Cross/vector product

We now look at the cross or vector product of two vectors, say $\boldsymbol{a}$ and $\boldsymbol{b}$ denoted by $\boldsymbol{a} \times \boldsymbol{b}$ . We note that the vectors need to be three-dimensional. While the result of a dot product is a number, the result of the cross product is a vector.

Given two vectors $\boldsymbol{a}=\left(a_{1}, a_{2}, a_{3}\right)$ and $\boldsymbol{b}=\left(b_{1}, b_{2}, b_{3}\right)$ , the cross product is defined as,

\boldsymbol{a} \times \boldsymbol{b}=\left(a_{2} b_{3}-a_{3} b_{2}, a_{3} b_{1}-a_{1} b_{3}, a_{1} b_{2}-a_{2} b_{1}\right) .

The cross product is simply the determinant of a $3 \times 3$ matrix. While matrices are the subject of Chapter 10, we include the determinant from which we derive Eq. (9.21) here, for completeness (refer to Section 10.2 for a more formal definition of determinants). The product $\boldsymbol{a} \times \boldsymbol{b}$ is given by the determinant,

\boldsymbol{a} \times \boldsymbol{b}=\left|\begin{array}{ccc} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ a_{1} & a_{2} & a_{3} \\ b_{1} & b_{2} & b_{3} \end{array}\right|

where the first row gives the standard basis vectors and the next two rows give the components of vectors $\boldsymbol{a}$ and $\boldsymbol{b}$ , respectively. The above is equivalent to

\boldsymbol{a} \times \boldsymbol{b}=\left|\begin{array}{ll} a_{2} & a_{3} \\ b_{2} & b_{3} \end{array}\right| \mathbf{i}-\left|\begin{array}{ll} a_{1} & a_{3} \\ b_{1} & b_{3} \end{array}\right| \mathbf{j}+\left|\begin{array}{ll} a_{1} & a_{2} \\ b_{1} & b_{2} \end{array}\right| \mathbf{k},

where

\left|\begin{array}{ll} a & b \\ c & d \end{array}\right|=a d-b c

We now give the geometric interpretation of the cross product. Let $\theta$ be the angle between two vectors $\boldsymbol{a}$ and $\boldsymbol{b}$ , with $0 \leq \theta \leq \pi$ as shown in Fig. 9.5. Then

\boldsymbol{a} \times \boldsymbol{b}=|\boldsymbol{a}||\boldsymbol{b}| \sin \theta \hat{\boldsymbol{n}},

where $\hat{\boldsymbol{n}}$ is the unit vector perpendicular to the plane containing $\boldsymbol{a}$ and $\boldsymbol{b}$ in the direction given by the right hand rule. This says that if we take our right hand, start at $\boldsymbol{a}$ and rotate our fingers toward $\boldsymbol{b}$ , our thumb will point in the direction of the cross product (check using Figs. 9.5 and 9.6). Had we sketched $\boldsymbol{b} \times \boldsymbol{a}$ , we would have found the cross product pointing in the downward direction.

Figure 9.5: Geometric interpretation of the vector product.

In Physics, the right hand rule is used to give the force $\boldsymbol{F}$ in terms of the current and electromagnetic field. For a charged particle with charge $q$ moving with velocity $\boldsymbol{v}$ in a magnetic field $\boldsymbol{B}$ , the force exerted on the particle is given by

\boldsymbol{F}=q \boldsymbol{v} \times \boldsymbol{B}

Figure 9.6: The right hand rule for electromagnetism. Example 9.2 Consider a plane that contains the points $P, Q$ , and $R$ with coordinates $P=(1,0,0), Q=(1,1,1)$ , and $R=(2,-1,3)$ . Find a vector that is perpendicular to the plane.

Solution We know that if we take the cross product of two vectors in the plane, then the resulting vector would be perpendicular to the plane. Since $P, Q$ , and $R$ are points in the plane, any vector between them also lies in the plane. There is more than one way to obtain vectors between the points. We consider,

\overrightarrow{P Q}=(0,1,1) \text { and } \quad \overrightarrow{P R}=(1,-1,3)

We now determine the cross product

\overrightarrow{P Q} \times \overrightarrow{P R}=\left|\begin{array}{ccc} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 0 & 1 & 1 \\ 1 & -1 & 3 \end{array}\right|,

which, using Eq. (9.21), is calculated as $(4,1,-1)$ or $4 \mathbf{i}+\mathbf{j}-\mathbf{k}$ . This vector is perpendicular to the plane containing the three points.

So far we have seen a case where the cross product is perpendicular to both the original vectors. For the most part this is true but there are some exceptions. Suppose the two vectors $\boldsymbol{a}$ and $\boldsymbol{b}$ are parallel; the angle between them is either 0 or 180 degrees. From Eq. (9.22), we have

|\boldsymbol{a} \times \boldsymbol{b}|=0,

which implies $\boldsymbol{a} \times \boldsymbol{b}=\mathbf{0}$ i.e. the result is the zero vector and it cannot be orthogonal to the original vectors. This leads to the following statement:

\text { if } \boldsymbol{a} \times \boldsymbol{b}=\mathbf{0} \text { then } \boldsymbol{a} \text { and } \boldsymbol{b} \text { are parallel vectors. }

Additionally,

\text { if } \boldsymbol{a} \times \boldsymbol{b} \neq \mathbf{0} \text { then } \boldsymbol{a} \times \boldsymbol{b} \text { is perpendicular to both } \boldsymbol{a} \text { and } \boldsymbol{b} .

Properties

The right hand rule implies

\boldsymbol{a} \times \boldsymbol{b}=-\boldsymbol{b} \times \boldsymbol{a} .

For scalars $\lambda$ and $\mu$ , we have

(\lambda \boldsymbol{a}) \times(\mu \boldsymbol{b})=\lambda \mu(\boldsymbol{a} \times \boldsymbol{b}) .

The cross product is distributive over addition of vectors,

\boldsymbol{a} \times(\boldsymbol{b}+\boldsymbol{c})=(\boldsymbol{a} \times \boldsymbol{b})+(\boldsymbol{a} \times \boldsymbol{c}) .

The cross product for components can be summarised as,

\begin{aligned} & \mathbf{i} \times \mathbf{i}=0, \quad \mathbf{j} \times \mathbf{j}=0, \quad \mathbf{k} \times \mathbf{k}=0 \\ & \mathbf{i} \times \mathbf{j}=\mathbf{k}, \quad \mathbf{j} \times \mathbf{k}=\mathbf{i}, \quad \mathbf{k} \times \mathbf{i}=\mathbf{j} \end{aligned}

Note that the order of the basis vectors matters; it is negative if the order of the vector is in the opposite order, for instance $\mathbf{j} \times \mathbf{i}=-\mathbf{k}$ .

As with the dot product,

\boldsymbol{a} \times \boldsymbol{b}=\boldsymbol{a} \times \boldsymbol{c}

does not necessarily mean that $\boldsymbol{b}=\boldsymbol{c}$ . From the distributive law for the cross product we have

a \times(b-c)=0 ;

this can imply that either $\boldsymbol{b}=\boldsymbol{c}$ or $\boldsymbol{a}$ is parallel to $\boldsymbol{b}-\boldsymbol{c}$ .

Applications of the cross product

(i) The area of a triangle: To obtain the area of the triangle, we use

\frac{1}{2}(\text { base } \times \text { height })

which using Fig. 9.7 is given by

\frac{1}{2}|\boldsymbol{a}| h

We notice that $h=|\boldsymbol{b}| \sin \theta$ ; from Eq. (9.22) we have

|\boldsymbol{a}| \underbrace{|\boldsymbol{b}| \sin \theta}_{h}=|\boldsymbol{a} \times \boldsymbol{b}|

and it follows that the area of the triangle from the cross product is given by

\frac{1}{2}|\boldsymbol{a} \times \boldsymbol{b}|

Figure 9.7: The area of a triangle from the cross product.

(ii) The volume of the parallelepiped: This is calculated using both the scalar and the vector product using what is known as the triple scalar product given as follows,

\boldsymbol{a} \cdot(\boldsymbol{b} \times \boldsymbol{c})=\boldsymbol{b} \cdot(\boldsymbol{c} \times \boldsymbol{a})=\boldsymbol{c} \cdot(\boldsymbol{a} \times \boldsymbol{b}) .

Swapping the order of the cross product negates the triple product, i.e.

\boldsymbol{a} \cdot(\boldsymbol{b} \times \boldsymbol{c})=-\boldsymbol{a} \cdot(\boldsymbol{c} \times \boldsymbol{b})

as expected from the properties of the cross product listed above.

In what follows, we show that the volume of the parallelepiped is given as

V=|\boldsymbol{a} \cdot(\boldsymbol{b} \times \boldsymbol{c})|

Figure 9.8 shows the vectors $\boldsymbol{a}, \boldsymbol{b}$ , and $\boldsymbol{c}$ defining a parallelepiped. The height $h$ is the perpendicular distance between the base and the opposite face, $\theta$ is the angle between $\boldsymbol{b}$ and $\boldsymbol{c}$ and $\phi$ is the internal angle between $\boldsymbol{a}$ and the distance $h$ . The volume of the parallelepiped is calculated from

\text { area of base } \times \text { height. }

To find the area of the base, i.e. the area of the parallelogram comprising the base, we make use of the cross product. With $\boldsymbol{b}$ and $\boldsymbol{c}$ representing the edges of the base as shown in Fig. 9.8, we have

\text { Area of base }=|\boldsymbol{b}||\boldsymbol{c}| \sin \theta=|\boldsymbol{b} \times \boldsymbol{c}| .

The height is given as

h=|\boldsymbol{a}| \cos \phi .

The volume is therefore given by

|\boldsymbol{b} \times \boldsymbol{c} \| \boldsymbol{a}| \cos \phi

By the scalar product theorem [see Eq. (9.16)], the volume of the parallelepiped is calculated from

|(b \times c) \cdot a|

since the scalar product is commutative, this gives the expression in Eq. (9.24).

Figure 9.8: Vectors defining a parallelepiped. Its volume is given by the triple scalar product. Example 9.3 Calculate the volume of the parallelepiped which has vertices with position vectors $(1,2,1),(2,3,2),(2,4,4)$ , and $(2,1,2)$ and three of its edges between the first vertex and the remaining three.

Solution Let $A$ be the point with coordinates $(1,2,1)$ . Also define $B=(2,3,2)$ , $C=(2,4,4)$ , and $D=(2,1,2)$ . The first vertex acts as the origin of the parallelepiped and we can then define the three direction vectors as

\boldsymbol{a}=\overrightarrow{A B}=(1,1,1), \quad \boldsymbol{b}=\overrightarrow{A C}=(1,2,3), \quad \boldsymbol{c}=\overrightarrow{A D}=(1,-1,1)

We use

(\boldsymbol{a} \times \boldsymbol{b}) \cdot \boldsymbol{c}

to calculate the volume of the parallelepiped. Starting with the cross product, we have

\boldsymbol{a} \times \boldsymbol{b}=\left|\begin{array}{ccc} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 1 & 1 \\ 1 & 2 & 3 \end{array}\right|=\mathbf{i}-2 \mathbf{j}+\mathbf{k}

The dot product of the result with $c$ gives

(\boldsymbol{a} \times \boldsymbol{b}) \cdot \boldsymbol{c}=(1,-2,1) \cdot(1,-1,1)=4 .

Exercises

Find the following cross products: (a) $(1,2,3) \times(-2,2,1)$ (b) $(2,-1,2) \times(-1,-1,3)$ (c) $(1,1,1) \times(2,-2,1)$
Find the area of the triangle with vertices at $(0,0,0),(1,2,1)$ , and $(1,1,-1)$ .
Find the volume of the parallelepiped with vertices $(0,0,0),(1,1,1),(1,-1,1)$ , and $(2,-1,-3)$ , where the three main edges are between the first vertex and the latter three.

Dot/scalar product Vector geometry