A LevelFurther MathsEdexcelFurther Pure 1Revision NotesFurther VectorsFurther VectorsVector Geometry with Points, Lines & Planes

Vector Geometry with Points, Lines & Planes (Edexcel A Level Further Maths: Further Pure 1): Revision Note

Exam code: 9FM0

Written by: Mark Curtis

Updated on 22 January 2026

Vector geometry with points, lines & planes

How do I write the equation of a line using vector products?

The vector equation of the line $r = a + λ b$ can be written in a vector-product form given by
- $(r - a) \times b = 0$
- where
  - $a$ is the position vector of a point, A, on the line
  - $b$ is a direction vector of the line
The vector-product form has no scalar, $λ$
- and can be derived by first rearranging $r = a + λ b$
  - $(r - a) = λ b$
- then crossing both sides with $b$
  - $(r - a) \times b = λ b \times b = 0$
  - as $b \times b = 0$

Diagram showing a dotted straight line with an origin O away from the line. The point A is on the line and the position vector from O to A is shown and labelled 'a'. The line is labelled with the equation (r-a)xb=0.

Examiner Tips and Tricks

You must learn the vector-product form of the equation of a line, as it is not given in the formula booklet.

Worked Example

Find the Cartesian equation of the line

$(r - (2 i - 3 j + k)) \times (5 i + 6 j + 8 k) = 0$

Answer:

The line is given in the form $(r - a) \times b = 0$

Rewrite it as $r = a + λ b$

$r = (2 i - 3 j + k) + λ (5 i + 6 j + 8 k)$

Recall that the Cartesian form of $r = a + λ b$ is $\frac{x - a_{1}}{b_{1}} = \frac{y - a_{2}}{b_{2}} = \frac{z - a_{3}}{b_{3}}$

$\frac{x - 2}{5} = \frac{y + 3}{6} = \frac{z - 1}{8}$

Examiner Tips and Tricks

The Cartesian form of the vector equation of a line is given in the formula booklet.

How do I find the intersection of two lines?

Diagram showing intersecting lines at point P, skew lines with perpendicular distance, and parallel lines with arrows indicating direction.

Two intersecting lines

Recall that the point of intersection of two intersecting lines $r = a_{1} + λ b_{1}$ and $r = a_{2} + μ b_{2}$ can be found by
- setting equal $a_{1} + λ b_{1} = a_{2} + μ b_{2}$
- solving simultaneously two of the three equations for $λ$ and $μ$
  - and checking $λ$ and $μ$ also satisfy the third equation
- then substitute $λ$ (or $μ$ ) into $r = a_{1} + λ b_{1}$ (or $r = a_{2} + μ b_{2}$ )
  - to find the point of intersection

Two skew lines

If the lines do not intersect (and are not parallel) then they are said to be skew
- This is when $λ$ and $μ$ do not satisfy the third equation

Two parallel lines

If the two lines are parallel (and not the same line)
- then the simultaneous equations for $λ$ and $μ$ cannot be solved
  - The equations lead to untrue statements like $2 = 1$
If the two lines are in fact equations for the same line
- then the simultaneous equations for $λ$ and $μ$ cancel to $0 = 0$
  - which is true for all $λ$ and $μ$ values

What is the vector equation of a plane?

Recall that the vector equation of a plane through a point with position vector $a$ and two distinct non-parallel direction vectors $b$ and $c$ is
- $r = a + s b + t c$

A plane with an origin, O, outside of the plane. The point A lies in the plane and the position vector from O to A is shown and labelled 'a'. Two direction vectors within the plane, b and c, are shown.

If, instead, $a$ , $b$ and $c$ represent the position vectors of three points on a plane
- then first create two direction vectors within the plane, e.g.
  - $b - a$
  - $c - a$
- so the vector equation of the plane is
  - $r = a + λ (b - a) + μ (c - a)$

A plane with an origin, O, outside of the plane. The point A lies in the plane and the position vector from O to A is shown and labelled 'a'. The points B and C also lie in the plane and their position vectors from O are shown and labelled 'a' and 'b'. The vectors AB and AC are shown, both inside the plane, labelled b-a and c-a. The plane is labelled r=a+lambda*(b-a)+mu*(c-a).

Examiner Tips and Tricks

These two forms of the vector equation of a plane are given in the formula booklet.

How do I find the Cartesian equation of a plane using vector products?

Recall that the Cartesian equation of a plane through $A$ with position vector $a$ and normal vector $n = n_{1} i + n_{2} j + n_{3} k$ is
- $n_{1} x + n_{2} y + n_{3} z + d = 0$
- where
  - $d = - a \cdot n$
Recall that this can also be written in scalar product form
- $r \cdot n = a \cdot n$
- where
  - $r = (\begin{matrix} x \\ y \\ z \end{matrix})$

Examiner Tips and Tricks

The Cartesian equation of a plane is given in the formula booklet, but the scalar-product form is not.

If the equation of a plane is given in vector form $r = a + s b + t c$ where $a$ is a position vector and $b$ and $c$ are direction vectors
- it can be converted into Cartesian form
  - from its scalar-product form $r \cdot n = a \cdot n$
- by first finding the normal to both direction vectors $b$ and $c$
- using the vector product
  - $n = b \times c$

How do I find the intersection of a line and a plane?

Diagram illustrating lines with a plane; one line intersects, another is parallel, and a third is inside the plane. Labels specify each line's relationship.

Line intersecting plane

To find where the line $r = a + λ b$ intersects the plane $n_{1} x + n_{2} y + n_{3} z + d = 0$
- Let $r = (\begin{matrix} x \\ y \\ z \end{matrix})$ in $r = a + λ b$
- Substitute expressions for $x$ , $y$ and $z$ into $n_{1} x + n_{2} y + n_{3} z + d = 0$
- Solve the equation for $λ$
- Substitute the value of $λ$ back into $r = a + λ b$
  - to find the point of intersection

Line parallel to plane

If the line is parallel to the plane but not inside the plane then
- no $λ$ can be found
  - The equation formed for $λ$ is untrue, e.g. $2 = 1$

Line inside plane

If the line lies fully inside the plane then
- the equation formed for $λ$ cancels to $0 = 0$
  - which is true for all values of $λ$

How do I find the intersection of two planes using vector products?

If two non-parallel planes, $r \cdot n_{1} = a_{1} \cdot n_{1}$ and $r \cdot n_{2} = a_{2} \cdot n_{2}$ , intersect
- then the line of intersection has a direction vector perpendicular to both $n_{1}$ and $n_{2}$
  - i.e. in the direction $n_{1} \times n_{2}$
The equation of the line of intersection is
- $r = a + λ (n_{1} \times n_{2})$
- where
  - $r = (\begin{matrix} x \\ y \\ z \end{matrix})$
  - $a$ is a point common to both planes

Two intersecting planes. The horizontal plane has a normal of n_1. The inclined plane has a normal of n_2. The dotted line of intersection is shown passing through the point A (with position vector 'a' shown from the origin O), where A is a point common to both planes. The line of intersection has a direction vector of n_1 x n_2 and the line is labelled with the equation r=a+lambda*(n_1xn_2).

If you are not given $a$ in the question, you must find a common point yourself
- Set one of either $x$ , $y$ or $z$ to be equal to zero
  - as long as its component in $n_{1} \times n_{2}$ is non-zero
- Substitute that into $r \cdot n_{1} = a_{1} \cdot n_{1}$ and $r \cdot n_{2} = a_{2} \cdot n_{2}$
  - This gives two simultaneous equations
  - in the other two variables
- Solve the simultaneous equations
  - The solutions $x$ , $y$ and $z$ are the components of $a$

Examiner Tips and Tricks

Always check first that the planes are not parallel by inspecting their equations, e.g.:

$x + 2 y + 3 z = 1$ is parallel to $2 x + 4 y + 6 z = 7$
$x + 2 y + 3 z = 1$ is not parallel to $2 x + 4 y + 3 z = 7$

Worked Example

Find the equation of the line of intersection of the two planes $2 x - y + 3 z = 7$ and $x - 3 y + 4 z = 11$ , giving your answer in the form $r = a + λ b$ .

Answer:

The equation of the line of intersection is $r = a + λ (n_{1} \times n_{2})$ where $a$ is a point common to both planes

Write down the normal vectors for each plane

$n_{1} = (\begin{matrix} 2 \\ - 1 \\ 3 \end{matrix})$ and $n_{2} = (\begin{matrix} 1 \\ - 3 \\ 4 \end{matrix})$

Work out $n_{1} \times n_{2}$

$(\begin{matrix} 2 \\ - 1 \\ 3 \end{matrix}) \times (\begin{matrix} 1 \\ - 3 \\ 4 \end{matrix}) = (\begin{matrix} 5 \\ - 5 \\ - 5 \end{matrix})$

To find $a$ , set one out of $x$ , $y$ or $z$ to be zero (as long as their component in $n_{1} \times n_{2}$ is non-zero), for example:

$z = 0$

Substitute $z = 0$ into $2 x - y + 3 z = 7$ and $x - 3 y + 4 z = 11$ to get two simultaneous equations

$2 x - y = 7$ and $x - 3 y = 11$

Solve these equations simultaneously

$x = 2$ and $y = - 3$

The components $x = 2$ , $y = - 3$ and $z = 0$ form the vector $a$

$(\begin{matrix} 2 \\ - 3 \\ 0 \end{matrix})$

Write out the answer in the form $r = a + λ (n_{1} \times n_{2})$

$r = (\begin{matrix} 2 \\ - 3 \\ 0 \end{matrix}) + λ (\begin{matrix} 5 \\ - 5 \\ - 5 \end{matrix})$

Examiner Tips and Tricks

There are many different answers to this worked example (depending on how you find $a$ or use multiples of $n_{1} \times n_{2}$ ) and all correct answers are accepted.

What angles do I need to be able to calculate?

You must recall how to find the angle $θ$ between two lines, $r = a_{1} + λ b_{1}$ and $r = a_{2} + μ b_{2}$
- by solving $\cos θ = \frac{b_{1} \cdot b_{2}}{|b_{1}| |b_{2}|}$ for $θ$

Intersecting lines with angles theta and 180 minus theta, labelled as r equals a1 plus lambda b1 and r equals a2 plus mu b2 in blue and red text.

You must recall how to find the angle $θ$ between the line $r = a + λ b$ and the plane $r \cdot n = d$
- by first solving $\cos α = \frac{| b \cdot n |}{| b | | n |}$ for $α$
- then using $θ = 90 ° - α$

Diagram of a line intersecting a plane showing angles, with normal vector labels, a right angle, and notes about the direction vector and intersection points.

You must recall how to find the angle $θ$ between the two planes $r \cdot n_{1} = d_{1}$ and $r \cdot n_{2} = d_{2}$
- by solving $\cos θ = \frac{n_{1} \cdot n_{2}}{|n_{1}| |n_{2}|}$ for $θ$

Diagram showing two intersecting planes with angle θ between them and their normal vectors. Labels explain the angle between planes and normal vectors.

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Vector Geometry with Points, Lines & Planes (Edexcel A Level Further Maths: Further Pure 1): Revision Note

Vector geometry with points, lines & planes

How do I write the equation of a line using vector products?

How do I find the intersection of two lines?

Two intersecting lines

Two skew lines

Two parallel lines

What is the vector equation of a plane?

How do I find the Cartesian equation of a plane using vector products?

How do I find the intersection of a line and a plane?

Line intersecting plane

Line parallel to plane

Line inside plane

How do I find the intersection of two planes using vector products?

What angles do I need to be able to calculate?

Unlock more, it's free!

Join the 100,000+ Students that ❤️ Save My Exams

Author: Mark Curtis