−1, 2, 2), C(1, 1, −3), D(4, 0, 3). Find the volume of tetrahedron ABCD.

AD⃗ = (3, 0, 1) AC⃗ = (0, 1, −5) AB⃗ = (−2, 2, 0)
{AD⃗, AC⃗, AB⃗} = | 3 0 1 ; 0 1 −5 ; −2 2 0 |
= 3(1×0 − (−5)(2)) − 0 + 1(0×2 − 1×(−2))
= 3(0+10) + 1(0+2) = 30 + 2 = 32
Volume = ⅙ × 32 = 5⅓

Practice: Find the volume of the parallelepiped with edges a = (1,2,0), b = (0,1,3), c = (2,0,1) from the same vertex.

Equations of Straight Lines

A line in 3-D needs a point it passes through and a direction to travel in. There are three equivalent ways of writing this down — you should be comfortable converting between all three.

1. Vector equation of a line

Vector form

r = a + λb

A line through the point with position vector a, parallel to direction vector b. λ is a scalar parameter — as it varies over all real numbers, r sweeps out every point on the line.

2. Cartesian equation of a line in 3-D

Write r = (x, y, z), a = (l, m, n), b = (α, β, γ), expand r = a + λb component-wise, and eliminate λ from all three equations:

Cartesian form

(x − l)/α = (y − m)/β = (z − n)/γ (= λ)

A line through (l, m, n), parallel to (α, β, γ). This "equation" is secretly the intersection of two planes: (x−l)/α = (y−m)/β, and (y−m)/β = (z−n)/γ.

3. Vector product form

Since AP⃗ = r − a lies along the line, it must be parallel to b — meaning (r − a) × b = 0.

Vector product form

(r − a) × b = 0 ⟺ r × b = a × b = c

This is a line parallel to b passing through the point whose position vector satisfies r × b = c. Useful when a line's equation needs to be combined with cross-product working elsewhere in a question.

Worked Example — Converting Between Forms

A line has Cartesian equation x = (2y+4)/5 = (3−z)/2. Write it (i) in the form r = a + λb, and (ii) in the form r × b = c.

Rewrite in standard form: (x−0)/1 = (y−(−2))/2.5 = (z−3)/(−2)
So the line passes through A(0, −2, 3), direction b = (1, 2.5, −2), or scaled: (2, 5, −4).

(i) r = (0, −2, 3) + λ(2, 5, −4)

(ii) c = a × b = (0,−2,3) × (2,5,−4)
i: (−2)(−4) − (3)(5) = 8 − 15 = −7
j: −[(0)(−4) − (3)(2)] = −[−6] = 6
k: (0)(5) − (−2)(2) = 4
c = (−7, 6, 4) ⇒ r × (2, 5, −4) = (−7, 6, 4)

Practice: Write the Cartesian line (x+2)/3 = (y−1)/(−1) = z/4 in vector form r = a + λb.

Equations of a Plane

1. Scalar product form

Let n be a vector perpendicular ("normal") to the plane, and A a fixed point in the plane. For any general point P = (x, y, z) also in the plane, AP⃗ lies in the plane so it's perpendicular to n, meaning AP⃗ · n = 0.

Scalar product form

(r − a) · n = 0 ⟺ r · n = a · n = d

r · n = d is the equation of a plane perpendicular to n, where d = a·n is just a constant number (the same for every point in the plane).

2. Cartesian form

Cartesian form

ax + by + cz = d

where n = (a, b, c) is the normal vector. This drops straight out of r·n = d by writing everything in components.

Worked Example — Finding the Equation of a Plane

Find the Cartesian equation of the plane through A(3, 2, 5), B(−1, 0, 3), C(2, 1, −2).

AB⃗ = (−4, −2, −2) AC⃗ = (−1, −1, −7)
n = AB⃗ × AC⃗ = | i j k ; −4 −2 −2 ; −1 −1 −7 |
i: (−2)(−7) − (−2)(−1) = 14 − 2 = 12
j: −[(−4)(−7) − (−2)(−1)] = −[28 − 2] = −26
k: (−4)(−1) − (−2)(−1) = 4 − 2 = 2
n = (12, −26, 2) = 2(6, −13, 1) — use the smaller version
Plane: 6x − 13y + z = d. Substitute A(3,2,5): d = 18 − 26 + 5 = −3
⇒ 6x − 13y + z = −3

3. Vector equation of a plane

Vector form

r = a + λb + μc

A plane through A, parallel to two non-parallel vectors b and c lying in the plane. Two independent parameters λ, μ are needed because a plane is 2-dimensional.

💡 Converting Cartesian → vector form of a plane Given ax+by+cz=d: (1) find any one point on the plane by setting two variables and solving for the third; (2) find two vectors perpendicular to n = (a,b,c) — inspect pairs of coordinates and use the fact that a vector ⊥ n has zero dot product with it. If a coordinate of n is zero this gets fiddly — instead set x = λ, y = μ and solve for z directly, which always works.

Practice: Find the vector equation r = a + λb + μc of the plane through A(1,4,−2), B(1,5,3), C(4,7,2).

Distance & Reflection Toolkit

This is the section students mix up the most, because there are several distance formulas that all similar. The trick is realising they're really all the same idea — projecting a connecting vector onto a unit vector — dressed up for different situations. Learn the underlying idea once, and every formula becomes obvious rather than memorised.

🎯 The one idea behind every formula here If n̂ is a unit vector, then for any vector v, the quantity |v · n̂| is the length of the "shadow" you get by shining light perpendicular to n̂ and projecting v onto the n̂ direction. Every distance formula below is really just: find a connecting vector between the two objects, then project it onto the right unit vector.

Distance from a point to a plane (two methods)

Method A — intersection method: find where the perpendicular line through P meets the plane, then measure the length.

Worked Example — Point to Plane (Intersection Method)

Find the distance from P(−2, 3, 5) to the plane 4x − 3y + 12z = 21.

Line through P parallel to n = (4,−3,12): r = (−2+4λ, 3−3λ, 5+12λ)
Substitute into the plane: 4(−2+4λ) − 3(3−3λ) + 12(5+12λ) = 21
−8+16λ −9+9λ +60+144λ = 21 ⇒ 169λ + 43 = 21 ⇒ λ = −22/169
PM⃗ = λ·n = (−22/169)(4,−3,12)
|PM⃗| = (22/169)√(16+9+144) = (22/169)(13) = 22/13

Method B — the projection shortcut (much faster, and this is the one worth memorising):

Distance from a point to a plane

s = |n₁α + n₂β + n₃γ + d| / √(n₁² + n₂² + n₃²)

for point P(α, β, γ) and plane n₁x + n₂y + n₃z + d = 0. Equivalently: pick point B on the plane, and s = |AB⃗ · n̂| where n̂ = n/|n|.

Worked Example — Point to Plane (Projection Method)

Same question — distance from A(−2, 3, 5) to 4x − 3y + 12z = 21.

Spot a point on the plane by inspection: B(0, −7, 0) works since 4(0)−3(−7)+12(0)=21 ✓
AB⃗ = (0,−7,0) − (−2,3,5) = (2, −10, −5)
|n| = √(16+9+144) = 13, so n̂ = (4,−3,12)/13
distance = |AB⃗ · n̂| = |(2)(4/13) + (−10)(−3/13) + (−5)(12/13)|
= |8/13 + 30/13 − 60/13| = |−22/13| = 22/13

✅ Same answer, way less algebra Both methods give 22/13 — but Method B skips solving for λ entirely. Once you're confident, always reach for the projection method first.

Projections — the geometric picture

Imagine a distant light shining perpendicular to a line l, casting a shadow of a rod AB onto l. The shadow's length is |BA·n̂|, where n̂ is a unit vector along l. This is exactly the "projection" idea used throughout this section — modulus signs are needed because n̂ might point the "wrong" way.

Shortest distance from a point to a plane (projection form)

AM = |AB⃗ · n̂|

B is any point on the plane, n̂ is the unit normal. This is exactly Method B above, generalised.

Reflection of a point in a plane

Find the foot of the perpendicular P from A to the plane (same method as distance-to-plane), then double the vector AP⃗ to jump right through the plane to the mirror image A′.

Reflection formula

OA′⃗ = OA⃗ + 2AP⃗

where P is the foot of the perpendicular from A onto the plane.

Worked Example — Reflection in a Plane

Find the reflection of A(10, 1, 7) in the plane r·(3, −2, 1) = 7.

Line through A ⊥ plane: r = (10,1,7) + λ(3,−2,1)
Meets plane: 3(10+3λ) −2(1−2λ) + (7+λ) = 7
30+9λ −2+4λ +7+λ = 7 ⇒ 14λ + 35 = 7 ⇒ λ = −2
AP⃗ = λ(3,−2,1) = (−6, 4, −2)
OA′⃗ = (10,1,7) + 2(−6,4,−2) = (10−12, 1+8, 7−4) = (−2, 9, 3)
A′ = (−2, 9, 3)

Distance between parallel planes

Two parallel planes share the same normal n. Pick any point on one plane, then measure its projected distance to the other — exactly the point-to-plane formula in disguise.

Distance between parallel planes

shortest distance = |(b − a) · n̂|

A on plane 1 (position vector a), B on plane 2 (position vector b), n̂ the shared unit normal.

Worked Example — Parallel Planes

Find the distance between π₁: 2x − 6y + 3z = 9 and π₂: 2x − 6y + 3z = 5.

Point on π₁: A(0, 0, 3). n = (2, −6, 3), |n| = √(4+36+9) = 7
Using the point-to-plane formula with A and π₂:
s = |2(0) − 6(0) + 3(3) − 5| / 7 = |9−5|/7 = 4/7

Shortest distance from a point to a line

Here you know the perpendicular direction in advance, so a bit of lateral thinking is needed: project AP⃗ onto the line's direction to find how far along the line the closest point is, then use Pythagoras.

Shortest distance, point to line

AX = |AP⃗ · b̂|, then PX = √(AP² − AX²)

A is a known point on the line l: r = a + λb, b̂ is the unit direction vector, P is the external point.

Worked Example — Point to Line

Find the shortest distance from P(3,−2,4) to the line r = (−2,3,0) + λ(2,−3,6).

|b| = √(4+9+36) = 7, so b̂ = (2,−3,6)/7
AP⃗ = (3,−2,4) − (−2,3,0) = (5,−5,4)
AX = |AP⃗ · b̂| = |(5·2 + (−5)(−3) + 4·6)/7| = |(10+15+24)/7| = 49/7 = 7
AP² = 25+25+16 = 66
PX = √(66 − 49) = √17

🧠 Alternative: the "plane through P" trick You can also construct the plane through P perpendicular to l (using l's direction as the plane's normal), find where l crosses that plane, then measure the straight-line distance from P to that intersection point. Both methods appear in the source notes and give identical answers — use whichever you find more natural.

Shortest distance between two skew lines

Two skew lines don't meet and aren't parallel. There's a unique line segment joining them that's perpendicular to — its direction is n = b × d, since that's automatically perpendicular to both direction vectors. Once you have that direction, treat the two lines as living in two parallel planes and measure between them exactly as before.

Shortest distance between skew lines

d = |(c − a) · (b × d)| / |b × d|

for l₁: r = a + λb and l₂: r = c + μd. This formula is not in the formula booklet — you must memorise it.

Worked Example — Skew Lines

Find the shortest distance between l₁: r = (1,0,0) + λ(1,1,0) and l₂: r = (0,1,2) + μ(0,1,1).

b × d = (1,1,0) × (0,1,1)
i: (1)(1)−(0)(1) = 1
j: −[(1)(1)−(0)(0)] = −1
k: (1)(1)−(1)(0) = 1
b × d = (1,−1,1), |b×d| = √3
c − a = (0,1,2) − (1,0,0) = (−1,1,2)
(c−a)·(b×d) = (−1)(1)+(1)(−1)+(2)(1) = −1−1+2 = 0
d = |0|/√3 = 0 — (a good reminder to always check whether "skew" lines really are skew before trusting a nonzero answer.)

Practice: Find the shortest distance from P(1, 2, −1) to the plane 2x + y − 2z = 6.

Intersections & Angles

Line of intersection of two planes

Two non-parallel planes meet in a line. Eliminate one variable between the two Cartesian equations to relate the other two, then set the remaining variable equal to a parameter λ and back-substitute to get a vector equation.

Worked Example — Line of Intersection

Find the line of intersection of x + y + 2z = 4 and 2x − y + 3z = 4.

Add the equations: 3x + 5z = 8
Let z = λ ⇒ x = (8 − 5λ)/3
Sub into first equation: y = 4 − x − 2z = 4 − (8−5λ)/3 − 2λ = (4−λ)/3
r = (8/3, 4/3, 0) + λ(−5/3, −1/3, 1), or scale the direction: (−5, −1, 3)

Angle between a line and a plane

First find the angle θ between the line's direction and the plane's (using a standard dot product). The angle you actually want, φ, between the line and the plane itself, is then either 90° − θ or θ − 90°, depending on which side of the plane the normal points.

Angle between line and plane

cos θ = (b · n) / (|b||n|), then φ = |90° − θ|

b = line direction, n = plane normal. Taking |90° − θ| avoids worrying about which case you're in — it always gives the correct acute angle φ between line and plane.

Worked Example — Line/Plane Angle

Find the angle between (x+1)/2 = (y−2)/1 = (z−3)/(−2) and the plane 2x + 3y − 7z = 5.

Line direction: (2,1,−2). Normal: (2,3,−7).
cos θ = [(2)(2)+(1)(3)+(−2)(−7)] / (√(4+1+4)·√(4+9+49))
= (4+3+14)/(3·√62) = 21/(3√62) = 7/√62
θ = 27.3° ⇒ φ = 90° − 27.3° = 62.7°

Angle between two planes

Viewed "end-on," the angle between two planes equals the angle between their normal vectors — no 90°-subtraction needed this time.

Angle between two planes

cos θ = (n₁ · n₂) / (|n₁||n₂|)

Just the standard angle-between-vectors formula, applied directly to the two normals.

Worked Example — Angle Between Planes

Find the angle between 2x + y + 3z = 5 and 2x + 3y + z = 7.

n₁ = (2,1,3), n₂ = (2,3,1)
n₁·n₂ = 4+3+3 = 10
|n₁| = |n₂| = √(4+1+9) = √14
cos θ = 10/14 = 5/7 ⇒ θ = 44.4°

Practice: Find the angle between the planes x − 2y + 2z = 4 and 2x + y − z = 3.

💡 Negative cosine? Take the acute angle. Normal vectors can point either "into" or "out of" a plane, so a dot product between two normals can come out negative even when the planes meet at a small acute angle. If you're asked for "the angle between two planes" and get an obtuse answer, it's fine (and often expected) to quote 180° minus it, or just use |n₁·n₂| from the start.

What to Memorise

Vector product

a × b = ab sin θ n̂; anticommutative (b×a = −a×b); zero iff parallel.

Base vector rule

i×j=k, j×k=i, k×i=j (cyclic); reverse order flips the sign.

Areas

Triangle OAB = ½|a×b|; parallelogram OADB = |a×b|.

Triple scalar product

{a,b,c} = a·(b×c) = (a×b)·c = volume of parallelepiped; cyclic swaps preserve value, single swaps flip sign.

Tetrahedron volume

V = ⅙|{a,b,c}| using three edge vectors from one vertex.

Line — three forms

Vector: r=a+λb. Cartesian: (x−l)/α=(y−m)/β=(z−n)/γ. Vector product: r×b = a×b.

Plane — three forms

Scalar product: r·n=d. Cartesian: ax+by+cz=d. Vector: r=a+λb+μc.

Point-to-plane distance

s = |n₁α+n₂β+n₃γ+d| / √(n₁²+n₂²+n₃²), or |AB⃗·n̂| for any B on the plane.

Reflection in a plane

OA′⃗ = OA⃗ + 2AP⃗, where P is the foot of the perpendicular from A.

Point-to-line distance

AX = |AP⃗·b̂|, then PX = √(AP²−AX²) by Pythagoras.

Skew lines distance (memorise — not in booklet)

d = |(c−a)·(b×d)| / |b×d|

Line/plane angle

Find θ between line direction and normal, then φ = |90° − θ|.

Angle between planes

cos θ = (n₁·n₂)/(|n₁||n₂|) — direct angle between the normals.

Concepts Checklist

I can compute a vector product using the determinant/component method.
I know the vector product is anticommutative and can spot when two vectors are parallel from a×b = 0.
I can find the area of a triangle or parallelogram given its vertices.
I can compute the triple scalar product and use cyclic/swap sign rules.
I can find the volume of a parallelepiped and a tetrahedron.
I can convert a line between vector, Cartesian, and vector-product forms.
I can convert a plane between scalar product, Cartesian, and vector forms.
I can find the distance from a point to a plane (both the intersection method and the projection shortcut).
I can reflect a point in a plane.
I can find the distance between two parallel planes.
I can find the shortest distance from a point to a line, using projection + Pythagoras.
I understand the projection interpretation |v·n̂| behind all of these distance formulas.
I can find the shortest distance between two skew lines using b×d.
I can find the line of intersection of two planes.
I can find the angle between a line and a plane, and know when to use 90°−θ.
I can find the angle between two planes directly from their normals.

Exam Tips

Order matters in the cross product. a × b ≠ b × a — they're negatives of each other. If your answer to a "find a×b" question looks right but with every sign flipped, you've probably swapped the order.

The skew-lines distance formula isn't in the formula booklet — the source notes flag this explicitly. Learn it properly: d = |(c−a)·(b×d)| / |b×d|.

Always simplify direction/normal vectors before using them (e.g. (12,−26,2) → (6,−13,1)). It keeps determinants and dot products manageable and reduces arithmetic slips.

Check skew lines are actually skew before applying the skew-lines distance formula — if the shortest distance comes out to 0, the lines intersect, and if the direction vectors are parallel, they're not skew at all (they're either parallel or the same line).

For line/plane angle questions, always sanity check with 90°−θ. Examiners often want the acute angle between the line and the plane itself — not the angle to the normal, which is a very common point lost.

When converting a Cartesian plane equation to vector form and a coordinate of the normal is zero, don't try to spot perpendicular vectors by inspection — it's fragile. Instead set x=λ, y=μ and solve for z; this method never breaks.

M(b × c) ≠ M(b) × M(c) for a linear transformation M — cross products don't transform simply, so when finding the image of a plane under a transformation, always build the plane's vector equation first and then transform each of a, b, c separately.

Show full determinant working for the cross product and triple scalar product — these carry method marks even if your final numeric answer has an arithmetic slip.

What's inside

📖 Revision notes ✦ AI flashcards ✓ Instant AI marking

Read the full Vectors notes free

That's the preview — create a free account to read the rest, plus flashcards and practice questions with instant AI marking. No credit card.

Unlock the full notes free →

Vectors

Vectors

More Further Pure Mathematics 3 topics