The vector (cross) product in three dimensions (WACE Year 12 Mathematics Specialist Unit 3)

What this dot point is asking

SCSA wants you to evaluate the cross product, recognise it as perpendicular to both inputs, and use its magnitude to find parallelogram and triangle areas.

The determinant definition

For $\mathbf{a} = (a_1, a_2, a_3)$ and $\mathbf{b} = (b_1, b_2, b_3)$,

Note the middle component carries a minus sign from the determinant expansion. Unlike the dot product, the cross product is anticommutative: $\mathbf{b}\times\mathbf{a} = -(\mathbf{a}\times\mathbf{b})$.

The scalar triple product and volume

A natural extension of the cross product is the scalar triple product $\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c})$, a single number whose absolute value equals the volume of the parallelepiped with edges $\mathbf{a}$, $\mathbf{b}$ and $\mathbf{c}$. It can be computed as the $3 \times 3$ determinant with the three vectors as rows. When the scalar triple product is zero, the three vectors are coplanar, because the parallelepiped collapses to zero volume; this gives a clean test for whether three vectors (or four points) lie in a common plane. The tetrahedron with those three edges has one sixth of the parallelepiped's volume.

Perpendicularity and direction

The result is perpendicular to both $\mathbf{a}$ and $\mathbf{b}$, which you can confirm because $\mathbf{a}\cdot(\mathbf{a}\times\mathbf{b}) = 0$. Its direction follows the right-hand rule: point the fingers from $\mathbf{a}$ toward $\mathbf{b}$ and the thumb gives $\mathbf{a}\times\mathbf{b}$. This makes the cross product the natural way to find a normal vector to a plane through two direction vectors.

Magnitude as area

If $\mathbf{a}$ and $\mathbf{b}$ are parallel then $\sin\theta = 0$ and the cross product is the zero vector, so a zero cross product is a test for parallel vectors.

The cross product as a plane normal

The single most common SCSA use of the cross product is to find a vector normal to a plane. A plane is determined by two non-parallel direction vectors lying in it; their cross product is perpendicular to both and therefore normal to the whole plane. Given three points $P$, $Q$, $R$ in the plane, form the edge vectors $\vec{PQ}$ and $\vec{PR}$ and take $\vec{PQ} \times \vec{PR}$. This normal is exactly what the cartesian equation of a plane requires, which is why this dot point feeds directly into the vector equations of planes.

Properties and the geometric definitions

The cross product satisfies several rules worth knowing. It is anticommutative, $\mathbf{a} \times \mathbf{b} = -(\mathbf{b} \times \mathbf{a})$, distributes over addition, $\mathbf{a} \times (\mathbf{b} + \mathbf{c}) = \mathbf{a} \times \mathbf{b} + \mathbf{a} \times \mathbf{c}$, and gives $\mathbf{a} \times \mathbf{a} = \mathbf{0}$ for any vector. The standard basis vectors cycle: $\mathbf{i} \times \mathbf{j} = \mathbf{k}$, $\mathbf{j} \times \mathbf{k} = \mathbf{i}$ and $\mathbf{k} \times \mathbf{i} = \mathbf{j}$, with a sign change if the order is reversed. These products give a quick mental check on a computed answer and confirm the right-hand orientation of the result.