Last video, I've talked about the dot product.

Showing both the standard introduction to the topic,

as well as a deeper view of how it relates to linear transformations.

I'd like to do the same thing for cross products,

which also have a standard introduction

along with a deeper understanding in the light of linear transformations.

But this time I am dividing it into two separate videos.

Here i'll try to hit the main points

that students are usually shown about the cross product.

And in the next video,

I'll be showing a view which is less commonly taught, but really satisfying when you learn

it.

We'll start in two dimensions.

If you have two vectors v̅ and w̅,

think about the parallelogram that they span out

What i mean by that is,

that if you take a copy of v̅

and move its tail to the tip of w̅,

and you take a copy of w̅

And move its tail to the tip of v̅,

the four vectors now on the screen enclose a certain parallelogram.

The cross product of v̅ and w̅,

written with the X-shaped multiplication symbol,

is the area of this parallelogram.

Well, almost. We also need to consider

orientation. Basically, if v̅ is on the

right of w̅, then v̅×w̅ is positive

and equal to the area of the

parallelogram. But if v̅ is on the left of w̅,

then the cross product is negative,

namely the negative area of that

parallelogram. Notice this means that

order matters. If you swapped v̅ and w̅

instead taking w̅×v̅, the cross

product would become the negative of

whatever it was before. The way I always

remember the ordering here is that when

you take the cross product of the two

basis vectors in order, î×ĵ,

the results should be positive. In fact,

the order of your basis vectors is what

defines orientation so since î is on

the right of ĵ, I remember that v̅×w̅

has to be positive whenever v̅ is

on the right of w̅.

So, for example with the vector shown

here, I'll just tell you that the area of

that parallelogram is 7. And since v̅

is on the left of w̅, the cross product

should be negative so v̅×w̅ is -7.

But of course you want to be able to

compute this without someone telling you

the area. This is where the determinant comes in.

So, if you didn't see Chapter 5 of this

series, where I talk about the

determinant now would be a really good

time to go take a look.

Even if you did see it, but it was a while

ago. I'd recommend taking another look

just to make sure those ideas are fresh in your mind.

For the 2-D cross-product v̅×w̅,

what you do is you write the coordinates

of v̅ as the first column of the matrix

and you take the coordinates of w̅ and

make them the second column then you

just compute the determinant.

This is because a matrix whose columns

represent v̅ and w̅ corresponds with a

linear transformation that moves the

basis vectors î and ĵ to v̅ and w̅.

The determinant is all about measuring

how areas change due to a transformation.

And the prototypical area that we look

at is the unit square resting on î and ĵ.

After the transformation,

that square gets turned into the

parallelogram that we care about.

So the determinant which generally

measures the factor by which areas are

changed, gives the area of this

parallelogram; since it evolved from a

square that started with area 1.

What's more if v̅ is on the left of w̅, it

means that orientation was flipped

during that transformation, which is what

it means for the determinant to be negative.

As an example let's say v̅ has

coordinates negative (-3,1) and w̅ has

coordinates (2,1). The determinant of the

matrix with those coordinates as columns

is (-3·1) - (2·1),

which is -5. So evidently the

area of the parallelogram we define is 5

and since v̅ is on the left of w̅, it

should make sense that this value is

negative. As with any new operation you learn

I'd recommend playing around with this

notion of it in your head just to get

kind of an intuitive feel for what the

cross product is all about.

For example you might notice that when

two vectors are perpendicular or at

least close to being perpendicular their

cross product is larger than it would be

if they were pointing in very similar

directions. Because the area of that

parallelogram is larger when the sides

are closer to being perpendicular.

Something else you might notice is that

if you were to scale up one of those

vectors, perhaps multiplying v̅ by three

then the area of that parallelogram is

also scaled up by a factor of three.

So what this means for the operation is

that 3v̅×w̅ will be exactly three

times the value of v̅×w̅ .

Now, even though all of this is a

perfectly fine mathematical operation

what i just described is technically not

the cross-product. The true cross product

is something that combines two different

3D vectors to get a new 3D vector. Just as before,

we're still going to consider the

parallelogram defined by the two vectors

that were crossing together. And the area

of this parallelogram is still going to

play a big role. To be concrete let's say

that the area is 2.5 for the vectors

shown here but as I said the cross

product is not a number it's a vector.

This new vector's length will be the area

of that parallelogram which in this case

is 2.5. And the direction of that new

vector is going to be perpendicular to

the parallelogram. But which way!, right?

I mean there are two possible vectors with

length 2.5 that are perpendicular to a given plane.

This is where the right hand rule comes

in. Put the fore finger of your right hand

in the direction of v̅ then stick out

your middle finger in the direction of w̅.

Then when you point up your thumb, that's the

direction of the cross product.

For example let's say that v̅ was a

vector with length 2 pointing straight

up in the Z direction and w̅ is a vector

with length 2 pointing in the pure Y

direction. The parallelogram that they

define in this simple example is

actually a square, since they're

perpendicular and have the same length.

And the area of that square is 4. So

their cross product should be a vector

with length 4. Using the right hand

rule, their cross product should point in the negative X direction.

So the cross product of these two

vectors is -4·î.

For more general computations,

there is a formula that you could

memorize if you wanted but it's common

and easier to instead remember a certain

process involving the 3D determinant.

Now, this process looks truly strange at

first. You write down a 3D matrix where

the second and third columns contain the

coordinates of v̅ and w̅. But for that

first column you write the basis vectors

î, ĵ and k̂. Then you compute

the determinant of this matrix. The

silliness is probably clear here.

What on earth does it mean to put in a

vector as the entry of a matrix?

Students are often told that this is

just a notational trick. When you carry

out the computations as if î, ĵ and k̂

were numbers, then you get some

linear combination of those basis vectors.

And the vector

defined by that linear combination, students

are told to just believe, is the unique

vector perpendicular to v̅ and w̅ whose

magnitude is the area of the appropriate

parallelogram and whose direction obeys

the right hand rule.

And, sure!. In some sense this is just a

notational trick. But there is a reason

for doing in.

It's not just a coincidence that the

determinant is once again important. And

putting the basis vectors in those slots

is not just a random thing to do. To

understand where all of this comes from

it helps to use the idea of duality that

I introduced in the last video.

This concept is a little bit heavy

though, so I'm putting it in a separate

follow-on video for any of you who are

curious to learn more.

Arguably it falls outside the essence of

linear algebra. The important part here

is to know what that cross product

vector geometrically represents. So if

you want to skip that next video, feel

free. But for those of you who are

willing to go a bit deeper and who are

curious about the connection between

this computation and the underlying

geometry, the ideas that I will talk about

in the next video or just a really

elegant piece of math.