The vector (cross) product
Master the vector (cross) product in Year 12 VCE Specialist Mathematics. 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 \(|\mathbf{a}||\mathbf{b}|\sin\theta\). It sits in the Space and measurement area of study of the VCE Mathematics Study Design (VCAA), within the Vectors topic of Unit 3.
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: 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: \(\hat{\mathbf{n}}=\dfrac{\mathbf{a}\times\mathbf{b}}{|\mathbf{a}\times\mathbf{b}|}\).