Vector product

This note covers the vector product (cross product) of two vectors in three dimensions. It gives a new vector perpendicular to both originals, and its magnitude gives areas of parallelograms and triangles. It is used for finding perpendicular directions and areas in 3D geometry.

Key ideas

The vector product \(\mathbf{u}\times\mathbf{v}\) is defined only for three-dimensional vectors, and the result is a vector. For \(\mathbf{u}=\begin{pmatrix}u_1\\u_2\\u_3\end{pmatrix}\) and \(\mathbf{v}=\begin{pmatrix}v_1\\v_2\\v_3\end{pmatrix}\), \[\mathbf{u}\times\mathbf{v}=\begin{pmatrix}u_2v_3-u_3v_2\\u_3v_1-u_1v_3\\u_1v_2-u_2v_1\end{pmatrix}.\] A reliable way to remember this is the "cover and cross" pattern. For the first component, ignore the first row and take the cross-product of the other two rows: \(u_2v_3-u_3v_2\). For the second component, ignore the second row and use the pattern \(u_3v_1-u_1v_3\). Note the order: it wraps round from 3 back to 1. For the third component, ignore the third row: \(u_1v_2-u_2v_1\).

Direction. The vector \(\mathbf{u}\times\mathbf{v}\) is perpendicular to both \(\mathbf{u}\) and \(\mathbf{v}\). You can check any answer by taking its scalar product with each original vector, which must be zero. Its direction is given by the right-hand rule: if the fingers of your right hand curl from \(\mathbf{u}\) towards \(\mathbf{v}\), the thumb points along \(\mathbf{u}\times\mathbf{v}\).

Magnitude. If \(\theta\) is the angle between the vectors, then \[|\mathbf{u}\times\mathbf{v}|=|\mathbf{u}||\mathbf{v}|\sin\theta.\] This is the area of the parallelogram with \(\mathbf{u}\) and \(\mathbf{v}\) as adjacent sides. A triangle with those two sides has half that area.

Properties.

• Not commutative: \(\mathbf{v}\times\mathbf{u}=-(\mathbf{u}\times\mathbf{v})\). Swapping the order reverses the direction.
• \(\mathbf{u}\times\mathbf{u}=\mathbf{0}\), and in general \(\mathbf{u}\times\mathbf{v}=\mathbf{0}\) when the vectors are parallel, because \(\sin0^\circ=\sin180^\circ=0\).
• Distributive: \(\mathbf{u}\times(\mathbf{v}+\mathbf{w})=\mathbf{u}\times\mathbf{v}+\mathbf{u}\times\mathbf{w}\).
• Scalar multiples: \((k\mathbf{u})\times\mathbf{v}=k(\mathbf{u}\times\mathbf{v})\).

Using the vector product.

• A vector perpendicular to two given vectors: take their vector product. For a unit vector, divide by its magnitude.
• Area of a parallelogram: \(|\mathbf{u}\times\mathbf{v}|\), where \(\mathbf{u}\) and \(\mathbf{v}\) are two adjacent sides.
• Area of a triangle \(ABC\): \(\tfrac12|\overrightarrow{AB}\times\overrightarrow{AC}|\).

Choosing between the two products. The scalar product gives a number and uses \(\cos\theta\). The vector product gives a vector and uses \(\sin\theta\). To find an angle, the scalar product is usually safer, because cosine distinguishes acute from obtuse angles, but sine gives the same value for \(\theta\) and \(180^\circ-\theta\). If you use the vector product's magnitude to find an angle, you must decide between acute and obtuse by other means, such as the sign of the scalar product.

Calculator use. Some GDCs have a built-in cross-product function for vectors. Check that your answer is perpendicular to both vectors, even if you use the calculator, as an input slip is easy to make.

Why the area result holds. A parallelogram with sides \(\mathbf{u}\) and \(\mathbf{v}\) has base \(|\mathbf{u}|\) and perpendicular height \(|\mathbf{v}|\sin\theta\). Its area is base times height, which is \(|\mathbf{u}||\mathbf{v}|\sin\theta=|\mathbf{u}\times\mathbf{v}|\). Cutting the parallelogram along a diagonal gives two congruent triangles, which explains the factor \(\tfrac12\) for a triangle.

Perpendicular to a plane. If three points \(A\), \(B\), \(C\) are given, the vector \(\overrightarrow{AB}\times\overrightarrow{AC}\) is perpendicular to the flat surface containing all three points. This is how you find a direction at right angles to a flat surface in three dimensions, such as a roof panel or a mirror.

Standard unit vectors. The vector products of the axis vectors follow a cycle: \(\mathbf{i}\times\mathbf{j}=\mathbf{k}\), \(\mathbf{j}\times\mathbf{k}=\mathbf{i}\) and \(\mathbf{k}\times\mathbf{i}=\mathbf{j}\). Going round the cycle in the other direction gives the negative, so \(\mathbf{j}\times\mathbf{i}=-\mathbf{k}\).

Definitions and formulae

Worked examples

Common errors

• Getting the sign wrong in the middle component. It is \(u_3v_1-u_1v_3\), not \(u_1v_3-u_3v_1\).
• Swapping the order of the vectors, which reverses the sign of the whole answer.
• Forgetting the factor \(\tfrac12\) when finding the area of a triangle.
• Using \(\sin\theta\) to find an angle and not checking whether it should be acute or obtuse.
• Writing the vector product of two-dimensional vectors. It is defined only in three dimensions.

Exam tip: After finding a vector product, take its scalar product with one of the original vectors. A result of zero confirms your arithmetic in seconds.

Check yourself

1. Find \(\begin{pmatrix}1\\0\\0\end{pmatrix}\times\begin{pmatrix}0\\1\\0\end{pmatrix}\).
   Answer

   \(\begin{pmatrix}0\cdot0-0\cdot1\\0\cdot0-1\cdot0\\1\cdot1-0\cdot0\end{pmatrix}=\begin{pmatrix}0\\0\\1\end{pmatrix}\), that is \(\mathbf{i}\times\mathbf{j}=\mathbf{k}\).
2. Find \(\begin{pmatrix}2\\4\\-6\end{pmatrix}\times\begin{pmatrix}-1\\-2\\3\end{pmatrix}\) and explain the result.
   Answer

   The result is \(\begin{pmatrix}12-12\\6-6\\-4+4\end{pmatrix}=\begin{pmatrix}0\\0\\0\end{pmatrix}\). The vectors are parallel, since the first is \(-2\) times the second.
3. Find the area of the parallelogram with adjacent sides \(\begin{pmatrix}3\\0\\0\end{pmatrix}\) and \(\begin{pmatrix}0\\4\\0\end{pmatrix}\).
   Answer

   The cross product is \(\begin{pmatrix}0\\0\\12\end{pmatrix}\), with magnitude 12. The area is 12.
4. If \(|\mathbf{a}|=5\), \(|\mathbf{b}|=6\) and the angle between them is \(30^\circ\), find \(|\mathbf{a}\times\mathbf{b}|\).
   Answer

   \(5\times6\times\sin30^\circ=30\times0.5=15\).