A LevelFurther MathsEdexcelFurther Pure 1Revision NotesFurther VectorsFurther VectorsAreas & Volumes using Vector Rules

Areas & Volumes using Vector Rules (Edexcel A Level Further Maths: Further Pure 1): Revision Note

Exam code: 9FM0

Written by: Mark Curtis

Updated on 20 January 2026

Areas & volumes using vector rules

How do I find the area of a triangle using the vector product?

The area, $A$ , of a triangle with two sides given by the vectors $a$ and $b$ is
- $A = \frac{1}{2} | a \times b |$

A triangle formed out of the two vectors 'a' and 'b' with the area formula A = 1/2 *|a x b|.

Examiner Tips and Tricks

You must learn this formula, as it is not given in the formula booklet.

Worked Example

Find the area of the triangle whose vertices are given by $P (1, 2, 4)$ , $Q (0, 5, 8)$ and $R (7, - 6, - 3)$ .

Give your answer correct to 3 significant figures.

Answer:

The area $A$ of a triangle is $A = \frac{1}{2} | a \times b |$ where $a$ and $b$ are different sides from the same vertex

Form two vectors coming from the same vertex, e.g. $\vec{P Q}$ and $\vec{P R}$

$\begin{array}{rcl} a & = & \vec{P Q} \\ = & (\begin{matrix} 0 \\ 5 \\ 8 \end{matrix}) - (\begin{matrix} 1 \\ 2 \\ 4 \end{matrix}) \\ = & (\begin{matrix} - 1 \\ 3 \\ 4 \end{matrix}) \end{array}$

and

$\begin{array}{rcl} b & = & \vec{P R} \\ = & (\begin{matrix} 7 \\ - 6 \\ - 3 \end{matrix}) - (\begin{matrix} 1 \\ 2 \\ 4 \end{matrix}) \\ = & (\begin{matrix} 6 \\ - 8 \\ - 7 \end{matrix}) \end{array}$

Find $a \times b$

$\begin{array}{rcl} (\begin{matrix} - 1 \\ 3 \\ 4 \end{matrix}) \end{array} \times \begin{array}{rcl} (\begin{matrix} 6 \\ - 8 \\ - 7 \end{matrix}) \end{array} = \begin{array}{rcl} (\begin{matrix} 11 \\ 17 \\ - 10 \end{matrix}) \end{array}$

Substitute this into $A = \frac{1}{2} | a \times b |$

$\begin{array}{rcl} A & = & \frac{1}{2} \sqrt{11^{2} + 17^{2} + {(- 10)}^{2}} \\ = & \frac{1}{2} \sqrt{510} \end{array}$

Give your answer to 3 s.f.

The area of triangle PQR is 11.3 square units

Examiner Tips and Tricks

The method in the worked example also finds the area of 2D triangles with given vertices; just add a third coordinate of zero to each vertex.

E.g. write P(1, 2) as P(1, 2, 0).

How do I prove the area formula for a triangle?

You can show that $A = \frac{1}{2} | a \times b |$ is equivalent to the formula $A = \frac{1}{2} a b \sin C$ from trigonometry
- Recall that $a \times b = | a | | b | \sin θ \hat{n}$
- where
  - $θ$ is the angle between $a$ and $b$
  - $\hat{n}$ is a unit normal vector
  - so $| \hat{n} | = 1$
- Take the modulus of both sides and use $| \hat{n} | = 1$
  - $| a \times b | = | | a | | b | \sin θ \hat{n} | = | a | | b | | \sin θ | | \hat{n} | = | a | | b | | \sin θ |$
- $θ$ is acute or obtuse in a triangle
  - so $\sin θ > 0$
  - i.e. $| \sin θ | = \sin θ$
- Therefore $A = \frac{1}{2} | a \times b | = \frac{1}{2} | a | | b | \sin θ$
  - This is the same form as the area rule $A = \frac{1}{2} a b \sin C$

Examiner Tips and Tricks

If you need to rearrange the equation, remember that there are actually two equations when you remove the modulus signs, $\frac{1}{2} a \times b = \pm A$ .

How do I find the area of a parallelogram using the vector product?

The area, $A$ , of a parallelogram with two non-parallel sides given by the vectors $a$ and $b$ is
- $A = | a \times b |$
This is because it can be formed using two triangles
- then double the triangle formula given above

A parallelogram formed out of the two vectors 'a' and 'b' with the area formula A = |a x b|.

Examiner Tips and Tricks

You must learn this formula, as it is not given in the formula booklet.

How do I find the volume of a tetrahedron using the scalar triple product?

The volume, $V$ , of a tetrahedron with a triangular base formed by vectors $a$ and $b$ and a slanted height formed by vector $c$ is
- $V = \frac{1}{6} | c \cdot (a \times b) |$

A tetrahedron with a base formed out of the two vectors 'a' and 'b' and a slanted height formed out of the vector 'c', with the volume formula V = 1/6 * |c . (a x b)|.

In reality, due to the modulus signs, the vectors $a$ , $b$ and $c$ can be swapped in any order
- e.g. $a$ and $b$ don't have to be the 'base'
  - as long as $a$ , $b$ and $c$ are different edges from the same vertex

Examiner Tips and Tricks

You must learn this formula, as it is not given in the formula booklet.

Worked Example

Find the volume of the tetrahedron whose vertices are given by $P (1, 2, 4)$ , $Q (0, 5, 8)$ , $R (7, - 6, - 3)$ and $S (10, 1, 0)$ .

Give your answer correct to 3 significant figures.

Answer:

The volume $V$ of a tetrahedron is $V = \frac{1}{6} | c \cdot (a \times b) |$ where $a$ , $b$ and $c$ are different edges from the same vertex

Form three vectors coming from the same vertex, e.g. $\vec{P Q}$ , $\vec{P R}$ and $\vec{P S}$

$\begin{array}{rcl} a & = & \vec{P Q} \\ = & (\begin{matrix} 0 \\ 5 \\ 8 \end{matrix}) - (\begin{matrix} 1 \\ 2 \\ 4 \end{matrix}) \\ = & (\begin{matrix} - 1 \\ 3 \\ 4 \end{matrix}) \end{array}$

and

$\begin{array}{rcl} c & = & \vec{P S} \\ = & (\begin{matrix} 10 \\ 1 \\ 0 \end{matrix}) - (\begin{matrix} 1 \\ 2 \\ 4 \end{matrix}) \\ = & (\begin{matrix} 9 \\ - 1 \\ - 4 \end{matrix}) \end{array}$

Find $a \times b$

Now 'dot' this result with $c$

$\begin{array}{rcl} c \cdot (a \times b) & = & (\begin{matrix} 9 \\ - 1 \\ - 4 \end{matrix}) \cdot (\begin{matrix} 11 \\ 17 \\ - 10 \end{matrix}) \\ = & 9 \times 11 + (- 1) \times 17 + (- 4) \times (- 10) \\ = & 122 \end{array}$

Substitute this into $V = \frac{1}{6} | c \cdot (a \times b) |$

$\begin{array}{rcl} V & = & \frac{1}{6} | 122 | \\ = & \frac{61}{3} \end{array}$

Give your answer to 3 s.f.

The volume of the tetrahedron PQRS is 20.3 cubic units

How do I prove the volume formula for a tetrahedron?

A tetrahedron with a base formed out of the two vectors 'a' and 'b' and a slanted height formed out of the vector 'c', with the volume formula V = 1/6 * |c . (a x b)|. The vertical vector a x b is shown perpendicular to the horizontal base and the angle alpha is the angle between 'c' and the vertical.

The formula $V = \frac{1}{6} | c \cdot (a \times b) |$ comes from the fact that the volume $V$ of any pyramid is always $\frac{1}{3} \times base area \times vertical height$
- Let the triangular base be horizontal
  - It has an area of $\frac{1}{2} | a \times b |$ using the formula given above
- The vertical height is $| c | \cos α$
  - where $α$ is the angle between $c$ and the vertical vector $a \times b$
- So $V = \frac{1}{3} \times \frac{1}{2} | a \times b | \times | c | \cos α = \frac{1}{6} | c | | a \times b | \cos α$
  - The right-hand side looks like a scalar product, $p \cdot q = | p | | q | \cos α$
  - where $p = c$ and $q = a \times b$
  - meaning $V = \frac{1}{6} c \cdot (a \times b)$
- Final modulus signs are required to make sure $V$ is always positive
  - as $c \cdot (a \times b)$ could be negative, e.g. $α$ obtuse ( $\cos α < 0$ ) with $c$ pointing downwards
  - This gives $V = \frac{1}{6} | c \cdot (a \times b) |$

Examiner Tips and Tricks

If you need to rearrange the equation, remember that there are actually two equations when you remove the modulus signs, $\frac{1}{6} c \cdot (a \times b) = \pm V$ .

How do I find the volume of a parallelepiped using the scalar triple product?

The volume, $V$ , of a parallelepiped with a parallelogram base formed by vectors $a$ and $b$ and a slanted height formed by vector $c$ is
- $V = | c \cdot (a \times b) |$

A parallelepiped with a base formed out of the two vectors 'a' and 'b' and a slanted height formed out of the vector 'c', with the volume formula V = |c . (a x b)|.

In reality, due to the modulus signs, the vectors $a$ , $b$ and $c$ can be swapped in any order
- e.g. $a$ and $b$ don't have to be the 'base'
  - as long as $a$ , $b$ and $c$ are different edges from the same vertex

Examiner Tips and Tricks

You must learn this formula, as it is not given in the formula booklet.

The formula comes from $V = base area \times vertical height$
- Let the base be a horizontal parallelogram with area $| a \times b |$
  - from the formula given above
- and vertical height is $| c | \cos α$
  - where $α$ is the angle between $c$ and the the vertical vector $a \times b$
  - see the tetrahedron diagram (the proof proceeds in similar fashion)

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Areas & Volumes using Vector Rules (Edexcel A Level Further Maths: Further Pure 1): Revision Note

Areas & volumes using vector rules

How do I find the area of a triangle using the vector product?

How do I prove the area formula for a triangle?

How do I find the area of a parallelogram using the vector product?

How do I find the volume of a tetrahedron using the scalar triple product?

How do I prove the volume formula for a tetrahedron?

How do I find the volume of a parallelepiped using the scalar triple product?

Unlock more, it's free!

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Author: Mark Curtis