The cross product

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The cross product is a special way to multiply two vectors in three-dimensional space.

There is no useful way to “multiply” two vectors and obtain another vector in $\mathbb {R}^n$ for arbitrary $n$. However, in the special case of $\mathbb {R}^3$, there is an important multiplication operation called “the cross product.”

The cross product is linked inextricably to the determinant, so we will first introduce the determinant before introducing this new operation.

1 Determinants

Given a $2\times 2$ matrix, the determinant is given by

\[ \det \begin{bmatrix} a & b\\ c & d \end{bmatrix} = \begin{vmatrix} a & b\\ c & d \end{vmatrix} = ad -bc. \]

Why would anyone ever be interested in this? Well to start, given nonzero vectors

\[ \overset {\boldsymbol {\rightharpoonup }}{\mathbf {v}} = \left < a,b \right > \quad \text {and}\quad \overset {\boldsymbol {\rightharpoonup }}{\mathbf {w}} = \left < c,d \right > \]

we can form a parallelogram as below.

The determinant gives the (signed) area of this parallelogram, where the sign of the area is given by the sign of the angle $\theta $ (drawn counterclockwise) between $\overset {\boldsymbol {\rightharpoonup }}{\mathbf {v}}$ and $\overset {\boldsymbol {\rightharpoonup }}{\mathbf {w}}$. To understand why this is true, make a rectangle around the parallelogram as below.

Now, we see that the area of the parallelogram is

\[ \text {Area of rectangle} - \text {Area of other regions} \]

and this is \begin{align*} (a+c)(b+d) - \left (cd + ab + 2bc \right )&= ab + ad + bc + cd - cd - ab-2bc\\ &= ad - bc\\ &= \det \begin{bmatrix} a & b\\ c & d \end{bmatrix}. \end{align*}

Typically, when one computes the determinant of a $2\times 2$ matrix, we think of the terms as follows.

Given a $3\times 3$ matrix, the determinant is denoted by the following.

\[ \det \begin{bmatrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{bmatrix} = \begin{vmatrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{vmatrix} \]

The determinant can be computed using the following pattern.

This equals

\[ a_1(b_2c_3-b_3c_2)- a_2(b_1c_3-b_3c_1)+ a_3(b_1c_2-b_2c_1). \]

Compute.

\[ \det \begin{bmatrix} 1 & 4 & 7\\ 2 & 5 & 8\\ 3 & 6 & 9\\ \end{bmatrix} \begin{prompt} = \answer {0} \end{prompt} \]

2 Cross products

Determinants in $\mathbb {R}^n$ have many uses. In $\mathbb {R}^3$, one of the uses is the definition of the cross product.

The cross product of two vectors is given by

\[ \left < a_1,a_2,a_3 \right > \boldsymbol \times \left < b_1,b_2,b_3 \right > = \det \begin{bmatrix} \boldsymbol {\hat {\imath }} & \boldsymbol {\hat {\jmath }} & \boldsymbol {\hat {k}}\\ a_1 & a_2 & a_3\\ b_1 & b_2 & b_3 \end{bmatrix}. \]

The first thing you should notice about the the cross product is that

\[ \mathbf {vector} \boldsymbol \times \mathbf {vector} = \mathbf {vector}. \]

In particular, the cross product is not precisely a determinant, since a determinant would be a number, not a vector. But, the determinant provides us with a useful memory tool for computing the cross product. The pattern

is efficient, because you immediately have the components of the desired vector without additional simplification.

Compute $\left < 2,3,1 \right > \boldsymbol \times \left < 1,1,1 \right >$. \begin{align*} \left < 2,3,1 \right > \boldsymbol \times \left < 1,1,1 \right > &= \det \begin{bmatrix} \boldsymbol {\hat {\imath }} & \boldsymbol {\hat {\jmath }} & \boldsymbol {\hat {k}} \\ \answer {2} & \answer {3} & \answer {1}\\ \answer {1} & \answer {1} & \answer {1} \end{bmatrix}\\ &= \left < \answer {2},\answer {-1},\answer {-1} \right > \end{align*}

Compute $\left < 1,1,1 \right > \boldsymbol \times \left < 2,3,1 \right >$. \begin{align*} \left < 1,1,1 \right > \boldsymbol \times \left < 2,3,1 \right > &=\det \begin{bmatrix} \boldsymbol {\hat {\imath }} & \boldsymbol {\hat {\jmath }} & \boldsymbol {\hat {k}} \\ \answer {1} & \answer {1} & \answer {1}\\ \answer {2} & \answer {3} & \answer {1} \end{bmatrix}\\ &= \left < \answer {-2},\answer {1},\answer {1} \right > \end{align*}

Notice that the cross product is not commutative! In fact, it is anticommutative, meaning the statement below.

The Cross Product is Anticommutative Given two vectors $\overset {\boldsymbol {\rightharpoonup }}{\mathbf {a}}$ and $\overset {\boldsymbol {\rightharpoonup }}{\mathbf {b}}$ in $\mathbb {R}^3$

\[ \overset {\boldsymbol {\rightharpoonup }}{\mathbf {a}} \boldsymbol \times \overset {\boldsymbol {\rightharpoonup }}{\mathbf {b}} = -\overset {\boldsymbol {\rightharpoonup }}{\mathbf {b}} \boldsymbol \times \overset {\boldsymbol {\rightharpoonup }}{\mathbf {a}} \]

The anticommutative property of the cross product demonstrates that $\overset {\boldsymbol {\rightharpoonup }}{\mathbf {a}} \boldsymbol \times \overset {\boldsymbol {\rightharpoonup }}{\mathbf {b}}$ and $\overset {\boldsymbol {\rightharpoonup }}{\mathbf {b}} \boldsymbol \times \overset {\boldsymbol {\rightharpoonup }}{\mathbf {a}}$ differ only by a sign. These vectors have the same magnitude but point in opposite directions.

Let’s examine the cross product on famous unit vectors.

Compute

\[ \boldsymbol {\hat {\imath }} \boldsymbol \times \boldsymbol {\hat {\jmath }}. \begin{prompt} = \left < \answer {0},\answer {0},\answer {1} \right > \end{prompt} \]

Compute

\[ \boldsymbol {\hat {\jmath }} \boldsymbol \times \boldsymbol {\hat {k}}. \begin{prompt} = \left < \answer {1},\answer {0},\answer {0} \right > \end{prompt} \]

Compute

\[ \boldsymbol {\hat {k}} \boldsymbol \times \boldsymbol {\hat {\imath }}. \begin{prompt} = \left < \answer {0},\answer {1},\answer {0} \right > \end{prompt} \]

One way to remember the cross products of the unit vectors $\boldsymbol {\hat {\imath }}$, $\boldsymbol {\hat {\jmath }}$, and $\boldsymbol {\hat {k}}$ is to use the diagram below.

Since we see that the cross product of two basic unit vectors produces a vector orthogonal to both unit vectors, we are led to our next theorem (which could be verified through brute force computations).

The vector $\overset {\boldsymbol {\rightharpoonup }}{\mathbf {a}} \boldsymbol \times \overset {\boldsymbol {\rightharpoonup }}{\mathbf {b}}$ is orthogonal to both $\overset {\boldsymbol {\rightharpoonup }}{\mathbf {a}}$ and $\overset {\boldsymbol {\rightharpoonup }}{\mathbf {b}}$.

In three-dimensional space, when seeking a vector perpendicular to both $\overset {\boldsymbol {\rightharpoonup }}{\mathbf {a}}$ and $\overset {\boldsymbol {\rightharpoonup }}{\mathbf {b}}$, we could choose one of two directions: the direction of $\overset {\boldsymbol {\rightharpoonup }}{\mathbf {a}} \boldsymbol \times \overset {\boldsymbol {\rightharpoonup }}{\mathbf {b}}$, or the direction of $\overset {\boldsymbol {\rightharpoonup }}{\mathbf {b}} \boldsymbol \times \overset {\boldsymbol {\rightharpoonup }}{\mathbf {a}}$. The direction of the cross product is given by the right-hand rule. Given $\overset {\boldsymbol {\rightharpoonup }}{\mathbf {a}}$ and $\overset {\boldsymbol {\rightharpoonup }}{\mathbf {b}}$ in $\mathbb {R}^3$ with the same initial point, point the index finger of your right hand in the direction of $\overset {\boldsymbol {\rightharpoonup }}{\mathbf {a}}$ and let your middle finger point in the direction of $\overset {\boldsymbol {\rightharpoonup }}{\mathbf {b}}$ (much as we did when establishing the right-hand rule for the 3-dimensional coordinate system). Your thumb will naturally extend in the direction of $\overset {\boldsymbol {\rightharpoonup }}{\mathbf {a}} \boldsymbol \times \overset {\boldsymbol {\rightharpoonup }}{\mathbf {b}}$. If you switch your fingers, pointing the index finder in the direction of $\overset {\boldsymbol {\rightharpoonup }}{\mathbf {b}}$ and the middle finger in the direction of $\overset {\boldsymbol {\rightharpoonup }}{\mathbf {a}}$, your thumb will now point in the opposite direction, allowing you to “visualize” the anticommutative property of the cross product.

Consider the vectors below.

Does $\overset {\boldsymbol {\rightharpoonup }}{\mathbf {a}} \boldsymbol \times \overset {\boldsymbol {\rightharpoonup }}{\mathbf {b}}$ point toward you or away from you?

toward me away from me

3 The geometry of the cross product

Just as we related the angle between two vectors and their dot product, there is a similar relationship relating the cross product of two vectors to the angle between them. Before we get started, we need an identity.

Lagrange’s Idenitity Given two vectors $\overset {\boldsymbol {\rightharpoonup }}{\mathbf {a}}$ and $\overset {\boldsymbol {\rightharpoonup }}{\mathbf {b}}$,

\[ |\overset {\boldsymbol {\rightharpoonup }}{\mathbf {a}} \boldsymbol \times \overset {\boldsymbol {\rightharpoonup }}{\mathbf {b}}|^2 = |\overset {\boldsymbol {\rightharpoonup }}{\mathbf {a}}|^2 |\overset {\boldsymbol {\rightharpoonup }}{\mathbf {b}}|^2 - (\overset {\boldsymbol {\rightharpoonup }}{\mathbf {a}} \bullet \overset {\boldsymbol {\rightharpoonup }}{\mathbf {b}})^2. \]

Set $\overset {\boldsymbol {\rightharpoonup }}{\mathbf {a}} = \left < a_1,a_2,a_3 \right >$ and $\overset {\boldsymbol {\rightharpoonup }}{\mathbf {b}} = \left < b_1,b_2,b_3 \right >$. Computing the left-hand side and the right-hand side separately, we see that they are equal. We leave both of these computations to the intrepid young mathematician.

Geometric Interpretation of the Cross Product For any two vectors $\overset {\boldsymbol {\rightharpoonup }}{\mathbf {a}}$ and $\overset {\boldsymbol {\rightharpoonup }}{\mathbf {b}}$,

\[ |\overset {\boldsymbol {\rightharpoonup }}{\mathbf {a}} \boldsymbol \times \overset {\boldsymbol {\rightharpoonup }}{\mathbf {b}}| = |\overset {\boldsymbol {\rightharpoonup }}{\mathbf {a}}||\overset {\boldsymbol {\rightharpoonup }}{\mathbf {b}}|\sin (\theta ), \]

where $0\le \theta \le \pi $ is the angle between $\overset {\boldsymbol {\rightharpoonup }}{\mathbf {a}}$ and $\overset {\boldsymbol {\rightharpoonup }}{\mathbf {b}}$.

By Lagrange’s identity, we have that

However, we also know that

\[ \overset {\boldsymbol {\rightharpoonup }}{\mathbf {a}} \bullet \overset {\boldsymbol {\rightharpoonup }}{\mathbf {b}} = |\overset {\boldsymbol {\rightharpoonup }}{\mathbf {a}}| |\overset {\boldsymbol {\rightharpoonup }}{\mathbf {b}}|\cos (\theta ). \]

So, write with me. \begin{align*} |\overset {\boldsymbol {\rightharpoonup }}{\mathbf {a}} \boldsymbol \times \overset {\boldsymbol {\rightharpoonup }}{\mathbf {b}}|^2 &= |\overset {\boldsymbol {\rightharpoonup }}{\mathbf {a}}|^2 |\overset {\boldsymbol {\rightharpoonup }}{\mathbf {b}}|^2 - (\overset {\boldsymbol {\rightharpoonup }}{\mathbf {a}} \bullet \overset {\boldsymbol {\rightharpoonup }}{\mathbf {b}})^2\\ &= |\overset {\boldsymbol {\rightharpoonup }}{\mathbf {a}}|^2 |\overset {\boldsymbol {\rightharpoonup }}{\mathbf {b}}|^2 - |\overset {\boldsymbol {\rightharpoonup }}{\mathbf {a}}|^2 |\overset {\boldsymbol {\rightharpoonup }}{\mathbf {b}}|^2\cos ^2(\theta )\\ &= |\overset {\boldsymbol {\rightharpoonup }}{\mathbf {a}}|^2 |\overset {\boldsymbol {\rightharpoonup }}{\mathbf {b}}|^2(1 - \cos ^2(\theta ))\\ &= |\overset {\boldsymbol {\rightharpoonup }}{\mathbf {a}}|^2 |\overset {\boldsymbol {\rightharpoonup }}{\mathbf {b}}|^2\sin ^2(\theta ) \end{align*}

Hence, $|\overset {\boldsymbol {\rightharpoonup }}{\mathbf {a}} \boldsymbol \times \overset {\boldsymbol {\rightharpoonup }}{\mathbf {b}}| = |\overset {\boldsymbol {\rightharpoonup }}{\mathbf {a}}||\overset {\boldsymbol {\rightharpoonup }}{\mathbf {b}}|\sin (\theta )$.

The theorems above help us make a strong connection between the cross product and geometry.

Given two vectors $\overset {\boldsymbol {\rightharpoonup }}{\mathbf {a}}$ and $\overset {\boldsymbol {\rightharpoonup }}{\mathbf {b}}$ in $\mathbb {R}^3$,

$|\overset {\boldsymbol {\rightharpoonup }}{\mathbf {a}} \boldsymbol \times \overset {\boldsymbol {\rightharpoonup }}{\mathbf {b}}|$ computes the area of the parallelogram spanned by $\overset {\boldsymbol {\rightharpoonup }}{\mathbf {a}}$ and $\overset {\boldsymbol {\rightharpoonup }}{\mathbf {b}}$.

It is a fact from geometry that the area of a parallelogram

\[ A= bh, \]

where $b$ is the length of the base and $h$ is the height of the parallelogram. Two vectors $\overset {\boldsymbol {\rightharpoonup }}{\mathbf {a}}$ and $\overset {\boldsymbol {\rightharpoonup }}{\mathbf {b}}$ define a parallelogram when drawn from the same initial point, as follows.

Trigonometry tells us that $h = |\overset {\boldsymbol {\rightharpoonup }}{\mathbf {a}}| \sin (\theta )$, and so the area of the parallelogram is

\[ A = |\overset {\boldsymbol {\rightharpoonup }}{\mathbf {a}}||\overset {\boldsymbol {\rightharpoonup }}{\mathbf {b}}|\sin (\theta ) = |\overset {\boldsymbol {\rightharpoonup }}{\mathbf {a}} \boldsymbol \times \overset {\boldsymbol {\rightharpoonup }}{\mathbf {b}}|. \]

What is the area of the parallelogram spanned by $\left < 1,2,4 \right >$ and $\left < 2,3,1 \right >$?

\[ \text {Area} = \answer {\sqrt {150}} \]

What is the area of the triangle spanned by $\left < 1,2,4 \right >$ and $\left < 2,3,1 \right >$?

\[ \text {Area} = \answer {\sqrt {150}/2} \]

This is half of the area of the parallelogram spanned by $\left < 1,2,4 \right >$ and $\left < 2,3,1 \right >$.

Note that if $\overset {\boldsymbol {\rightharpoonup }}{\mathbf {a}}$ and $\overset {\boldsymbol {\rightharpoonup }}{\mathbf {b}}$ are in $\mathbb {R}^2$, we can still use the cross product to compute the area of the parallelogram spanned by $\overset {\boldsymbol {\rightharpoonup }}{\mathbf {a}}$ and $\overset {\boldsymbol {\rightharpoonup }}{\mathbf {b}}$. We just add a $z$-component of $0$ to each vector.

What is the area of the parallelogram spanned by $\left < 1,2 \right >$ and $\left < 3,4 \right >$?

\[ \text {Area} = \answer {2} \]

If $\overset {\boldsymbol {\rightharpoonup }}{\mathbf {a}}$ and $\overset {\boldsymbol {\rightharpoonup }}{\mathbf {b}}$ are two nonzero vectors, and $\theta $ is the angle between them,

\[ \overset {\boldsymbol {\rightharpoonup }}{\mathbf {a}} \boldsymbol \times \overset {\boldsymbol {\rightharpoonup }}{\mathbf {b}} = \overset {\boldsymbol {\rightharpoonup }}{\mathbf {0}} \text { if and only if $\theta = 0$ or $\theta =\pi $}. \]

This follows from our work with parallelograms, since the area of the parallelogram spanned by parallel vectors is zero.

4 Applications

In addition to the geometric applications we have already seen, we can also use the cross product in some physical applications.

4.1 Torque

Imagine turning a wrench. The wrench originates at a point $O$ and terminates at a point $P$. Let $\overset {\boldsymbol {\rightharpoonup }}{\mathbf {r}} = \overrightarrow {OP}$. You apply a force $\overset {\boldsymbol {\rightharpoonup }}{\mathbf {F}}$ to the end of the wrench. If $\overset {\boldsymbol {\rightharpoonup }}{\mathbf {F}}$ points in the same direction as $\overset {\boldsymbol {\rightharpoonup }}{\mathbf {r}}$, the bolt will not twist at all, since you will just be pulling on the handle. If $\overset {\boldsymbol {\rightharpoonup }}{\mathbf {F}}$ is perpendicular to the handle, then we expect quite a bit of twisting to occur.

The torque $\overset {\boldsymbol {\rightharpoonup }}{\mathbf {t}}$ obtained by applying a force $\overset {\boldsymbol {\rightharpoonup }}{\mathbf {F}}$ to a lever arm with position vector $\overset {\boldsymbol {\rightharpoonup }}{\mathbf {r}}$ is given by

\[ \overset {\boldsymbol {\rightharpoonup }}{\mathbf {t}} = \overset {\boldsymbol {\rightharpoonup }}{\mathbf {r}} \boldsymbol \times \overset {\boldsymbol {\rightharpoonup }}{\mathbf {F}}. \]

A pipe of length $3 \mathrm {m}$ has a force of magnitude $2 \mathrm {N}$ applied to it as shown below.

Assuming the $z$-axis is coming towards you (out of the page), is the torque produced in the positive or negative $z$ direction?

positive negative

\[ |\overset {\boldsymbol {\rightharpoonup }}{\mathbf {t}}| = \answer {3\sqrt {3}} \]

4.2 Magnetism

When a charged particle moves through a magnetic field, it experiences a force. If the charge is $q$, the velocity of the particle is $\overset {\boldsymbol {\rightharpoonup }}{\mathbf {v}}$, and the magnetic field is $\overset {\boldsymbol {\rightharpoonup }}{\mathbf {B}}$, then the force is given by

\[ \overset {\boldsymbol {\rightharpoonup }}{\mathbf {F}} = q \overset {\boldsymbol {\rightharpoonup }}{\mathbf {v}} \boldsymbol \times \overset {\boldsymbol {\rightharpoonup }}{\mathbf {B}}. \]

A particle with negative charge $-2$ enters a constant magnetic field given by $\overset {\boldsymbol {\rightharpoonup }}{\mathbf {B}} = \boldsymbol {\hat {\imath }}+2\boldsymbol {\hat {\jmath }}$. The velocity vector of the particle is $\overset {\boldsymbol {\rightharpoonup }}{\mathbf {v}} = \boldsymbol {\hat {k}}+\boldsymbol {\hat {\imath }}$. What is the force acting on the particle?

\[ \overset {\boldsymbol {\rightharpoonup }}{\mathbf {F}} = \answer {4}\boldsymbol {\hat {\imath }}+\answer {-2}\boldsymbol {\hat {\jmath }}+\answer {-4}\boldsymbol {\hat {k}} \]

5 The algebra of the cross product

Below, we summarize some rules for working with cross products.

The following are true for all scalars $s$ and $t$, and vectors $\overset {\boldsymbol {\rightharpoonup }}{\mathbf {u}}$, $\overset {\boldsymbol {\rightharpoonup }}{\mathbf {v}}$, and $\overset {\boldsymbol {\rightharpoonup }}{\mathbf {w}}$ in $\mathbb {R}^3$.

Anticommutative:: $\overset {\boldsymbol {\rightharpoonup }}{\mathbf {v}} \boldsymbol \times \overset {\boldsymbol {\rightharpoonup }}{\mathbf {w}} = -\overset {\boldsymbol {\rightharpoonup }}{\mathbf {w}} \boldsymbol \times \overset {\boldsymbol {\rightharpoonup }}{\mathbf {v}}$
Left-distributive:: $\overset {\boldsymbol {\rightharpoonup }}{\mathbf {u}} \boldsymbol \times (\overset {\boldsymbol {\rightharpoonup }}{\mathbf {v}} +\overset {\boldsymbol {\rightharpoonup }}{\mathbf {w}}) = \overset {\boldsymbol {\rightharpoonup }}{\mathbf {u}} \boldsymbol \times \overset {\boldsymbol {\rightharpoonup }}{\mathbf {v}}+\overset {\boldsymbol {\rightharpoonup }}{\mathbf {u}} \boldsymbol \times \overset {\boldsymbol {\rightharpoonup }}{\mathbf {w}}$
Right-distributive:: $(\overset {\boldsymbol {\rightharpoonup }}{\mathbf {v}}+\overset {\boldsymbol {\rightharpoonup }}{\mathbf {w}}) \boldsymbol \times \overset {\boldsymbol {\rightharpoonup }}{\mathbf {u}} = \overset {\boldsymbol {\rightharpoonup }}{\mathbf {v}} \boldsymbol \times \overset {\boldsymbol {\rightharpoonup }}{\mathbf {u}}+\overset {\boldsymbol {\rightharpoonup }}{\mathbf {w}} \boldsymbol \times \overset {\boldsymbol {\rightharpoonup }}{\mathbf {u}}$
Respects scalar multiplication:: $s\overset {\boldsymbol {\rightharpoonup }}{\mathbf {v}} \boldsymbol \times t\overset {\boldsymbol {\rightharpoonup }}{\mathbf {w}} = st \overset {\boldsymbol {\rightharpoonup }}{\mathbf {v}} \boldsymbol \times \overset {\boldsymbol {\rightharpoonup }}{\mathbf {w}}$
Relation to parallel vectors:: $\overset {\boldsymbol {\rightharpoonup }}{\mathbf {v}} \boldsymbol \times \overset {\boldsymbol {\rightharpoonup }}{\mathbf {v}} = \overset {\boldsymbol {\rightharpoonup }}{\mathbf {0}}$
Orientation:: $\boldsymbol {\hat {\imath }} \boldsymbol \times \boldsymbol {\hat {\jmath }} = \boldsymbol {\hat {k}}$, $\boldsymbol {\hat {\jmath }} \boldsymbol \times \boldsymbol {\hat {k}} = \boldsymbol {\hat {\imath }}$, and $\boldsymbol {\hat {k}} \boldsymbol \times \boldsymbol {\hat {\imath }} = \boldsymbol {\hat {\jmath }}$.

Moreover, these properties determine the cross product uniquely.

We will not prove that the cross product is the only function with these properties, but that is an important point. If you ever wondered where this crazy formula came from, the uniqueness of the cross product is your explanation. If you want these properties, there is only one operation which gives them to you, and it is the cross product. We leave you with the following curious fact. The cross product only exists in $\mathbb {R}^3$ and $\mathbb {R}^7$. While a proof of this fact is beyond the scope of this course, we hope that this mystery encourages you to travel deeper into your studies.

For some interesting extra reading check out:

2025-01-06 20:26:10