Further Pure — Vectors and 3D geometry: A-level Further Maths OCR revision notes

4.04

Equations of lines

Equations of a line in three dimensions 4.04

Key results

Vector equation: $r = a + λ d$ , where $a$ is the position vector of a point on the line and $d$ a direction vector.
Cartesian equation: $\frac{x - a _{1}}{d _{1}} = \frac{y - a _{2}}{d _{2}} = \frac{z - a _{3}}{d _{3}}$ $(= λ)$ .
The line through points A and B: $r = a + λ (b - a)$ .

Method

Cartesian to vector form: read the point from the numerators and the direction from the denominators, after making each coefficient of $x$ , $y$ , $z$ equal to 1. For example $\frac{2 x - 1}{4} = \frac{y + 3}{- 1} = z$ is $\frac{x - \frac{1}{2}}{2} = \frac{y + 3}{- 1} = \frac{z - 0}{1}$ .
If a component of $d$ is zero, that coordinate is constant: the line $r = (1, 2, 3) + λ (2, 0, 1)$ has Cartesian equations $\frac{x - 1}{2} = z - 3$ , $y = 2$ .

In practiceWrite the line through $A (2, - 1, 3)$ and $B (4, 1, 2)$ in vector and Cartesian form.

$d = A B = 4 - 2 1 + 1 2 - 3 = 22 - 1$
$r = 2 - 1 3 + λ 22 - 1$
$\frac{x - 2}{2} = \frac{y + 1}{2} = \frac{z - 3}{- 1}$ check B: each fraction equals 1

Notes

Neither form is unique: any point on the line and any non-zero multiple of $d$ give the same line.
Check that a point lies on a line by finding $λ$ from one coordinate and testing it in the other two.

Pairs of lines: intersecting, parallel or skew 4.04

Key results

Parallel: the direction vectors are multiples of each other.
Intersecting: there are values of $λ$ and $μ$ with $a + λ d_{1} = b + μ d_{2}$ .
Skew: not parallel and not intersecting. This can only happen in three dimensions.

Method

Equate the $x$ , $y$ and $z$ components of the two lines: three equations in two unknowns.
Solve two of them for $λ$ and $μ$ , then test the third. If it holds, the lines meet (substitute to find the point); if not, and they are not parallel, they are skew.

In practiceDo $r = (1, 0, 2) + λ (1, 1, - 1)$ and $r = (2, 3, 0) + μ (0, 1, 1)$ meet?

$1 + λ = 2, λ = 3 + μ, 2 - λ = μ$ equate the components
$λ = 1, μ = - 2$ from the first two equations
$z : 2 - 1 = 1 \neq = - 2$ the third equation fails
The directions $(1, 1, - 1)$ and $(0, 1, 1)$ are not parallel, so the lines are skew.

Notes

Always use the third equation as a check. Two equations in two unknowns can be solved almost always, which proves nothing.
The shortest distance between skew lines is $\frac{∣ ( b - a ) \cdot n ∣}{∣ n ∣}$ with $n = d_{1} \times d_{2}$ (see the last section).

Worked examples

Worked example

Show that the lines:

r = (1, 0, 2) + λ (1, 1, - 1)

and:

r = (2, 3, 0) + μ (0, 1, 1)

are skew, and find the shortest distance between them.

Show worked solution

The directions are not parallel.

Equating components: $1 + λ = 2$ , $λ = 3 + μ$ , $2 - λ = μ$ .

The first gives $λ = 1$ , the second $μ = - 2$ , and the third then reads $1 = - 2$ , which is false: the lines do not meet, so they are skew.

n = (1, 1, - 1) \times (0, 1, 1) = (2, - 1, 1)

and:

b - a = (1, 3, - 2)

so the distance is:

\frac{∣2 - 3 - 2∣}{6} = \frac{3}{6} = \frac{6}{2}

Worked example

Write the line through A $(1, 2, - 1)$ and B $(3, - 1, 4)$ in vector and Cartesian form.

Show worked solution

Direction:

A B = (2, - 3, 5)

so:

r = (1, 2, - 1) + λ (2, - 3, 5)

and:

\frac{x - 1}{2} = \frac{y - 2}{- 3} = \frac{z + 1}{5}

4.04

Equations of planes

Equations of a plane 4.04

Definitions

Normal vector: a non-zero vector perpendicular to every line in the plane.

Key results

Scalar product form: $r \cdot n = a \cdot n$ , where A is a point of the plane and $n$ a normal.
Cartesian form: $a x + b y + cz = d$ , where $(a, b, c)$ is a normal vector.
Parametric (vector) form: $r = a + λ b + μ c$ , with $b$ , $c$ non-parallel vectors in the plane.

Method

Plane through three points A, B, C: find $A B$ and $A C$ , take $n = A B \times A C$ , and find $d = n \cdot a$ . Check with B and C.
Parametric to Cartesian: $n = b \times c$ , then $d = n \cdot a$ .

In practiceFind the plane through $A (1, 0, 2)$ , $B (2, 1, 0)$ and $C (0, 2, 1)$ .

$A B = 11 - 2, A C = - 1 2 - 1$
$n = A B \times A C = 1 (- 1) - (- 2) (2) (- 2) (- 1) - 1 (- 1) 1 (2) - 1 (- 1) = 333$ use the simpler normal $(1, 1, 1)$
$x + y + z = n \cdot a = 1 + 0 + 2 = 3$ check B: $2 + 1 + 0 = 3$ ; C: $0 + 2 + 1 = 3$

Notes

Reading a normal straight off $a x + b y + cz = d$ is the quickest step in the topic: $3 x - y + 2 z = 7$ has normal $(3, - 1, 2)$ .
If $n$ is a unit vector, $∣ d ∣$ in $r \cdot n = d$ is the distance of the plane from the origin.

Worked example

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

Show worked solution

A B = (1, 1, - 2)

and:

A C = (- 1, 3, - 1)

n = A B \times A C = (1 \cdot (- 1) - (- 2) \cdot 3, (- 2) (- 1) - 1 \cdot (- 1), 1 \cdot 3 - 1 \cdot (- 1)) = (5, 3, 4)

Then:

d = n \cdot a = 5 + 0 + 8 = 13

so the plane is:

5 x + 3 y + 4 z = 13

Check: B gives $10 + 3 + 0 = 13$ and C gives $0 + 9 + 4 = 13$ .

4.04

The vector product

The vector product 4.04

Key results

$a \times b = a_{2} b_{3} - a_{3} b_{2} a_{3} b_{1} - a_{1} b_{3} a_{1} b_{2} - a_{2} b_{1}$ , also found as the determinant $i a_{1} b_{1} j a_{2} b_{2} k a_{3} b_{3}$ .
$a \times b$ is perpendicular to both $a$ and $b$ , with $∣ a \times b ∣ = ∣ a ∣∣ b ∣ sin θ$ .
$b \times a = - a \times b$ , and $a \times b = 0$ exactly when $a$ and $b$ are parallel.
Area of triangle ABC $= \frac{1}{2} ∣ A B \times A C ∣$ .

Notes

Check a vector product by dotting it with both original vectors: each answer must be 0.
The direction follows the right-hand rule, so the order matters for the sign, though not for which line the result lies along.

4.04

Intersections, angles and distances

Where a line meets a plane 4.04

Key results

A line is parallel to a plane exactly when $d \cdot n = 0$ .
The line of intersection of two planes has direction $n_{1} \times n_{2}$ . A point on it is found by setting one coordinate to a convenient value and solving the two plane equations.

Method

Write a general point of the line in terms of $λ$ , substitute it into the Cartesian equation of the plane, and solve for $λ$ . Substitute back for the point.
If the equation reduces to a false statement ( $0 = 5$ ), the line is parallel to the plane and does not meet it; if it reduces to $0 = 0$ , the line lies in the plane.

In practiceFind where $r = (1, 2, 0) + λ (2, - 1, 3)$ meets the plane $x + y + z = 9$ .

$(1 + 2 λ) + (2 - λ) + 3 λ = 9$ substitute a general point of the line
$3 + 4 λ = 9 ⟹ λ = \frac{3}{2}$
$r = 1 + 3 2 - \frac{3}{2} 0 + \frac{9}{2} = 4 \frac{1}{2} \frac{9}{2}$ check: $4 + \frac{1}{2} + \frac{9}{2} = 9$

Notes

The foot of the perpendicular from a point P to a plane is where the line $r = p + λ n$ meets the plane. Reflecting P in the plane uses $λ$ twice as large.

Angles between lines and planes 4.04

Key results

Between two lines: $cos θ = \frac{∣ d _{1} \cdot d _{2} ∣}{∣ d _{1} ∣∣ d _{2} ∣}$ , the acute angle between the directions.
Between a line and a plane: $sin θ = \frac{∣ d \cdot n ∣}{∣ d ∣∣ n ∣}$ , since the angle with the plane is $9 0^{\circ}$ minus the angle with the normal.
Between two planes: $cos θ = \frac{∣ n _{1} \cdot n _{2} ∣}{∣ n _{1} ∣∣ n _{2} ∣}$ , the acute angle between the normals.

Method

Take the direction vector of each line and the normal of each plane, then compute the scalar product and the moduli.
Use $cos$ for two lines or two planes, and $sin$ for a line and a plane, because the scalar product gives the angle with the normal.

In practiceFind the angle between the line $r = (0, 1, 2) + λ (1, 2, 2)$ and the plane $2 x - y + 2 z = 5$ .

$d \cdot n = (1) (2) + (2) (- 1) + (2) (2) = 4, ∣ d ∣ = ∣ n ∣ = 3$
$sin θ = \frac{∣ d \cdot n ∣}{∣ d ∣∣ n ∣} = \frac{4}{9}$ sine, not cosine, for a line and a plane
$θ = 26. 4^{\circ}$ to 1 decimal place

Notes

The commonest error is using $cos$ for a line and a plane. The scalar product gives the angle with the normal; the angle with the plane is its complement.
The modulus in the numerator gives the acute angle. If the question asks for the obtuse angle between two planes, subtract from $18 0^{\circ}$ .

Shortest distances 4.04

Key results

Point $(x_{1}, y_{1}, z_{1})$ to plane $a x + b y + cz = d$ : $\frac{∣ a x _{1} + b y _{1} + c z _{1} - d ∣}{a ^{2} + b ^{2} + c ^{2}}$ .
Between skew lines $r = a + λ d_{1}$ and $r = b + μ d_{2}$ : $\frac{∣ ( b - a ) \cdot ( d _{1} \times d _{2} ) ∣}{∣ d _{1} \times d _{2} ∣}$ .
Point P to a line $r = a + λ d$ : find $λ$ so that $(a + λ d - p) \cdot d = 0$ , then the distance is the length of that vector; equivalently $\frac{∣ ( p - a ) \times d ∣}{∣ d ∣}$ .

Method

Point to plane: substitute the point into $a x + b y + cz - d$ , take the modulus and divide by $∣ n ∣$ .
Skew lines: find the common perpendicular $n = d_{1} \times d_{2}$ , then project the vector joining a point of each line onto it.

In practiceFind the shortest distance between the skew lines $r = (1, 0, 2) + λ (1, 1, - 1)$ and $r = (2, 3, 0) + μ (0, 1, 1)$ .

$n = 11 - 1 \times 011 = 2 - 1 1$ check: $n \cdot d_{1} = 2 - 1 - 1 = 0$
$b - a = 13 - 2, (b - a) \cdot n = 2 - 3 - 2 = - 3$
$d = \frac{∣ - 3∣}{6} = \frac{6}{2} \approx 1.22$

Notes

Every shortest distance runs along a common perpendicular: to the plane, or to both skew lines. The formulae are projections onto that perpendicular direction.
The point-to-plane formula is in the formula booklet; knowing why it works lets you adapt it, for instance to find the distance between two parallel planes.

Worked examples

Worked example

The line:

r = (1, 1, 1) + t (1, 2, - 1)

meets the plane:

2 x - y + 3 z = 10

Find the point of intersection and the acute angle between the line and the plane.

Show worked solution

Substituting:

2 (1 + t) - (1 + 2 t) + 3 (1 - t) = 10

so $4 - 3 t = 10$ and $t = - 2$ : the point is $(- 1, - 3, 3)$ .

With:

d = (1, 2, - 1)

and:

n = (2, - 1, 3)

$d \cdot n = - 3$ , so:

sin θ = \frac{3}{6 14} = 0.327

and $θ = 19. 1^{\circ}$ .

Worked example

Find the distance from P $(3, 1, - 2)$ to the plane $2 x - y + 2 z = 5$ , and the acute angle between the planes $x + y + z = 1$ and $2 x - y + z = 3$ .

Show worked solution

Distance:

= \frac{∣2 ( 3 ) - 1 + 2 ( - 2 ) - 5∣}{4 + 1 + 4} = \frac{∣ - 4∣}{3} = \frac{4}{3}

For the angle:

cos θ = \frac{∣ ( 1 , 1 , 1 ) \cdot ( 2 , - 1 , 1 ) ∣}{3 6} = \frac{2}{18} = 0.471

so $θ = 61. 9^{\circ}$ .