The Vector Product Revision Notes for AQA A-Level Further Maths

The Vector Product

What is the vector product?

The vector product (also called the cross product) is a way of multiplying two vectors together that produces a third vector. Unlike the scalar product which gives a number, the vector product gives a new vector that is perpendicular to both of the original vectors.

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Key Difference from Scalar Product:

While the scalar product gives a scalar (number) as a result, the vector product produces a completely new vector that is perpendicular to both input vectors. This makes it particularly useful for finding normal vectors and calculating areas.

Definition

The vector product $\mathbf{a} \times \mathbf{b}$ of two vectors $\mathbf{a}$ and $\mathbf{b}$ is defined as:

$\mathbf{a} \times \mathbf{b} = |\mathbf{a}||\mathbf{b}|\sin\theta\,\hat{\mathbf{n}}$

where:

$\theta$ is the angle between vectors $\mathbf{a}$ and $\mathbf{b}$
$\hat{\mathbf{n}}$ is a unit vector that is perpendicular to both $\mathbf{a}$ and $\mathbf{b}$

The direction of $\hat{\mathbf{n}}$ is determined by the right-hand rule: point your index finger along $\mathbf{a}$ , your middle finger along $\mathbf{b}$ , and your thumb will point in the direction of $\mathbf{a} \times \mathbf{b}$ .

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Understanding the Magnitude:

The magnitude of the vector product $|\mathbf{a} \times \mathbf{b}|$ represents the area of the parallelogram formed by the two vectors. This geometric interpretation is crucial for many applications, particularly in calculating areas.

Properties of the vector product

Anticommutative property

The vector product is not commutative. This means the order matters:

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

Swapping the order of the vectors changes the sign of the result. The magnitude stays the same, but the direction reverses.

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Remember: Swap the Order, Flip the Sign

Unlike ordinary multiplication, switching the order of vectors in a cross product changes the sign of the result. Always pay attention to which vector comes first!

Distributive property

The vector product distributes over addition:

$\mathbf{a} \times (\mathbf{b} + \mathbf{c}) = \mathbf{a} \times \mathbf{b} + \mathbf{a} \times \mathbf{c}$

This property is useful when expanding brackets in vector product expressions.

Vector products of unit vectors

The standard unit vectors $\mathbf{i}$ , $\mathbf{j}$ , and $\mathbf{k}$ have special cross product relationships:

$\mathbf{i} \times \mathbf{j} = \mathbf{k}$ $\mathbf{j} \times \mathbf{k} = \mathbf{i}$ $\mathbf{k} \times \mathbf{i} = \mathbf{j}$

These follow a cyclic pattern. If you reverse the order, the sign changes:

$\mathbf{j} \times \mathbf{i} = -\mathbf{k}$ $\mathbf{k} \times \mathbf{j} = -\mathbf{i}$ $\mathbf{i} \times \mathbf{k} = -\mathbf{j}$

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Memory Aid: The Cyclic Pattern

Think of the unit vectors in a cycle: $\mathbf{i} \to \mathbf{j} \to \mathbf{k} \to \mathbf{i}$

Going forward in the cycle gives a positive result: $\mathbf{i} \times \mathbf{j} = \mathbf{k}$

Going backward gives a negative result: $\mathbf{j} \times \mathbf{i} = -\mathbf{k}$

Also, any vector crossed with itself gives the zero vector:

$\mathbf{i} \times \mathbf{i} = \mathbf{j} \times \mathbf{j} = \mathbf{k} \times \mathbf{k} = \mathbf{0}$

Parallel vectors

When two vectors are parallel, their cross product is the zero vector:

$\mathbf{a} \times \mathbf{b} = \mathbf{0} \text{ if and only if } \mathbf{a} \text{ and } \mathbf{b} \text{ are parallel}$

This follows from the definition because parallel vectors have $\theta = 0°$ or $\theta = 180°$ , and $\sin 0° = \sin 180° = 0$ .

An important consequence is that any vector crossed with itself equals zero:

$\mathbf{a} \times \mathbf{a} = \mathbf{0}$

Calculating the vector product

Using component form

When vectors are expressed in component form as $\mathbf{a} = a_1\mathbf{i} + a_2\mathbf{j} + a_3\mathbf{k}$ and $\mathbf{b} = b_1\mathbf{i} + b_2\mathbf{j} + b_3\mathbf{k}$ , the vector product is:

$\mathbf{a} \times \mathbf{b} = (a_2b_3 - a_3b_2)\mathbf{i} - (a_1b_3 - a_3b_1)\mathbf{j} + (a_1b_2 - a_2b_1)\mathbf{k}$

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Watch Out for the Negative Sign!

Notice the middle term has a negative sign. This is a common source of errors. The $\mathbf{j}$ component always has a minus sign in front when using the component formula.

This formula can be difficult to remember, so the determinant method is often preferred.

Using the determinant method

A more systematic way to calculate the vector product uses a $3 \times 3$ 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 first row:

$\mathbf{a} \times \mathbf{b} = \mathbf{i}\begin{vmatrix} a_2 & a_3 \\ b_2 & b_3 \end{vmatrix} - \mathbf{j}\begin{vmatrix} a_1 & a_3 \\ b_1 & b_3 \end{vmatrix} + \mathbf{k}\begin{vmatrix} a_1 & a_2 \\ b_1 & b_2 \end{vmatrix}$

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Why Use the Determinant Method?

The determinant method is more systematic and less prone to sign errors than trying to remember the component formula. By always expanding along the first row, you have a consistent procedure to follow.

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Critical: The $\mathbf{j}$ Component Always Has a Minus Sign

When expanding the determinant, be very careful with negative signs. The $\mathbf{j}$ component always has a minus sign in front, even though you're expanding in a pattern of $+, -, +$ along the first row. This is the most common mistake in vector product calculations!

Finding angles between vectors

The formula for the vector product can be rearranged to find the angle between two vectors:

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

This gives an alternative to using the scalar product formula with cosine. Calculate the cross product, find its magnitude, then divide by the product of the magnitudes of the original vectors.

Applications of the vector product

Finding perpendicular vectors

The vector product always produces a vector perpendicular to both original vectors. This is useful when you need to find a normal vector to a plane or a direction perpendicular to two given directions.

To find a vector perpendicular to both $\mathbf{a}$ and $\mathbf{b}$ , simply calculate $\mathbf{a} \times \mathbf{b}$ .

Calculating areas of triangles

The magnitude of the vector product equals the area of the parallelogram formed by the two vectors. Therefore, the area of a triangle is half this value.

For a triangle with vertex at the origin and two sides represented by vectors $\mathbf{a}$ and $\mathbf{b}$ :

$\text{Area of triangle} = \frac{1}{2}|\mathbf{a} \times \mathbf{b}|$

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Why the Factor of $\frac{1}{2}$ ?

The magnitude $|\mathbf{a} \times \mathbf{b}|$ gives the area of the entire parallelogram formed by the two vectors. Since a triangle is exactly half a parallelogram, we multiply by $\frac{1}{2}$ to get the triangle's area.

For a triangle $ABC$ with vertices at specific points, first find the vectors representing two sides:

$\text{Area of triangle } ABC = \frac{1}{2}|\overrightarrow{AB} \times \overrightarrow{AC}|$

Calculate $\overrightarrow{AB}$ and $\overrightarrow{AC}$ by subtracting position vectors, then use the vector product formula.

Equation of a line in vector product form

A straight line passing through a point with position vector $\mathbf{a}$ and parallel to vector $\mathbf{b}$ can be expressed as:

$(\mathbf{r} - \mathbf{a}) \times \mathbf{b} = \mathbf{0}$

This works because any point $\mathbf{r}$ on the line creates a vector $(\mathbf{r} - \mathbf{a})$ that is parallel to $\mathbf{b}$ , and parallel vectors have a zero cross product.

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Understanding the Line Equation Form:

Expanding the brackets gives: $\mathbf{r} \times \mathbf{b} - \mathbf{a} \times \mathbf{b} = \mathbf{0}$

Rearranging: $\mathbf{r} \times \mathbf{b} = \mathbf{a} \times \mathbf{b}$

This form is useful because the right-hand side is a constant vector that can be calculated once, and the left-hand side varies with $\mathbf{r}$ .

Worked examples

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Worked Example 1: Proving Unit Vector Cross Products

Show that $\mathbf{i} \times \mathbf{j} = \mathbf{k}$ and $\mathbf{j} \times \mathbf{i} = -\mathbf{k}$

Solution:

Since $\mathbf{i}$ and $\mathbf{j}$ are unit vectors perpendicular to each other:

$|\mathbf{i}| = |\mathbf{j}| = 1$
The angle between them is $90°$

Using the formula: $\mathbf{i} \times \mathbf{j} = |\mathbf{i}||\mathbf{j}|\sin 90°\,\hat{\mathbf{n}}$ $= 1 \times 1 \times 1 \times \hat{\mathbf{n}}$ $= \hat{\mathbf{n}}$

The unit vector perpendicular to both $\mathbf{i}$ and $\mathbf{j}$ (using the right-hand rule) is $\mathbf{k}$ .

Therefore: $\mathbf{i} \times \mathbf{j} = \mathbf{k}$ ✓

For $\mathbf{j} \times \mathbf{i}$ , the angle from $\mathbf{j}$ to $\mathbf{i}$ is $-90°$ (or equivalently $+270°$ ): $\mathbf{j} \times \mathbf{i} = 1 \times 1 \times \sin(-90°) \times \mathbf{k}$ $= -\mathbf{k}$ ✓

This demonstrates the anticommutative property.

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Worked Example 2: Finding a Perpendicular Vector and Angle

Given $\mathbf{a} = \begin{pmatrix} 1 \\ -3 \\ -2 \end{pmatrix}$ and $\mathbf{b} = \begin{pmatrix} -2 \\ 0 \\ 4 \end{pmatrix}$

a) Find a vector perpendicular to both $\mathbf{a}$ and $\mathbf{b}$

b) Calculate the acute angle between $\mathbf{a}$ and $\mathbf{b}$

Solution:

Part a:

Calculate $\mathbf{a} \times \mathbf{b}$ :

$\mathbf{a} \times \mathbf{b} = \begin{pmatrix} 1 \\ -3 \\ -2 \end{pmatrix} \times \begin{pmatrix} -2 \\ 0 \\ 4 \end{pmatrix}$

$= (-3 \times 4 - (-2) \times 0)\mathbf{i} - (1 \times 4 - (-2) \times (-2))\mathbf{j} + (1 \times 0 - (-3) \times (-2))\mathbf{k}$

$= -12\mathbf{i} - 0\mathbf{j} + (-6)\mathbf{k}$

$= -12\mathbf{i} - 6\mathbf{k}$

Or in column form: $\begin{pmatrix} -12 \\ 0 \\ -6 \end{pmatrix}$ (or any scalar multiple, such as $\begin{pmatrix} 12 \\ 0 \\ 6 \end{pmatrix}$ or $\begin{pmatrix} 2 \\ 0 \\ 1 \end{pmatrix}$ ) ✓

Part b:

First, find the magnitudes:

$|\mathbf{a}| = \sqrt{1^2 + (-3)^2 + (-2)^2} = \sqrt{1 + 9 + 4} = \sqrt{14}$

$|\mathbf{b}| = \sqrt{(-2)^2 + 0^2 + 4^2} = \sqrt{4 + 16} = \sqrt{20} = 2\sqrt{5}$

$|\mathbf{a} \times \mathbf{b}| = \sqrt{(-12)^2 + 0^2 + (-6)^2} = \sqrt{144 + 36} = \sqrt{180} = 6\sqrt{5}$

Now use the rearranged formula:

$\sin\theta = \frac{|\mathbf{a} \times \mathbf{b}|}{|\mathbf{a}||\mathbf{b}|} = \frac{6\sqrt{5}}{\sqrt{14} \times 2\sqrt{5}} = \frac{6\sqrt{5}}{2\sqrt{70}} = \frac{3}{\sqrt{14}}$

$\theta = \sin^{-1}\left(\frac{3}{\sqrt{14}}\right) = 53.3°$ (to 1 d.p.) ✓

Note: Always check you're calculating the correct angle (acute or obtuse) as required by the question.

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Worked Example 3: Using the Determinant Method

Find the vector product of $2\mathbf{i} + \mathbf{j} - \mathbf{k}$ and $\mathbf{i} - 3\mathbf{j} - 2\mathbf{k}$

Solution:

Set up the determinant:

$\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & 1 & -1 \\ 1 & -3 & -2 \end{vmatrix}$

Expand along the first row:

$= \mathbf{i}\begin{vmatrix} 1 & -1 \\ -3 & -2 \end{vmatrix} - \mathbf{j}\begin{vmatrix} 2 & -1 \\ 1 & -2 \end{vmatrix} + \mathbf{k}\begin{vmatrix} 2 & 1 \\ 1 & -3 \end{vmatrix}$

$= \mathbf{i}[(1)(-2) - (-1)(-3)] - \mathbf{j}[(2)(-2) - (-1)(1)] + \mathbf{k}[(2)(-3) - (1)(1)]$

$= \mathbf{i}[-2 - 3] - \mathbf{j}[-4 + 1] + \mathbf{k}[-6 - 1]$

$= -5\mathbf{i} - (-3)\mathbf{j} + (-7)\mathbf{k}$

$= -5\mathbf{i} + 3\mathbf{j} - 7\mathbf{k}$ ✓

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Worked Example 4: Calculating the Area of a Triangle

Calculate the area of triangle $ABC$ with vertices at $A(2, 5, -1)$ , $B(-4, 9, -6)$ and $C(3, -4, 8)$

Solution:

Step 1: Find vectors $\overrightarrow{AB}$ and $\overrightarrow{AC}$ by subtracting position vectors:

$\overrightarrow{AB} = \begin{pmatrix} -4 \\ 9 \\ -6 \end{pmatrix} - \begin{pmatrix} 2 \\ 5 \\ -1 \end{pmatrix} = \begin{pmatrix} -6 \\ 4 \\ -5 \end{pmatrix}$

$\overrightarrow{AC} = \begin{pmatrix} 3 \\ -4 \\ 8 \end{pmatrix} - \begin{pmatrix} 2 \\ 5 \\ -1 \end{pmatrix} = \begin{pmatrix} 1 \\ -9 \\ 9 \end{pmatrix}$

Step 2: Calculate the cross product:

$\overrightarrow{AB} \times \overrightarrow{AC} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -6 & 4 & -5 \\ 1 & -9 & 9 \end{vmatrix}$

$= \mathbf{i}[(4)(9) - (-5)(-9)] - \mathbf{j}[(-6)(9) - (-5)(1)] + \mathbf{k}[(-6)(-9) - (4)(1)]$

$= \mathbf{i}[36 - 45] - \mathbf{j}[-54 + 5] + \mathbf{k}[54 - 4]$

$= -9\mathbf{i} + 49\mathbf{j} + 50\mathbf{k}$

Step 3: Find the magnitude:

$|\overrightarrow{AB} \times \overrightarrow{AC}| = \sqrt{(-9)^2 + 49^2 + 50^2} = \sqrt{81 + 2401 + 2500} = \sqrt{4982}$

Step 4: Calculate the area:

$\text{Area} = \frac{1}{2}|\overrightarrow{AB} \times \overrightarrow{AC}| = \frac{1}{2}\sqrt{4982} = 35.3 \text{ square units}$ (to 3 s.f.) ✓

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Worked Example 5: Finding the Equation of a Line

Find the equation of the line through points $(2, 0, -1)$ and $(-3, 1, 4)$ in the form $\mathbf{r} \times \mathbf{b} = \mathbf{c}$

Solution:

Step 1: Find a vector parallel to the line.

Subtract the position vectors of the two points:

$\mathbf{b} = \begin{pmatrix} -3 \\ 1 \\ 4 \end{pmatrix} - \begin{pmatrix} 2 \\ 0 \\ -1 \end{pmatrix} = \begin{pmatrix} -3 - 2 \\ 1 - 0 \\ 4 - (-1) \end{pmatrix} = \begin{pmatrix} -5 \\ 1 \\ 5 \end{pmatrix}$

Step 2: Use the formula $(\mathbf{r} - \mathbf{a}) \times \mathbf{b} = \mathbf{0}$ where $\mathbf{a}$ is the position vector of a point on the line.

Using point $(2, 0, -1)$ :

$\left(\mathbf{r} - \begin{pmatrix} 2 \\ 0 \\ -1 \end{pmatrix}\right) \times \begin{pmatrix} -5 \\ 1 \\ 5 \end{pmatrix} = \mathbf{0}$

Step 3: Expand and rearrange.

$\mathbf{r} \times \begin{pmatrix} -5 \\ 1 \\ 5 \end{pmatrix} - \begin{pmatrix} 2 \\ 0 \\ -1 \end{pmatrix} \times \begin{pmatrix} -5 \\ 1 \\ 5 \end{pmatrix} = \mathbf{0}$

Calculate the cross product on the right:

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

Therefore, the equation is:

$\mathbf{r} \times \begin{pmatrix} -5 \\ 1 \\ 5 \end{pmatrix} = \begin{pmatrix} 1 \\ -5 \\ 2 \end{pmatrix}$ ✓

Or equivalently: $\mathbf{r} \times (-5\mathbf{i} + \mathbf{j} + 5\mathbf{k}) = \mathbf{i} - 5\mathbf{j} + 2\mathbf{k}$

Note: When finding line equations in vector product form, always check your arithmetic carefully, especially when calculating the cross product. The negative signs are easy to get wrong.

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Key Points to Remember:

The vector product $\mathbf{a} \times \mathbf{b} = |\mathbf{a}||\mathbf{b}|\sin\theta\,\hat{\mathbf{n}}$ produces a vector perpendicular to both $\mathbf{a}$ and $\mathbf{b}$
Order matters: $\mathbf{a} \times \mathbf{b} = -\mathbf{b} \times \mathbf{a}$ (anticommutative property) - swap the order, flip the sign!
Parallel vectors have zero cross product: $\mathbf{a} \times \mathbf{b} = \mathbf{0}$ if and only if $\mathbf{a}$ and $\mathbf{b}$ are parallel
Use the determinant method for systematic calculation: $\mathbf{a} \times \mathbf{b} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \end{vmatrix}$
Remember: the $\mathbf{j}$ component always has a minus sign when expanding the determinant
Area of a triangle = $\frac{1}{2}|\mathbf{a} \times \mathbf{b}|$ where $\mathbf{a}$ and $\mathbf{b}$ are vectors representing two sides
For triangle with vertices: Area of $ABC = \frac{1}{2}|\overrightarrow{AB} \times \overrightarrow{AC}|$
Line equation in vector product form: $(\mathbf{r} - \mathbf{a}) \times \mathbf{b} = \mathbf{0}$ or $\mathbf{r} \times \mathbf{b} = \mathbf{a} \times \mathbf{b}$

The Vector Product (AQA A-Level Further Maths): Revision Notes