Last video, I 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’m 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
We’ll start in two dimensions. If you have two vectors 𝐕 and 𝐖,
think about the parallelogram that they span out. What I mean by that is that if you
take a copy of 𝐕 and move its tail to the tip of 𝐖 and you take a copy of 𝐖 and
move its tail to the tip of 𝐕, the four vectors now on the screen enclose a certain
The cross-product of 𝐕 and 𝐖,
written with the x-shaped multiplication symbol, is the area of this parallelogram,
well almost. We also need to consider
Basically, if 𝐕 is on the right of
𝐖, then 𝐕 cross 𝐖 is positive and equal to the area of the parallelogram. But if 𝐕 is on the left of 𝐖,
then the cross-product is negative, namely, the negative area of that
Notice this means that order
matters. If you swapped 𝐕 and 𝐖, instead
taking 𝐖 cross 𝐕, 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, 𝑖–hat cross 𝑗-hat, the result should be positive.
In fact, the order of your basis
vectors is what defines orientation. So since 𝑖–hat is on the right of
𝑗-hat, I remember that 𝐕 cross 𝐖 has to be positive whenever 𝐕 is on the right
So, for example, with the vector
shown here, I’ll just tell you that the area of that parallelogram is seven. And since 𝐕 is on the left of 𝐖,
the cross-product should be negative, so 𝐕 cross 𝐖 is negative seven. But of course you wanna be able to
compute this without someone telling you the area. This is where the determinant comes
So, if you didn’t see Chapter five
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 2D cross-product 𝐕 cross
𝐖, what you do is, you write the coordinates of 𝐕 as the first column of a matrix
and you take the coordinates of 𝐖 and make them the second column; then you just
compute the determinant.
This is because a matrix whose
columns represent 𝐕 and 𝐖 corresponds with a linear transformation that moves the
basis vectors 𝑖–hat and 𝑗-hat to 𝐕 and 𝐖. 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 𝑖–hat and 𝑗-hat.
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 one.
What’s more, if 𝐕 is on the left
of 𝐖, 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 𝐕 has
coordinates negative three, one and 𝐖 has coordinates two, one. The determinant of the matrix with
those coordinates as columns is negative three times one minus two times one, which
is negative five. So, evidently, the area of the
parallelogram we define is five.
And since 𝐕 is on the left of 𝐖,
it should make sense that this value is negative. As with any new operation you
learn, I’d recommend playing around with this notion a bit 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 𝐕 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 three 𝐕 cross 𝐖 will be exactly three times the value of 𝐕
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 gonna
consider the parallelogram defined by the two vectors that were crossing
together. And the area of this parallelogram
is still gonna 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 gonna 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. Point the forefinger of your right
hand in the direction of 𝐕; then stick out your middle finger in the direction of
𝐖. Then, when you point up your thumb,
that’s the direction of the cross-product.
For example, let’s say that 𝐕 was
a vector with length two pointing straight up in the 𝑧-direction and 𝐖 is a vector
with length two 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
four. So their cross-product should be a
vector with length four.
Using the right-hand rule, their
cross-product should point in the negative 𝑥-direction. So the cross-product of these two
vectors is negative four times 𝑖–hat.
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 𝐕 and 𝐖. But for that first column, you
write the basis vectors 𝑖–hat, 𝑗-hat, and 𝑘-hat. Then you compute the determinant of
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 𝑖–hat, 𝑗-hat, and 𝑘-hat 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 𝐕 and 𝐖 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
it. 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 wanna 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’ll talk about in the next
video are just a really elegant piece of math.