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## Multivariable calculus

### Course: Multivariable calculus>Unit 1

Lesson 2: Vectors and matrices

# Cross products

Learn about what the cross product means geometrically, along with the right-hand rule and how to compute a cross product.
Like the dot product, the cross product is an operation between two vectors. Before getting to a formula for the cross product, let's talk about some of its properties.

## Properties of the cross product

We write the cross product between two vectors as $\stackrel{\to }{a}×\stackrel{\to }{b}$ (pronounced "a cross b"). Unlike the dot product, which returns a number, the result of a cross product is another vector. Let's say that $\stackrel{\to }{a}×\stackrel{\to }{b}=\stackrel{\to }{c}$. This new vector $\stackrel{\to }{c}$ has a two special properties.
First, it is perpendicular to both $\stackrel{\to }{a}$ and $\stackrel{\to }{b}$. Phrasing this in terms of the dot product, we could say that $\stackrel{\to }{c}\cdot \stackrel{\to }{a}=\stackrel{\to }{c}\cdot \stackrel{\to }{b}=0$. This property alone makes the cross product quite useful. This is also why the cross product only works in three dimensions. In 2D, there isn't always a vector perpendicular to any pair of other vectors. In four and more dimensions, there are infinitely many vectors perpendicular to a given pair of other vectors.
Second, the length of $\stackrel{\to }{c}$ is a measure of how far apart $\stackrel{\to }{a}$ and $\stackrel{\to }{b}$ are pointing, augmented by their magnitudes.
$‖\stackrel{\to }{c}‖=‖\stackrel{\to }{a}‖‖\stackrel{\to }{b}‖\mathrm{sin}\left(\theta \right)$
It's similar to the dot product, but instead of $\mathrm{cos}\left(\theta \right)$ the cross product uses $\mathrm{sin}\left(\theta \right)$, where $\theta$ is the angle between $\stackrel{\to }{a}$ and $\stackrel{\to }{b}$. That way, when the angle is $90$ degrees, the cross product is at its largest. In this sense, the dot product and the cross product complement each other.
There's one interpretation of the length of $\stackrel{\to }{c}$ that is particularly useful. Think of the parallelogram formed by $\stackrel{\to }{a}$ and $\stackrel{\to }{b}$. The base of this parallelogram has length $‖\stackrel{\to }{a}‖$, and the height has length $‖\stackrel{\to }{b}‖\mathrm{sin}\left(\theta \right)$. That means the area of the parallelogram in total is precisely the magnitude of the cross product.

## The right-hand rule

Notice that in the image above the cross product is perpendicular to $\stackrel{\to }{a}$ and $\stackrel{\to }{b}$, as expected. But there are actually two vectors that could be perpendicular to $\stackrel{\to }{a}$ and $\stackrel{\to }{b}$. If $\stackrel{\to }{c}=\stackrel{\to }{a}×\stackrel{\to }{b}$, then these two choices are $\stackrel{\to }{c}$ and $-\stackrel{\to }{c}$. How do we decide which of the two perfectly valid choices is the cross product?
We have a convention called the right-hand rule to resolve this ambiguity. If you hold up your right hand, point your index finger in the direction of $\stackrel{\to }{a}$, and point your middle finger in the direction of $\stackrel{\to }{b}$, then your thumb will point in the direction of $\stackrel{\to }{a}×\stackrel{\to }{b}$.
It's arbitrary that we define the cross product with the right-hand rule instead of a left-hand rule, but by using this convention the cross product no longer has any ambiguity.

## The not so pretty formula

The most important takeaway from the cross product should be its properties, not its formula. But sometimes we do need to compute a cross product. Unfortunately, the formula for the cross product is not as nice as it was for the dot product. When we get to the article on determinants, we'll see a nicer way to remember the formula for the cross product. For now:
$\begin{array}{rl}\stackrel{\to }{a}& =\left({a}_{1},{a}_{2},{a}_{3}\right)\\ \\ \stackrel{\to }{b}& =\left({b}_{1},{b}_{2},{b}_{3}\right)\\ \\ \stackrel{\to }{a}×\stackrel{\to }{b}& =\left[\begin{array}{c}{a}_{2}{b}_{3}-{a}_{3}{b}_{2}\\ \\ {a}_{3}{b}_{1}-{a}_{1}{b}_{3}\\ \\ {a}_{1}{b}_{2}-{a}_{2}{b}_{1}\end{array}\right]\end{array}$
Let's try an example using the formula.
Problem 1
$\left(3,0,2\right)×\left(-1,4,2\right)=$
$\left($
$,$
$,$
$\right)$

Now let's see one of those properties we discussed in action.
Problem 2
One of the vectors we took the cross product of was $\stackrel{\to }{a}=\left(3,0,2\right)$. Let $\stackrel{\to }{c}$ be the result of the cross product from above.
What is $\stackrel{\to }{a}\cdot \stackrel{\to }{c}$?

## Cross product vs. dot product

When we compare the dot product and the cross product, there are three main differences.
1. The dot product returns a number, but the cross product returns a vector.
2. The dot product works in any number of dimensions, but the cross product only works in 3D.
3. The dot product measures how much two vectors point in the same direction, but the cross product measures how much two vectors point in different directions.
This is all we need to know about cross products for now. If you want to learn more, check out this video.

## What's next

Now that we have a solid foundation in vectors and the ways we can combine them, the last topic we'll cover is matrices. The next three articles will describe what matrices are, how to visualize them, and a useful property they have called the determinant.

## Want to join the conversation?

• at the problem 1, are we sure the 2nd value of the answer is -8? because it seems wrong, (I did a1*b3-(a3*b1) to find it)
• The 2nd value should be (a3*b1)-(a1*b3); you subtracted in the wrong order.

A useful way to think of the cross product <a1, a2, a3> x <b1, b2, b3> is the determinant of the 3 by 3 matrix

i j k
a1 a2 a3
b1 b2 b3

Note that the coefficient on j is -1 times the determinant of the 2 by 2 matrix
a1 a3
b1 b3

So the 2nd value is -[(a1*b3)-(a3*b1)] = (a3*b1)-(a1*b3).

Have a blessed, wonderful day!