A LevelFurther MathsEdexcelFurther Pure 1Revision NotesFurther VectorsFurther VectorsVector (Cross) Product

Vector (Cross) Product (Edexcel A Level Further Maths: Further Pure 1): Revision Note

Exam code: 9FM0

Written by: Mark Curtis

Updated on 23 March 2026

Vector (cross) product

What is the vector product?

The vector product is an operation which takes two vectors and outputs a vector that is
- perpendicular (normal) to both vectors
The vector product between two vectors $a$ and $b$ is written as $a \times b$
- This is why it is also called the cross product
The direction of the vector produced follows the right-hand rule
- Using your right hand:
  - Point your index finger in the direction of the first vector, $a$
  - Point your middle finger in the direction of the second vector, $b$
  - The direction of the vector product $a \times b$ is given by the direction of your thumb

Right hand showing vector cross product. Index finger points left for vector a, middle down for b, and thumb up for normal vector a×b. — **Diagram of the right-hand rule**

How do I calculate the vector product?

If $a = (\binom{a_{1}}{\begin{matrix} a_{2} \\ a_{3} \end{matrix}})$ and $b = (\binom{b_{1}}{\begin{matrix} b_{2} \\ b_{3} \end{matrix}})$ then the formula for the vector product is
- $a \times b = |\begin{matrix} i & j & k \\ a_{1} & a_{2} & a_{3} \\ b_{1} & b_{2} & b_{3} \end{matrix}| = (\begin{matrix} a_{2} b_{3} - a_{3} b_{2} \\ a_{3} b_{1} - a_{1} b_{3} \\ a_{1} b_{2} - a_{2} b_{1} \end{matrix})$
Another formula for the vector product is
- $a \times b = |a| |b| \sin θ \hat{n}$
- where
  - $θ$ is the angle between $a$ and $b$
  - $\hat{n}$ is a unit vector that is normal to the two vectors and follows the right-hand rule

Examiner Tips and Tricks

Both formulae for the vector product are given in the formula booklet.

What properties of the vector product do I need to know?

The vector product is not commutative
- $a \times b \neq b \times a$
Changing the order of the vectors reverses the direction of the vector product
- $a \times b = - b \times a$
The distributive law over addition can be used to expand brackets
- $a \times (b + c) = a \times b + a \times c$
The vector product is associative with respect to multiplication by a scalar
- $(k a) \times b = a \times (k b) = k (a \times b)$
The vector product between a vector and itself is equal to the zero vector
- $a \times a = 0$
  - as $θ = 0$ in $|a| |b| \sin θ \hat{n}$
The vector product of two parallel vectors is equal to the zero vector
- as $θ = 0$ in $|a| |b| \sin θ \hat{n}$
The converse is also true
- If $a \times b = 0$ for non-zero vectors
  - then $a$ and $b$ must be parallel
The absolute value of the vector product of two perpendicular vectors is equal to the product of their magnitudes
- $| a \times b | = | a | | b |$
  - as $θ = 90 °$ and $| \hat{n} | = 1$ in $| |a| |b| \sin θ \hat{n} |$

Worked Example

Given that $a = 2 i - 5 k$ and $b = 3 i - 2 j - k$ , calculate

(a) $a \times b$

(b) $| a \times b |$

Answer:

(a)

Method 1

Substitute $a = (\begin{matrix} 2 \\ 0 \\ - 5 \end{matrix})$ and $b = (\begin{matrix} 3 \\ - 2 \\ - 1 \end{matrix})$ into $a \times b = (\begin{matrix} a_{2} b_{3} - a_{3} b_{2} \\ a_{3} b_{1} - a_{1} b_{3} \\ a_{1} b_{2} - a_{2} b_{1} \end{matrix})$

$(\begin{matrix} 2 \\ 0 \\ - 5 \end{matrix}) \times (\begin{matrix} 3 \\ - 2 \\ - 1 \end{matrix}) = (\begin{matrix} 0 \times (- 1) - (- 5) \times (- 2) \\ (- 5) \times 3 - 2 \times (- 1) \\ 2 \times (- 2) - 0 \times 3 \end{matrix})$

Simplify

$(\begin{matrix} - 10 \\ - 13 \\ - 4 \end{matrix})$

Write the answer in $i$ , $j$ and $k$ notation

$- 10 i - 13 j - 4 k$

Method 2

Substitute $a = 2 i - 5 k$ and $b = 3 i - 2 j - k$ into $a \times b = |\begin{matrix} i & j & k \\ a_{1} & a_{2} & a_{3} \\ b_{1} & b_{2} & b_{3} \end{matrix}|$

$(\begin{matrix} 2 \\ 0 \\ - 5 \end{matrix}) \times (\begin{matrix} 3 \\ - 2 \\ - 1 \end{matrix}) = | \begin{matrix} i & j & k \\ 2 & 0 & - 5 \\ 3 & - 2 & - 1 \end{matrix} |$

Work out the determinant of the 3x3 matrix

$i (0 \times (- 1) - (- 2) \times (- 5)) - j (2 \times (- 1) - 3 \times (- 5)) + k (2 \times (- 2) - 3 \times 0)$

Simplify the coefficients

$- 10 i - 13 j - 4 k$

(b)

Find the magnitude of the answer in part (a)

$\sqrt{{(- 10)}^{2} + {(- 13)}^{2} + {(- 4)}^{2}}$

$| a \times b | = \sqrt{285}$

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