Vector Product (Cross Product) (AA HL)
The vector (cross) product takes two vectors and returns a third vector perpendicular to both - and its size tells you the area of the parallelogram they span. It's one of the more mark-heavy HL-only additions, showing up in area, volume, and plane-equation questions alike. This page covers the determinant method, worked examples, and the mistakes that cost marks most often. It's part of the broader Vectors, Lines & Planes topic.
23 questions on this sub-topic.
The key formulas
Covered under IB syllabus references AHL3.16 / AHL3.17: the vector (cross) product of two vectors, and its use to find the area of a parallelogram or triangle.
Vector (cross) product
\(\mathbf a\times\mathbf b=|\mathbf a||\mathbf b|\sin\theta\,\hat{\mathbf n}\)
Given in the formula booklet. In component form, expand it as a \(3\times3\) determinant with \(\mathbf i,\mathbf j,\mathbf k\) in the top row.
Cross product and area
The magnitude \(|\mathbf a\times\mathbf b|\) is the area of the parallelogram spanned by \(\mathbf a\) and \(\mathbf b\); halve it for the area of the triangle they form.
Need the scalar (dot) product too, or the full syllabus wording? See Vectors, Lines & Planes.
Worked examples
Given \(\mathbf{a} = \begin{pmatrix}2\\0\\1\end{pmatrix}\) and \(\mathbf{b} = \begin{pmatrix}0\\3\\1\end{pmatrix}\), find \(\mathbf{a} \times \mathbf{b}\).
Worked solution
\(\mathbf{a} \times \mathbf{b} = \begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\2&0&1\\0&3&1\end{vmatrix}\) M1
\(= \mathbf{i}(0 \cdot 1 - 1 \cdot 3) - \mathbf{j}(2 \cdot 1 - 1 \cdot 0) + \mathbf{k}(2 \cdot 3 - 0)\) \(= \begin{pmatrix}-3\\-2\\6\end{pmatrix}\) A1
Check: \(\mathbf{a} \cdot (\mathbf{a}\times\mathbf{b}) = -6 + 0 + 6 = 0\) ✓ A1
Vectors \(\mathbf{a} = \begin{pmatrix}1\\2\\3\end{pmatrix}\) and \(\mathbf{b} = \begin{pmatrix}4\\5\\6\end{pmatrix}\).
(a) Find \(\mathbf{a} \times \mathbf{b}\).
(b) Find the area of the parallelogram with sides \(\mathbf{a}\) and \(\mathbf{b}\).
Worked solution
(a) \(\mathbf{a} \times \mathbf{b} = \begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\1&2&3\\4&5&6\end{vmatrix}\) M1
\(= \mathbf{i}(12-15) - \mathbf{j}(6-12) + \mathbf{k}(5-8)\) M1
\(= \begin{pmatrix}-3\\6\\-3\end{pmatrix}\) A1
(b) Area \(= |\mathbf{a}\times\mathbf{b}| = \sqrt{9+36+9} = \sqrt{54}\) M1
\( = 3\sqrt{6}\) A1
A plane passes through \(A(2,0,1)\), \(B(1,3,0)\) and \(C(0,1,4)\).
(a) Find a normal vector to the plane.
(b) Hence find the Cartesian equation of the plane.
(c) Verify that \(B\) lies on the plane.
Worked solution
(a) \(\overrightarrow{AB} = \begin{pmatrix}-1\\3\\-1\end{pmatrix}\), \(\overrightarrow{AC} = \begin{pmatrix}-2\\1\\3\end{pmatrix}\) M1
\(\mathbf{n} = \overrightarrow{AB}\times\overrightarrow{AC} = \begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\-1&3&-1\\-2&1&3\end{vmatrix}\) M1
\(= \begin{pmatrix}9+1\\2-(-3)\\-1+6\end{pmatrix} = \begin{pmatrix}10\\5\\5\end{pmatrix}\), normal \(\begin{pmatrix}2\\1\\1\end{pmatrix}\) A1
(b) Using \(A(2,0,1)\): \(2(x-2)+1(y-0)+1(z-1)=0\) M1
\(2x + y + z = 5\) A1
(c) \(2(1)+3+0 = 5\) ✓ A1 AG
Common mistakes
- Writing the cross product as a number. The vector product always produces a vector - give the answer as \((x,y,z)\) or in \(\mathbf i,\mathbf j,\mathbf k\) form, never as a single scalar.
- Sign errors in the determinant expansion. The middle (\(\mathbf j\)) component of the expansion is negated - it's the step most students forget under time pressure.
- Forgetting to halve the area for a triangle. \(|\mathbf a\times\mathbf b|\) gives the parallelogram's area directly - a triangle question needs that value halved, and it's easy to submit the parallelogram figure by mistake.
Ready to practise properly?
25 cross-product questions, marked instantly like the real exam.
Quick answers
How do you find the cross product of two vectors?
Expand the \(3\times3\) determinant with \(\mathbf i,\mathbf j,\mathbf k\) in the top row and the two vectors' components in the second and third rows. Remember the middle (\(\mathbf j\)) component is negated.
What does the magnitude of a cross product represent?
\(|\mathbf a\times\mathbf b|\) equals the area of the parallelogram spanned by \(\mathbf a\) and \(\mathbf b\). Halve it to get the area of the triangle formed by the two vectors.