Linear Algebra · Book or PDF

A First Course in Linear Algebra

A.G. Hamilton · Study material generated by AI from this edition. Not affiliated with the publisher.

Chapter 11 · pp. 132–142

Cross Product

The cross product is a way to combine two 3-vectors into a new 3-vector. Its direction and length carry geometric information. It gives a vector perpendicular to both inputs, the area of a triangle or parallelogram, the volume of a box-like solid, and a test for whether three vectors lie in one plane.

Definition and the 3-vector restriction

For $\mathbf{a}=(a_1,a_2,a_3)$ and $\mathbf{b}=(b_1,b_2,b_3)$ , the cross product is

$\mathbf{a}\times\mathbf{b}=(a_2b_3-a_3b_2,\ a_3b_1-a_1b_3,\ a_1b_2-a_2b_1).$

The dot product gives a number. The cross product gives a vector. (p. 132)

A memory aid is to write each vector as $a_1\mathbf{i}+a_2\mathbf{j}+a_3\mathbf{k}$ . Then set up a "determinant" with $\mathbf{i},\mathbf{j},\mathbf{k}$ in one column and the components of $\mathbf{a}$ and $\mathbf{b}$ in the other two. Expanding it along the $\mathbf{i},\mathbf{j},\mathbf{k}$ column gives the same three components. (p. 132)

Example. Take $\mathbf{a}=(1,0,0)$ and $\mathbf{b}=(0,1,0)$ . The first component is $0\cdot 0-0\cdot 1=0$ . The second is $0\cdot 0-1\cdot 0=0$ . The third is $1\cdot 1-0\cdot 0=1$ . So $\mathbf{i}\times\mathbf{j}=\mathbf{k}=(0,0,1)$ . (p. 132)

The cross product exists only for 3-vectors. The formula needs exactly three components. The dot product works for vectors of any length, but the cross product does not. To use it on 2-vectors, you first have to treat them as 3-vectors, for example by adding a third component of 0. This is why it is a tool for three-dimensional geometry. (p. 132)

Geometric meaning: direction and length

Direction. The vector $\mathbf{a}\times\mathbf{b}$ is perpendicular to both $\mathbf{a}$ and $\mathbf{b}$ . To check this, expand $\mathbf{a}\cdot(\mathbf{a}\times\mathbf{b})$ and $\mathbf{b}\cdot(\mathbf{a}\times\mathbf{b})$ . In both cases all the terms cancel, so each dot product is 0, and a zero dot product means perpendicular. For instance, $\mathbf{k}=\mathbf{i}\times\mathbf{j}$ is perpendicular to both $\mathbf{i}$ and $\mathbf{j}$ . This is the most useful property of the cross product. (p. 134)

Length. If $\mathbf{a}$ and $\mathbf{b}$ are non-zero and $\theta$ is the angle between them, then

$|\mathbf{a}\times\mathbf{b}|=|\mathbf{a}||\mathbf{b}|\sin\theta.$

The proof works by expanding both sides in components. First, $|\mathbf{a}\times\mathbf{b}|^2=|\mathbf{a}|^2|\mathbf{b}|^2-(\mathbf{a}\cdot\mathbf{b})^2$ . Second, $|\mathbf{a}|^2|\mathbf{b}|^2\sin^2\theta=|\mathbf{a}|^2|\mathbf{b}|^2(1-\cos^2\theta)$ , and since $\mathbf{a}\cdot\mathbf{b}=|\mathbf{a}||\mathbf{b}|\cos\theta$ , this is the same expression. Everything involved is positive, so you can take square roots. (pp. 133–134)

Example. If $\mathbf{a}$ and $\mathbf{b}$ are perpendicular with lengths 2 and 3, then $\sin\theta=1$ and $|\mathbf{a}\times\mathbf{b}|=6$ . If they point in the same or opposite directions, $\sin\theta=0$ and the cross product is the zero vector. (pp. 133–134)

Algebraic rules

Order matters. $\mathbf{a}\times\mathbf{b}=-(\mathbf{b}\times\mathbf{a})$ . Swapping the vectors flips the sign, so the cross product is not commutative. For example, $\mathbf{i}\times\mathbf{j}=\mathbf{k}$ but $\mathbf{j}\times\mathbf{i}=-\mathbf{k}$ . (pp. 133–134)
Parallel vectors give zero. $\mathbf{a}\times\mathbf{a}=\mathbf{0}$ , and $\mathbf{a}\times(k\mathbf{a})=\mathbf{0}$ for any number $k$ . This is consistent with the length formula, since $\sin\theta=0$ here. Both can be checked straight from the definition. (p. 134)
Linearity. $\mathbf{a}\times(k\mathbf{b})=(k\mathbf{a})\times\mathbf{b}=k(\mathbf{a}\times\mathbf{b})$ , and $\mathbf{a}\times(\mathbf{b}+\mathbf{c})=(\mathbf{a}\times\mathbf{b})+(\mathbf{a}\times\mathbf{c})$ . Each follows by writing out components and regrouping. (pp. 133–134)
Cyclic rules. $\mathbf{i}\times\mathbf{j}=\mathbf{k}$ , $\mathbf{j}\times\mathbf{k}=\mathbf{i}$ , $\mathbf{k}\times\mathbf{i}=\mathbf{j}$ . Reversing any of these gives the negative. (p. 134)

Together these let you expand products of sums piece by piece, as long as you keep each product in its original order. (p. 134)

Areas and planes

Areas. Let $P$ , $A$ , $B$ be points with $\overrightarrow{PA}=\mathbf{a}$ and $\overrightarrow{PB}=\mathbf{b}$ . A triangle has area $\tfrac12|PA||PB|\sin\theta$ . By the length formula, this equals

$\text{area of triangle }PAB=\tfrac12|\mathbf{a}\times\mathbf{b}|.$

If $C$ completes the parallelogram $PACB$ , its area is $|\mathbf{a}\times\mathbf{b}|$ , which is twice the triangle. (pp. 135–136)

Example. Let $P=(0,0,0)$ , $A=(2,0,0)$ and $B=(0,3,0)$ . Then $\mathbf{a}\times\mathbf{b}=(0,0,6)$ , which has length 6. The triangle has area 3 and the parallelogram has area 6.

Plane through three points. The cross product gives a normal vector, so you can write an equation of the plane from three points $A$ , $B$ , $C$ on it. (pp. 135–136)

Find $\overrightarrow{AB}=\mathbf{b}-\mathbf{a}$ and $\overrightarrow{AC}=\mathbf{c}-\mathbf{a}$ .
Compute $\overrightarrow{AB}\times\overrightarrow{AC}$ . It is perpendicular to both vectors, so it is perpendicular to the plane.
Use this normal and one of the points in the point-normal method.

Example. For $A(5,3,-1)$ , $B(2,-2,0)$ , $C(3,1,1)$ , we get $\overrightarrow{AB}=(-3,-5,1)$ and $\overrightarrow{AC}=(-2,-2,2)$ . Their cross product is $(-8,4,-4)$ . Any multiple works as a normal, so use $(2,-1,1)$ . The plane is $2(x-5)-(y-3)+(z+1)=0$ , which simplifies to $2x-y+z-6=0$ . (pp. 135–136)

Volume and the scalar triple product

A parallelepiped is a solid with six faces in which opposite faces are congruent parallelograms. Suppose one vertex is $P$ and the three edges leaving it are $\mathbf{a}$ , $\mathbf{b}$ , $\mathbf{c}$ . Use the face spanned by $\mathbf{b}$ and $\mathbf{c}$ as the base. Its area is $|\mathbf{b}\times\mathbf{c}|$ . The vector $\mathbf{b}\times\mathbf{c}$ is perpendicular to the base, so the height is the part of $\mathbf{a}$ along that direction, $|\mathbf{a}|\cos\phi$ , where $\phi$ is the angle between $\mathbf{a}$ and $\mathbf{b}\times\mathbf{c}$ . Base times height gives

$V=|\mathbf{a}\cdot(\mathbf{b}\times\mathbf{c})|.$

The absolute value is needed because $\mathbf{b}\times\mathbf{c}$ may point to the other side of the base, which makes the dot product negative. (pp. 137–138)

The number $\mathbf{a}\cdot(\mathbf{b}\times\mathbf{c})$ is called the scalar triple product. It is a number, since it is a dot product. The solid is the same whatever order you list the edges, so reordering $\mathbf{a},\mathbf{b},\mathbf{c}$ never changes the absolute value. It can change only the sign. (p. 138)

Example. Let $P=(0,0,0)$ , $A=(2,1,0)$ , $B=(1,2,0)$ , $C=(3,3,2)$ . Then $\mathbf{b}\times\mathbf{c}=(4,-2,-3)$ . So $\mathbf{a}\cdot(\mathbf{b}\times\mathbf{c})=8-2+0=6$ , and the volume is 6 cubic units. (p. 137)

Coplanar vectors and linear dependence

Three vectors are coplanar if line segments from the origin representing them all lie in one plane. Then the parallelepiped they span is squashed flat and has volume 0. So $\mathbf{a},\mathbf{b},\mathbf{c}$ are coplanar exactly when $\mathbf{a}\cdot(\mathbf{b}\times\mathbf{c})=0$ . (p. 139)

This connects to determinants. If $A$ is the $3\times 3$ matrix with columns $\mathbf{a},\mathbf{b},\mathbf{c}$ , then

$\det A=\mathbf{a}\cdot(\mathbf{b}\times\mathbf{c}).$

You can check this by expanding the determinant along its first column. (pp. 139–140)

A determinant is 0 exactly when the columns are linearly dependent. So three 3-vectors are coplanar if and only if they are linearly dependent. The identity also gives a handy way to compute scalar triple products and volumes. (pp. 139–140)

To test whether four points are coplanar, pick one point and form the three vectors from it to the others. Then check whether those vectors are dependent, for instance by Gaussian elimination. If the elimination leaves a row of zeros, the vectors are dependent and the points are coplanar. (pp. 139–140)

Example. Take $O(0,0,0)$ , $A(1,3,0)$ , $B(0,1,1)$ , $C(-2,1,7)$ . Notice that $\overrightarrow{OC}=-2\overrightarrow{OA}+7\overrightarrow{OB}$ , so the vectors are dependent and the four points are coplanar. As a check, $\mathbf{b}\times\mathbf{c}=(6,-2,2)$ and $\mathbf{a}\cdot(6,-2,2)=6-6+0=0$ . (p. 139)

Check yourself

Compute $(1,2,0)\times(0,1,3)$ . Then show that your answer is perpendicular to both original vectors.
If $\mathbf{a}\times\mathbf{b}=(2,-1,4)$ , what is $\mathbf{b}\times\mathbf{a}$ ? What is $\mathbf{a}\times(3\mathbf{a})$ ?
Two vectors have lengths 4 and 5 and the angle between them is $30^\circ$ . What is the length of their cross product, and what is the area of the triangle they span?
Explain the steps for finding an equation of the plane through three given points. Why does the cross product help?
Why does $\mathbf{a}\cdot(\mathbf{b}\times\mathbf{c})=0$ mean that $\mathbf{a},\mathbf{b},\mathbf{c}$ are linearly dependent?