The vector (cross) product
Master the vector (cross) product for Year 12 Specialist Mathematics in Queensland (QCAA). The cross product \(\mathbf{a}\times\mathbf{b}\) returns a vector perpendicular to both inputs, found from a \(3\times3\) determinant, and its length is \(|\mathbf{a}||\mathbf{b}|\sin\theta\).
You will learn to compute the cross product, apply the right-hand rule, find normal and unit-normal vectors, and use \(|\mathbf{a}\times\mathbf{b}|\) for the areas of parallelograms and triangles — the groundwork for equations of planes later in Unit 3.
Theory
The vector (cross) product \(\mathbf{a}\times\mathbf{b}\) is a central tool of Year 12 Specialist Mathematics (QCAA, Queensland): unlike the dot product it returns a vector, one that is perpendicular to both \(\mathbf{a}\) and \(\mathbf{b}\). This page covers the determinant formula, the length rule \(|\mathbf{a}\times\mathbf{b}|=|\mathbf{a}||\mathbf{b}|\sin\theta\), the right-hand rule, and using the cross product to find normals and the areas of parallelograms and triangles.
The cross (vector) product of two three-dimensional vectors \(\mathbf{a}\) and \(\mathbf{b}\) is written \(\mathbf{a}\times\mathbf{b}\). Its defining feature is that the result is a vector perpendicular to both \(\mathbf{a}\) and \(\mathbf{b}\) — that is, normal to the plane they span. This is what makes it the natural tool for finding a normal vector.
In components, with \(\mathbf{a}=a_1\mathbf{i}+a_2\mathbf{j}+a_3\mathbf{k}\) and \(\mathbf{b}=b_1\mathbf{i}+b_2\mathbf{j}+b_3\mathbf{k}\), the cross product is evaluated as a \(3\times3\) determinant with \(\mathbf{i},\mathbf{j},\mathbf{k}\) across the top row. Expanding gives \(\mathbf{a}\times\mathbf{b}=(a_2b_3-a_3b_2)\mathbf{i}-(a_1b_3-a_3b_1)\mathbf{j}+(a_1b_2-a_2b_1)\mathbf{k}\).
The magnitude carries geometric meaning: \(|\mathbf{a}\times\mathbf{b}|=|\mathbf{a}||\mathbf{b}|\sin\theta\), where \(\theta\) is the angle between the vectors. This equals the area of the parallelogram with sides \(\mathbf{a}\) and \(\mathbf{b}\); half of it is the area of the triangle. The direction of \(\mathbf{a}\times\mathbf{b}\) is set by the right-hand rule.
The cross product is anti-commutative: \(\mathbf{b}\times\mathbf{a}=-(\mathbf{a}\times\mathbf{b})\), so swapping the order reverses the vector. A vector crossed with itself, or with any parallel vector, gives the zero vector: \(\mathbf{a}\times\mathbf{a}=\mathbf{0}\), since \(\sin 0^\circ=0\). Dividing \(\mathbf{a}\times\mathbf{b}\) by its magnitude gives a unit normal to both vectors.
For \(\mathbf{a}=a_1\mathbf{i}+a_2\mathbf{j}+a_3\mathbf{k}\) and \(\mathbf{b}=b_1\mathbf{i}+b_2\mathbf{j}+b_3\mathbf{k}\), evaluate the cross product as a determinant along the top row:
The magnitude gives the angle rule and the two area formulas:
A unit vector normal to both \(\mathbf{a}\) and \(\mathbf{b}\) is the cross product divided by its length:
Computing and using \(\mathbf{a}\times\mathbf{b}\)
- Set up the \(3\times3\) determinant with \(\mathbf{i},\mathbf{j},\mathbf{k}\) on the top row, the components of \(\mathbf{a}\) on the second row and \(\mathbf{b}\) on the third.
- Expand along the top row, remembering the minus sign on the \(\mathbf{j}\) term: \(\mathbf{i}(a_2b_3-a_3b_2)-\mathbf{j}(a_1b_3-a_3b_1)+\mathbf{k}(a_1b_2-a_2b_1)\).
- Interpret the result: it is a vector perpendicular to both. Take its magnitude for a parallelogram area (or half for a triangle), or divide by the magnitude for a unit normal.
- Check if useful: the dot product of \(\mathbf{a}\times\mathbf{b}\) with either \(\mathbf{a}\) or \(\mathbf{b}\) should be \(0\).
Set up the determinant and expand along the top row (note the minus on \(\mathbf{j}\)):
| \(\mathbf{a}\times\mathbf{b}\) | \(=\) | \(\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 3 & 1 & -2 \\ 1 & 2 & 1 \end{vmatrix}\) |
| \(=\) | \(\mathbf{i}(1\times1-(-2)\times2)-\mathbf{j}(3\times1-(-2)\times1)+\mathbf{k}(3\times2-1\times1)\) | |
| \(=\) | \(\mathbf{i}(1+4)-\mathbf{j}(3+2)+\mathbf{k}(6-1)\) | |
| \(=\) | \(5\mathbf{i}-5\mathbf{j}+5\mathbf{k}\) |
Check perpendicularity with the dot product against \(\mathbf{a}\):
| \((\mathbf{a}\times\mathbf{b})\cdot\mathbf{a}\) | \(=\) | \((5)(3)+(-5)(1)+(5)(-2)\) |
| \(=\) | \(15-5-10\) | |
| \(=\) | \(0\) |
\(\mathbf{a}\times\mathbf{b}=5\mathbf{i}-5\mathbf{j}+5\mathbf{k}\); the dot product is \(0\), so it is perpendicular to \(\mathbf{a}\).
Apply \(|\mathbf{a}\times\mathbf{b}|=|\mathbf{a}||\mathbf{b}|\sin\theta\):
| \(|\mathbf{a}\times\mathbf{b}|\) | \(=\) | \(|\mathbf{a}||\mathbf{b}|\sin\theta\) |
| \(=\) | \(4\times6\times\sin30^\circ\) | |
| \(=\) | \(24\times\dfrac{1}{2}\) | |
| \(=\) | \(12\) |
\(|\mathbf{a}\times\mathbf{b}|=12\).
The area is \(|\mathbf{a}\times\mathbf{b}|\); expand the determinant first:
| \(\mathbf{a}\times\mathbf{b}\) | \(=\) | \(\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 2 & 2 \\ 3 & 0 & 1 \end{vmatrix}\) |
| \(=\) | \(\mathbf{i}(2\times1-2\times0)-\mathbf{j}(1\times1-2\times3)+\mathbf{k}(1\times0-2\times3)\) | |
| \(=\) | \(2\mathbf{i}+5\mathbf{j}-6\mathbf{k}\) |
Now take the magnitude:
| \(\text{area}\) | \(=\) | \(\sqrt{2^2+5^2+(-6)^2}\) |
| \(=\) | \(\sqrt{4+25+36}\) | |
| \(=\) | \(\sqrt{65}\) |
Area \(=\sqrt{65}\) square units.
First cross the vectors to get a normal:
| \(\mathbf{a}\times\mathbf{b}\) | \(=\) | \(\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 2 & 2 \\ 2 & 1 & -2 \end{vmatrix}\) |
| \(=\) | \(\mathbf{i}(2\times(-2)-2\times1)-\mathbf{j}(1\times(-2)-2\times2)+\mathbf{k}(1\times1-2\times2)\) | |
| \(=\) | \(-6\mathbf{i}+6\mathbf{j}-3\mathbf{k}\) |
Find its magnitude:
| \(|\mathbf{a}\times\mathbf{b}|\) | \(=\) | \(\sqrt{(-6)^2+6^2+(-3)^2}\) |
| \(=\) | \(\sqrt{36+36+9}\) | |
| \(=\) | \(\sqrt{81}=9\) |
Divide the normal by its magnitude for the unit normal:
| \(\hat{\mathbf{n}}\) | \(=\) | \(\dfrac{1}{9}(-6\mathbf{i}+6\mathbf{j}-3\mathbf{k})\) |
| \(=\) | \(-\dfrac{2}{3}\mathbf{i}+\dfrac{2}{3}\mathbf{j}-\dfrac{1}{3}\mathbf{k}\) |
\(\hat{\mathbf{n}}=-\dfrac{2}{3}\mathbf{i}+\dfrac{2}{3}\mathbf{j}-\dfrac{1}{3}\mathbf{k}\) (its negative is also valid).
Common pitfalls
Frequently asked questions
What is the cross product of two vectors?
The cross product \(\mathbf{a}\times\mathbf{b}\) is a vector that is perpendicular to both \(\mathbf{a}\) and \(\mathbf{b}\). Its length is \(|\mathbf{a}||\mathbf{b}|\sin\theta\) and its direction follows the right-hand rule.
How do you calculate a cross product?
Write a \(3\times3\) determinant with \(\mathbf{i},\mathbf{j},\mathbf{k}\) on the top row and the two vectors' components below, then expand along the top row (subtracting the \(\mathbf{j}\) term).
What is the difference between the dot product and the cross product?
The dot product gives a scalar and uses \(\cos\theta\); the cross product gives a vector perpendicular to both and uses \(\sin\theta\). Use the cross product when you need a normal or an area.
How does the cross product give the area of a triangle?
The magnitude \(|\mathbf{a}\times\mathbf{b}|\) is the area of the parallelogram on \(\mathbf{a}\) and \(\mathbf{b}\); a triangle with those two sides has half of it, \(\tfrac12|\mathbf{a}\times\mathbf{b}|\).
Why is \(\mathbf{a}\times\mathbf{a}=\mathbf{0}\)?
The angle between a vector and itself is \(0^\circ\), and \(\sin 0^\circ=0\), so the magnitude is zero. The same holds for any two parallel vectors.
How do you find a unit vector perpendicular to two vectors?
Compute \(\mathbf{a}\times\mathbf{b}\) for a perpendicular vector, then divide it by its magnitude: \(\hat{\mathbf{n}}=\dfrac{\mathbf{a}\times\mathbf{b}}{|\mathbf{a}\times\mathbf{b}|}\).