Notes · Further MathematicsUK · A-Levels

Further vectors

Further vectors develops three-dimensional coordinate geometry using the scalar and vector products. It covers the equations of lines and planes in space, how they intersect, the angles between them, and the distances between points, lines and planes. The scalar product from A-Level Mathematics is assumed, and the new vector (cross) product provides normals and areas.

4 sections~11 min reading time3 competenciesLevel Standard 2 · Advanced 2

T·0666 / 13

Exam profile

AO1 · Compute scalar and vector products, equations of lines and planes, and distances accuratelyAO2 · Derive the equation of a plane from a normal and justify a distance formulaAO3 · Solve three-dimensional configuration problems involving intersections, angles and distances

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basic level

AS Further Mathematics covers the scalar product and the vector equation of a line in three dimensions.

higher level

The full A-Level adds the vector product, equations of planes, and the full range of intersection, angle and distance problems, including skew lines.

Depth

Text

Contents · 4 sections

Further vectors

Scalar and vector products#

StandardAQA 7367 §3.6DfE Further Mathematics — Further vectors

The vector product is perpendicular to both vectors

Fig. 1With

\mathbf{a}

along

x

and

\mathbf{b}

along

y

, the product

\mathbf{a}\times\mathbf{b}

points along

z

, perpendicular to both.

The scalar (dot) product

\mathbf{a}\cdot\mathbf{b} = a_1 b_1 + a_2 b_2 + a_3 b_3 = |\mathbf{a}||\mathbf{b}|\cos\theta

returns a number and is the tool for angles and perpendicularity. Rearranged, it gives

\cos\theta = \dfrac{\mathbf{a}\cdot\mathbf{b}}{|\mathbf{a}||\mathbf{b}|}

, and in particular two non-zero vectors are perpendicular exactly when their scalar product is zero. This single fact underlies every perpendicularity condition later in the topic.

The vector (cross) product

\mathbf{a}\times\mathbf{b}

is new to Further Mathematics and returns a vector. Its magnitude is

|\mathbf{a}\times\mathbf{b}| = |\mathbf{a}||\mathbf{b}|\sin\theta

, which equals the area of the parallelogram spanned by

\mathbf{a}

and

\mathbf{b}

(so half of it is the area of the triangle). Its direction is perpendicular to both

\mathbf{a}

and

\mathbf{b}

, in the sense given by the right-hand rule — which is precisely why the cross product is the standard way to produce a normal to a plane.

In components, the vector product is computed as a symbolic determinant

\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}

, expanding along the top row. Unlike the scalar product, the cross product is anticommutative:

\mathbf{b}\times\mathbf{a} = -\mathbf{a}\times\mathbf{b}

, so order matters and reverses the direction of the resulting normal.

The two products answer complementary geometric questions. The scalar product measures how much two vectors point the same way (it is zero at right angles, maximal when parallel); the vector product measures how much they fail to be parallel (it is zero when parallel, maximal at right angles). Together they equip you to find angles, test perpendicularity, compute areas of triangles and parallelograms, and construct normals — the foundational operations for the geometry of lines and planes that follows.

\mathbf{a}\cdot\mathbf{b} = |\mathbf{a}||\mathbf{b}|\cos\theta, \qquad \mathbf{a}\perp\mathbf{b} \iff \mathbf{a}\cdot\mathbf{b} = 0

Scalar product

Returns a number; zero exactly at right angles.

|\mathbf{a}\times\mathbf{b}| = |\mathbf{a}||\mathbf{b}|\sin\theta = \text{area of the parallelogram}, \qquad \mathbf{a}\times\mathbf{b} \perp \mathbf{a},\mathbf{b}

Vector product

Returns a vector perpendicular to both; its magnitude is the parallelogram area.

Worked example

Angle and triangle area

For $\mathbf{a} = \begin{pmatrix} 1 \\ 2 \\ 2 \end{pmatrix}$ and $\mathbf{b} = \begin{pmatrix} 2 \\ 3 \\ 6 \end{pmatrix}$ , find the angle between them and the area of the triangle they span.

Scalar product and magnitudes
$\mathbf{a}\cdot\mathbf{b} = 2 + 6 + 12 = 20$ ; $|\mathbf{a}| = \sqrt{1+4+4} = 3$ , $|\mathbf{b}| = \sqrt{4+9+36} = 7$ .
Angle
$\cos\theta = \dfrac{20}{3\times 7} = \dfrac{20}{21}$ , so $\theta = \arccos\tfrac{20}{21} \approx 17.8^{\circ}$ .
$\cos\theta = \frac{20}{21} \ \Rightarrow\ \theta \approx 17.8^{\circ}$
Vector product
$\mathbf{a}\times\mathbf{b} = \begin{pmatrix} 2\cdot 6 - 2\cdot 3 \\ 2\cdot 2 - 1\cdot 6 \\ 1\cdot 3 - 2\cdot 2 \end{pmatrix} = \begin{pmatrix} 6 \\ -2 \\ -1 \end{pmatrix}$ , with magnitude $\sqrt{36+4+1} = \sqrt{41}$ .
Triangle area
Area $= \tfrac{1}{2}|\mathbf{a}\times\mathbf{b}| = \tfrac{1}{2}\sqrt{41} \approx 3.20$ .

Result: $\theta = \arccos\tfrac{20}{21} \approx 17.8^{\circ}$ ; triangle area $\tfrac{1}{2}\sqrt{41} \approx 3.20$ .

Typical mistakes

Confusing the two products — using $\mathbf{a}\cdot\mathbf{b}$ (a scalar) where a normal vector is needed, or vice versa.
Sign or order slips in the cross product; remember $\mathbf{b}\times\mathbf{a} = -\mathbf{a}\times\mathbf{b}$ .

Active revision

Given $\mathbf{a} = \begin{pmatrix} 1 \\ 2 \\ 2 \end{pmatrix}$ and $\mathbf{b} = \begin{pmatrix} 2 \\ 3 \\ 6 \end{pmatrix}$ , find the angle between them and the area of the triangle with these as two sides.

Active recall

Recall the key points — then reveal.

Sources: Further mathematics: AS and A level content (GCE subject content) (Department for Education)

Lines in three dimensions#

StandardAQA 7367 §3.6DfE Further Mathematics — Further vectors

A line in three dimensions

Fig. 2The line

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

passes through

\mathbf{a}

in the direction

\mathbf{d}

A straight line in space is determined by one point on it and a direction. Its vector equation is

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

, where

\mathbf{a}

is the position vector of a known point,

\mathbf{d}

is the direction vector, and the parameter

\lambda

ranges over all real numbers, tracing out every point of the line. Any point on the line and any non-zero multiple of the direction serve equally well, so a line has many equivalent equations.

Eliminating the parameter gives the Cartesian form

\dfrac{x - a_1}{d_1} = \dfrac{y - a_2}{d_2} = \dfrac{z - a_3}{d_3}

, each fraction equalling

\lambda

. This form is convenient for reading off a point and a direction, though care is needed when a component of

\mathbf{d}

is zero, since then the corresponding coordinate is constant and that fraction is replaced by the statement that the coordinate is fixed.

Two lines in space are related in one of three ways: they are parallel (their direction vectors are scalar multiples), they intersect (there is a single common point), or they are skew (not parallel and never meeting). Deciding which requires testing: check whether the directions are proportional first, and if not, set the two vector equations equal and try to solve for consistent parameters. If the resulting equations are inconsistent, the lines are skew.

The parameter test for intersection is a small but disciplined piece of algebra. Setting

\mathbf{a}_1 + \lambda\mathbf{d}_1 = \mathbf{a}_2 + \mu\mathbf{d}_2

gives three scalar equations in the two unknowns

\lambda

and

\mu

; solve two of them and then check the third. If the third is satisfied, the lines meet and the common point follows by substitution; if it is not, the lines are skew. Distinguishing skew lines from intersecting ones by this check is a standard exam demand.

\mathbf{r} = \mathbf{a} + \lambda\mathbf{d}, \qquad \frac{x-a_1}{d_1} = \frac{y-a_2}{d_2} = \frac{z-a_3}{d_3}

Line: vector and Cartesian forms

A point $\mathbf{a}$ and a direction $\mathbf{d}$ determine the line; the Cartesian form eliminates $\lambda$ .

Worked example

Do two lines intersect?

Determine whether $\ell_1: \mathbf{r} = \begin{pmatrix} 1 \\ 0 \\ 2 \end{pmatrix} + \lambda\begin{pmatrix} 1 \\ 1 \\ -1 \end{pmatrix}$ and $\ell_2: \mathbf{r} = \begin{pmatrix} 2 \\ 1 \\ 0 \end{pmatrix} + \mu\begin{pmatrix} 2 \\ 1 \\ 1 \end{pmatrix}$ meet.

Set the equations equal
$1 + \lambda = 2 + 2\mu$ , $\ \lambda = 1 + \mu$ , $\ 2 - \lambda = \mu$ .
Solve two equations
From the second, $\lambda = 1 + \mu$ ; substitute into the third: $2 - (1+\mu) = \mu \Rightarrow 1 - \mu = \mu \Rightarrow \mu = \tfrac12$ , so $\lambda = \tfrac32$ .
Check the first equation
$1 + \lambda = 1 + \tfrac32 = \tfrac52$ ; $2 + 2\mu = 2 + 1 = 3$ . Since $\tfrac52 \neq 3$ , the first equation fails.
$\tfrac{5}{2} \neq 3 \ \Rightarrow\ \text{no consistent } (\lambda,\mu)$
Conclude
The directions $(1,1,-1)$ and $(2,1,1)$ are not proportional, so the lines are not parallel; and no common point exists, so they are skew.

Result: The lines are skew (not parallel and never meeting).

Typical mistakes

Assuming that non-parallel lines must intersect; in three dimensions they are usually skew.
Solving only two of the three component equations and forgetting to check consistency with the third.

Active revision

Determine whether the lines $\mathbf{r} = \begin{pmatrix} 1 \\ 0 \\ 2 \end{pmatrix} + \lambda\begin{pmatrix} 1 \\ 1 \\ -1 \end{pmatrix}$ and $\mathbf{r} = \begin{pmatrix} 2 \\ 1 \\ 0 \end{pmatrix} + \mu\begin{pmatrix} 2 \\ 1 \\ 1 \end{pmatrix}$ intersect, are parallel, or are skew.

Active recall

Recall the key points — then reveal.

Sources: AQA A-level Further Mathematics 7367 specification (AQA)

Planes in three dimensions#

AdvancedAQA 7367 §3.6DfE Further Mathematics — Further vectors

A plane and its normal vector

Fig. 3The plane

x + y + z = 3

has normal

\mathbf{n} = (1,1,1)

, perpendicular to every line in the plane.

A plane can be described in three equivalent ways. The vector (parametric) form

\mathbf{r} = \mathbf{a} + \lambda\mathbf{d}_1 + \mu\mathbf{d}_2

builds the plane from a point and two independent direction vectors lying in it. The scalar-product form

\mathbf{r}\cdot\mathbf{n} = d

uses a normal vector

\mathbf{n}

perpendicular to the plane, with

d = \mathbf{a}\cdot\mathbf{n}

. The Cartesian form

n_1 x + n_2 y + n_3 z = d

is just the scalar-product form written out, and reveals that the coefficients of

x, y, z

are the components of the normal.

The normal is the key to almost every plane calculation, and the vector product supplies it: given two directions

\mathbf{d}_1, \mathbf{d}_2

in the plane (for instance formed from three points as

\mathbf{d}_1 = \overrightarrow{AB}

\mathbf{d}_2 = \overrightarrow{AC}

), the normal is

\mathbf{n} = \mathbf{d}_1 \times \mathbf{d}_2

. Then

d

is found by substituting any known point into

\mathbf{r}\cdot\mathbf{n} = d

. This is the standard route from a parametric or three-point description to the Cartesian equation.

Reading a plane's equation geometrically is essential. In

\mathbf{r}\cdot\mathbf{n} = d

, the vector

\mathbf{n}

is perpendicular to the whole plane, and

\dfrac{d}{|\mathbf{n}|}

is the (signed) perpendicular distance of the plane from the origin. Two planes are parallel when their normals are parallel, and perpendicular when their normals are perpendicular — so questions about planes reduce to questions about their normals.

Moving fluently between the three forms is the examinable skill. A question may give three points and ask for the Cartesian equation (cross product then substitute), or give the Cartesian equation and ask for the perpendicular distance from a point, or ask where a line crosses the plane. Recognising which form makes a given task easiest — parametric for 'lies in the plane', scalar-product for angles and distances — turns a long calculation into a short one.

\mathbf{r}\cdot\mathbf{n} = d, \qquad n_1 x + n_2 y + n_3 z = d, \qquad \mathbf{n} = \overrightarrow{AB}\times\overrightarrow{AC}

Plane: scalar-product and Cartesian forms

The normal $\mathbf{n}$ is perpendicular to the plane; $d = \mathbf{a}\cdot\mathbf{n}$ .

Worked example

The plane through three points

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

Two directions in the plane
$\overrightarrow{AB} = \begin{pmatrix} -1 \\ 2 \\ 0 \end{pmatrix}$ , $\overrightarrow{AC} = \begin{pmatrix} -1 \\ 0 \\ 3 \end{pmatrix}$ .
Normal by cross product
$\mathbf{n} = \overrightarrow{AB}\times\overrightarrow{AC} = \begin{pmatrix} 2\cdot 3 - 0\cdot 0 \\ 0\cdot(-1) - (-1)\cdot 3 \\ (-1)\cdot 0 - 2\cdot(-1) \end{pmatrix} = \begin{pmatrix} 6 \\ 3 \\ 2 \end{pmatrix}$ .
$\mathbf{n} = \begin{pmatrix} 6 \\ 3 \\ 2 \end{pmatrix}$
Find d
Substitute $A(1,0,0)$ : $d = 6(1) + 3(0) + 2(0) = 6$ .
State the equation
The plane is $6x + 3y + 2z = 6$ . (Check $B$ : $0 + 6 + 0 = 6$ ✓; $C$ : $0 + 0 + 6 = 6$ ✓.)

Result: $6x + 3y + 2z = 6$ .

Typical mistakes

Taking the coefficients of a Cartesian plane equation as a point on the plane rather than as the components of its normal.
Forgetting to compute $d$ by substituting a known point after finding the normal.

Active revision

Find the Cartesian equation of the plane through the points $A(1,0,0)$ , $B(0,2,0)$ and $C(0,0,3)$ .

Active recall

Recall the key points — then reveal.

Sources: Further mathematics: AS and A level content (GCE subject content) (Department for Education)

Intersections, angles and distances#

AdvancedAQA 7367 §3.6DfE Further Mathematics — Further vectors

A line meeting a plane

Fig. 4The line crosses the plane

z = 0

at the single point

I(2,2,0)

Where a line meets a plane is found by substitution: put the line's parametric coordinates into the plane's Cartesian equation and solve for the parameter

\lambda

. A single value of

\lambda

gives a unique point of intersection; if the line's direction is perpendicular to the normal (

\mathbf{d}\cdot\mathbf{n} = 0

) the line is parallel to the plane, meeting it either everywhere (line in the plane) or nowhere. This mirrors the classification of systems of equations in the Matrices topic.

Angles between objects are computed with the scalar product, but with a twist for the line-and-plane case. The angle

\theta

between two planes equals the angle between their normals:

\cos\theta = \dfrac{|\mathbf{n}_1\cdot\mathbf{n}_2|}{|\mathbf{n}_1||\mathbf{n}_2|}

. The angle

\alpha

between a line and a plane, however, is the complement of the angle between the direction and the normal, so it uses sine:

\sin\alpha = \dfrac{|\mathbf{d}\cdot\mathbf{n}|}{|\mathbf{d}||\mathbf{n}|}

. Keeping the sine-versus-cosine distinction straight is essential and frequently examined.

The perpendicular distance from a point

P

with position vector

\mathbf{p}

to the plane

\mathbf{r}\cdot\mathbf{n} = d

\dfrac{|\mathbf{p}\cdot\mathbf{n} - d|}{|\mathbf{n}|}

. This measures how far

P

sits off the plane along the normal direction; the modulus makes the distance positive whichever side of the plane

P

lies on. It is the three-dimensional analogue of the familiar two-dimensional point-to-line distance formula.

The most demanding standard result is the shortest distance between two skew lines, which is the length of their common perpendicular. Its direction is

\mathbf{d}_1\times\mathbf{d}_2

(perpendicular to both directions), and the distance is

\dfrac{|(\mathbf{a}_2 - \mathbf{a}_1)\cdot(\mathbf{d}_1\times\mathbf{d}_2)|}{|\mathbf{d}_1\times\mathbf{d}_2|}

: project the vector joining the two lines onto the common-perpendicular direction. This blends both products — the cross product to build the common perpendicular, the scalar product to project onto it — and is the culmination of the topic.

\sin\alpha = \frac{|\mathbf{d}\cdot\mathbf{n}|}{|\mathbf{d}||\mathbf{n}|} \ (\text{line-plane}), \qquad \cos\theta = \frac{|\mathbf{n}_1\cdot\mathbf{n}_2|}{|\mathbf{n}_1||\mathbf{n}_2|} \ (\text{plane-plane})

Angles

Line-plane uses sine (complement of the normal angle); plane-plane uses the angle between normals.

\text{dist}(P,\text{plane}) = \frac{|\mathbf{p}\cdot\mathbf{n} - d|}{|\mathbf{n}|}, \qquad \text{dist(skew)} = \frac{|(\mathbf{a}_2-\mathbf{a}_1)\cdot(\mathbf{d}_1\times\mathbf{d}_2)|}{|\mathbf{d}_1\times\mathbf{d}_2|}

Distances

Point-to-plane, and the shortest distance between two skew lines along their common perpendicular.

Worked example

Intersection and a perpendicular distance

(a) Find where $\mathbf{r} = \begin{pmatrix} 1 \\ 0 \\ -1 \end{pmatrix} + \lambda\begin{pmatrix} 2 \\ 1 \\ 3 \end{pmatrix}$ meets $x + y + z = 6$ . (b) Find the distance from $P(4,1,3)$ to $x + 2y + 2z = 6$ .

Substitute the line
Coordinates $(1+2\lambda,\ \lambda,\ -1+3\lambda)$ ; substitute: $(1+2\lambda) + \lambda + (-1+3\lambda) = 6$ , so $6\lambda = 6$ , $\lambda = 1$ .
Intersection point
At $\lambda = 1$ : $(1+2,\ 1,\ -1+3) = (3, 1, 2)$ ; check $3+1+2 = 6$ ✓.
$\text{intersection } (3, 1, 2)$
Distance formula
For plane $x+2y+2z=6$ , $\mathbf{n}=(1,2,2)$ , $|\mathbf{n}|=3$ . At $P(4,1,3)$ : $\mathbf{p}\cdot\mathbf{n} - d = 4 + 2 + 6 - 6 = 6$ .
Evaluate
Distance $= \dfrac{|6|}{3} = 2$ .

Result: (a) The line meets the plane at $(3,1,2)$ ; (b) the distance is $2$ .

Typical mistakes

Using cosine for the line-plane angle (it needs sine, being the complement of the normal angle).
Omitting the modulus in a distance formula, or using $\mathbf{d}_1\times\mathbf{d}_2$ with the wrong magnitude in the denominator.

Active revision

Find the point of intersection of the line $\mathbf{r} = \begin{pmatrix} 1 \\ 0 \\ -1 \end{pmatrix} + \lambda\begin{pmatrix} 2 \\ 1 \\ 3 \end{pmatrix}$ with the plane $x + y + z = 6$ , and find the perpendicular distance from the point $(4,1,3)$ to the plane $x + 2y + 2z = 6$ .

Active recall

Recall the key points — then reveal.

Sources: Further mathematics: AS and A level content (GCE subject content) (Department for Education)

References & sources

Sources

Department for Education

Further mathematics: AS and A level content (GCE subject content)

AQA

AQA A-level Further Mathematics 7367 specification

Further vectors

Key points

Angle and triangle area

Key points

Do two lines intersect?

Key points

The plane through three points

Key points

Intersection and a perpendicular distance

Sources