Contents · 0% complete

Learning objectives
Core idea
Key formulae
Worked examples
Explore the graph
Common mistakes
Exam tip
Guided practice
Practice questions
Challenge
Summary
Quick quiz
Same topic, different board
Past paper questions

Learn7367Paper 2 · Compulsory contentFurther vectorsVector product and planes

Further vectors · Subtopic 1 of 2

The Vector Product and Planes

$\mathbf a\times\mathbf b$ is a vector perpendicular to both $\mathbf a$ and $\mathbf b$ , of length $|\mathbf a||\mathbf b|\sin\theta$ . A plane is fixed by a point and a normal $\mathbf n$ : $\mathbf r\cdot\mathbf n=\mathbf a\cdot\mathbf n$ , or $ax+by+cz=d$ . Substituting a line into a plane finds where they meet.

Cambridge 9231 Paper 1Edexcel YFM01 WFM03Edexcel 9FM0 Paper 1Edexcel 9FM0 Paper 2Edexcel 9FM0 OptionAQA 7367 Paper 2≈ 50 minRevision 1 · updated 2026-10-27

Practise this topicTimed test Ask Salam TutorSwitch board

Section 1 of 14

Learning objectives

Calculate $\mathbf a\times\mathbf b$ and use it for normals and areas.
Write planes in vector, scalar-product and Cartesian forms.
Find the intersection of a line and a plane, and of two planes.
Find angles between lines and planes, and between two planes.

Pause and think

A sheet of paper lies on a table. How many directions are at right angles to the sheet?

Section 2 of 14

Core idea

Start hereWhich way is "up" from a slope?

A tilted solar panel must face the Sun squarely. Engineers describe its tilt by a single arrow sticking straight out of it, its normal. Given two edges of the panel, one calculation, the cross product, produces that arrow.

The big idea

$\mathbf a\times\mathbf b\perp$ both; $|\mathbf a\times\mathbf b|=|\mathbf a||\mathbf b|\sin\theta$ . Plane: $\mathbf r\cdot\mathbf n=\mathbf a\cdot\mathbf n$ , i.e. $n_1x+n_2y+n_3z=d$ .

1. The vector product

\mathbf a\times\mathbf b=\begin{vmatrix}\mathbf i&\mathbf j&\mathbf k\\a_1&a_2&a_3\\b_1&b_2&b_3\end{vmatrix}=\begin{pmatrix}a_2b_3-a_3b_2\\a_3b_1-a_1b_3\\a_1b_2-a_2b_1\end{pmatrix}

$\mathbf i,\mathbf j,\mathbf k$: unit vectors along the axes.
direction: right-hand rule from a to b.

Checking perpendicularity

$\mathbf a\cdot(\mathbf a\times\mathbf b)=a_1(a_2b_3-a_3b_2)+a_2(a_3b_1-a_1b_3)+a_3(a_1b_2-a_2b_1)$
Dot with $\mathbf a$ .
$=0$
Every term cancels with another.

Similarly for

\mathbf b

: the cross product is a normal to the plane containing

\mathbf a

and

\mathbf b

2. Equations of planes

From a point and a normal

$(\mathbf r-\mathbf a)\cdot\mathbf n=0$
Any vector in the plane is perpendicular to $\mathbf n$ .
$\mathbf r\cdot\mathbf n=\mathbf a\cdot\mathbf n$
Expand.
$n_1x+n_2y+n_3z=d$
Write $\mathbf r=(x,y,z)$ ; $d=\mathbf a\cdot\mathbf n$ .

Worked examplePlane through three points

$A(1,0,0)$ , $B(0,2,0)$ , $C(0,0,3)$ .

Goal: Cartesian equation.
Strategy: Two edge vectors, cross for $\mathbf n$ , then a point.

$\overrightarrow{AB}=(-1,2,0),\ \overrightarrow{AC}=(-1,0,3)$
Edges.
$\mathbf n=(2\cdot3-0,\ 0\cdot(-1)-(-1)\cdot3,\ 0-2(-1))=(6,3,2)$
Cross product.
$6x+3y+2z=6$
$d=\mathbf n\cdot A=6$ .

Answer

6x+3y+2z=6

✓ Check

B

6

✓;

C

6

✓.

3. Intersections and angles

Worked exampleLine meets plane

$\mathbf r=(1,0,2)+t(1,1,1)$ and $x+2y+z=11$ .

Goal: Intersection; angle.
Strategy: Substitute; then $\sin\theta=\frac{|\mathbf d\cdot\mathbf n|}{|\mathbf d||\mathbf n|}$ .

$(1+t)+2t+(2+t)=11\Rightarrow t=2$
Substitute the line's coordinates.
$(3,2,4)$
Point.
$\sin\theta=\frac{4}{\sqrt3\sqrt6}\Rightarrow\theta=70.5^\circ$
Angle with the plane uses sine.

Answer

(3,2,4)

;

70.5^\circ

✓ Check

3+4+4=11

✓.

Common misconceptionCos for line–plane

\cos

with the normal gives the angle to the normal; the angle to the plane is

90^\circ

minus that.

Key takeawayTwo planes Their line of intersection has direction

\mathbf n_1\times\mathbf n_2

; find one common point by setting

z=0

Summary

Cross product: a normal and an area.
$\mathbf r\cdot\mathbf n=d$ .
Substitute lines into planes.
sin for line–plane, cos for plane–plane.

Normal: Vector perpendicular to a plane.
Vector product: $\mathbf a\times\mathbf b$ .

HistoryGibbs and Heaviside. Josiah Willard Gibbs and Oliver Heaviside independently created modern vector notation in the 1880s, splitting Hamilton's quaternion product into the dot and cross products.

Fun factRight-hand rule. Point your index finger along

\mathbf a

and middle finger along

\mathbf b

: your thumb points along

\mathbf a\times\mathbf b

. Swap them and it flips:

\mathbf b\times\mathbf a=-\mathbf a\times\mathbf b

Section 3 of 14

Key formulae

Vector product

\mathbf a\times\mathbf b=\begin{pmatrix}a_2b_3-a_3b_2\\a_3b_1-a_1b_3\\a_1b_2-a_2b_1\end{pmatrix},\quad|\mathbf a\times\mathbf b|=|\mathbf a||\mathbf b|\sin\theta

Plane

\mathbf r\cdot\mathbf n=\mathbf a\cdot\mathbf n\iff n_1x+n_2y+n_3z=d

Angles

\text{line–plane: }\sin\theta=\frac{|\mathbf d\cdot\mathbf n|}{|\mathbf d||\mathbf n|};\quad\text{plane–plane: }\cos\theta=\frac{|\mathbf n_1\cdot\mathbf n_2|}{|\mathbf n_1||\mathbf n_2|}

TrickCheck the cross product. Dot your answer with

\mathbf a

and with

\mathbf b

: both must be

0

. If not, a sign slipped.

Section 4 of 14

Worked examples

Example 1 · Plane through three points0 / 3 steps

Find the plane through

A(1,0,0)

B(0,2,0)

C(0,0,3)

Example 2 · Line meets plane0 / 3 steps

Find where

\mathbf r=(1,0,2)+t(1,1,1)

meets

x+2y+z=11

Section 5 of 14

Explore the graph

Area from the cross product

Loading the interactive graph…

Try this

The parallelogram on $(a,b,0)$ and $(c,d,0)$ has area $|\mathbf u\times\mathbf v|=|ad-bc|$ : the $z$ -component of the cross product.
Make the vectors parallel: the area (and cross product) becomes zero.

$|\mathbf u\times\mathbf v|$ = area of the parallelogram.

In the real worldTorque. Turning effect is

\boldsymbol\tau=\mathbf r\times\mathbf F

: largest when you push at right angles to the spanner, zero when you push along it.

Section 6 of 14

Common mistakes

Incorrect

\mathbf a\times\mathbf b=\mathbf b\times\mathbf a

Correct

\mathbf b\times\mathbf a=-\mathbf a\times\mathbf b

The cross product is anticommutative.

IncorrectLine–plane angle with cos

CorrectUse sin (or

90^\circ-

the normal angle)

The formula with

\mathbf n

gives the angle with the normal.

IncorrectMiddle component

a_1b_3-a_3b_1

Correct

a_3b_1-a_1b_3

The middle term carries a minus sign.

Two vectors in a plane

Cross product: let me point straight out of the page.

Section 7 of 14

Exam tip

Exam tipSimplify normals. Divide a normal by any common factor before writing the plane:

(6,3,0)\to(2,1,0)

. It keeps the arithmetic small.

Section 8 of 14

Guided practice

SA-FP-VP-G012 marksL1 FoundationNon-calculatorCross product

Find the

x

-component of

\begin{pmatrix}1\\2\\3\end{pmatrix}\times\begin{pmatrix}0\\1\\4\end{pmatrix}

Salam Academy Original Practice · Proposed mark scheme

Section 9 of 14

Practice questions

SA-FP-VP-0012 marksL2 StandardNon-calculatorPlanes

The plane through

(1,2,3)

with normal

\begin{pmatrix}2\\-1\\1\end{pmatrix}

2x-y+z=d

. Find

d

Salam Academy Original Practice · Proposed mark scheme

SA-FP-VP-0023 marksL2 StandardNon-calculatorIntersections

Find the

x

-coordinate where

\mathbf r=\begin{pmatrix}1\\0\\2\end{pmatrix}+t\begin{pmatrix}1\\1\\1\end{pmatrix}

meets

x+2y+z=11

Salam Academy Original Practice · Proposed mark scheme

SA-FP-VP-0033 marksL3 ExamCalculatorIntersections

Find the acute angle between the line with direction

\begin{pmatrix}1\\1\\1\end{pmatrix}

and the plane

x+2y+z=5

, in degrees to 1 d.p.

Salam Academy Original Practice · Proposed mark scheme

SA-FP-VP-0043 marksL3 ExamCalculatorCross product

Find the area of the triangle formed by vectors

\mathbf a=\begin{pmatrix}2\\1\\0\end{pmatrix}

and

\mathbf b=\begin{pmatrix}1\\3\\2\end{pmatrix}

from a common vertex (to 3 s.f.).

Salam Academy Original Practice · Proposed mark scheme

More questions on every skill in this subtopic are in the question bank under Practise.

Section 10 of 14

Challenge

SA-FP-VP-0054 marksL4 ChallengingNon-calculatorIntersections

The line of intersection of

x+y+z=6

and

x-y+2z=5

has direction

\begin{pmatrix}3\\b\\c\end{pmatrix}

. Find

b

and

c

Salam Academy Original Practice · Proposed mark scheme

Section 11 of 14

Summary

$\mathbf a\times\mathbf b\perp\mathbf a,\mathbf b$ ; length = parallelogram area.
Plane: $\mathbf r\cdot\mathbf n=d$ .
Line–plane: substitute; solve for $t$ .
Angles: sin for line–plane, cos for plane–plane.

Section 12 of 14

Quick quiz

Q1 · SA-FP-VP-Q011 markL1 FoundationNon-calculatorCross product

The vector

\mathbf a\times\mathbf b

Salam Academy Original Practice · Proposed mark scheme

Section 13 of 14

Same topic, different board

Specification: 7367 A-level Further Mathematics
Where: Paper 2 · Further vectors
Expected: Planes, intersections, angles, and the vector product.
Calculator: Permitted on all papers.
Typical style: AQA mixes short multiple-choice questions with longer structured ones.
Open in: 7367 · Paper 2

Section 14 of 14

Past paper questions

No past paper questions are tagged to this subtopic yet. We add papers every week.

Browse past papers

Salam Academy original content · Content status: reviewed and published. Syllabus mapping: AQA 7367, First teaching 2017 (official specification).