Compute the vector (cross) product in three dimensions and use it to find perpendicular vectors and areas

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: $\mathbf{b}\times\mathbf{a} = -(\mathbf{a}\times\mathbf{b})$.

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.