# Cross products | Essence of linear algebra, Chapter 10

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
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
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
kind of an intuitive feel for what the
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