# 14.3: The Vector Product

• • Contributed by Marcia Levitus

The vector product of two vectors is a vector defined as

$\mathbf{u}\times \mathbf{v}=|\mathbf{u}| |\mathbf{v}| \mathbf{n} \sin\theta \nonumber$

where $$\theta$$ is again the angle between the two vectors, and $$\mathbf{n}$$ is the unit vector perpendicular to the plane formed by $$\mathbf{u}$$ and $$\mathbf{v}$$. The direction of the vector $$\mathbf{n}$$ is given by the right-hand rule. Extend your right hand and point your index finger in the direction of $$\mathbf{u}$$ (the vector on the left side of the $$\times$$ symbol) and your forefinger in the direction of $$\mathbf{v}$$. The direction of $$\mathbf{n}$$, which determines the direction of $$\mathbf{u}\times \mathbf{v}$$, is the direction of your thumb. If you want to revert the multiplication, and perform $$\mathbf{v}\times \mathbf{u}$$, you need to point your index finger in the direction of $$\mathbf{v}$$ and your forefinger in the direction of $$\mathbf{u}$$ (still using the right hand!). The resulting vector will point in the opposite direction (Figure $$\PageIndex{1}$$).

The magnitude of $$\mathbf{u}\times \mathbf{v}$$ is the product of the magnitudes of the individual vectors times $$\sin \theta$$. This magnitude has an interesting geometrical interpretation: it is the area of the parallelogram formed by the two vectors (Figure $$\PageIndex{1}$$). Figure $$\PageIndex{1}$$: The vector product (CC BY-NC-SA; Marcia Levitus)

The cross product can also be expressed as a determinant:

$\mathbf{u}\times \mathbf{v}= \begin{vmatrix} \hat{\mathbf{i}}&\hat{\mathbf{j}}&\hat{\mathbf{k}}\\ u_x&u_y&u_z\\ v_x&v_y&v_z\\ \end{vmatrix} \nonumber$

Example $$\PageIndex{1}$$:

Given $$\mathbf{u}=-2 \hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}$$ and $$\mathbf{v}=3 \hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}}$$, calculate $$\mathbf{w}=\mathbf{u}\times \mathbf{v}$$ and verify that the result is perpendicular to both $$\mathbf{u}$$ and $$\mathbf{v}$$.

Solution

\begin{align*} \mathbf{u}\times \mathbf{v} &= \begin{vmatrix} \hat{\mathbf{i}}&\hat{\mathbf{j}}&\hat{\mathbf{k}}\\ u_x&u_y&u_z\\ v_x&v_y&v_z\\ \end{vmatrix}=\begin{vmatrix} \hat{\mathbf{i}}&\hat{\mathbf{j}}&\hat{\mathbf{k}}\\ -2&1&1\\ 3&-1&1\\ \end{vmatrix} \\[4pt] &=\hat{\mathbf{i}}(1+1)-\hat{\mathbf{j}}(-2-3)+\hat{\mathbf{k}}(2-3) \\[4pt] &=\displaystyle{\color{Maroon}2 \hat{\mathbf{i}}+5 \hat{\mathbf{j}}-\hat{\mathbf{k}}} \end{align*} \nonumber

To verify that two vectors are perpendicular we perform the dot product:

$\mathbf{u} \cdot \mathbf{w}=(-2)(2)+(1)(5)+(1)(-1)=0 \nonumber$

$\mathbf{v} \cdot \mathbf{w}=(3)(2)+(-1)(5)+(1)(-1)=0 \nonumber$

An important application of the cross product involves the definition of the angular momentum. If a particle with mass $$m$$ moves a velocity $$\mathbf{v}$$ (a vector), its (linear) momentum is $$\mathbf{p}=m\mathbf{v}$$. Let $$\mathbf{r}$$ be the position of the particle (another vector), then the angular momentum of the particle is defined as

$\mathbf{l}=\mathbf{r}\times\mathbf{p} \nonumber$

The angular momentum is therefore a vector perpendicular to both $$\mathbf{r}$$ and $$\mathbf{p}$$. Because the position of the particle needs to be defined with respect to a particular origin, this origin needs to be specified when defining the angular momentum. Figure $$\PageIndex{2}$$: The angular momentum of a particle of position $$\mathbf{r}$$ from the origin and momentum $$\mathbf{p}=m\mathbf{v}$$ (CC BY-NC-SA; Marcia Levitus)