### Video Transcript

In this video, weβre talking about
the vector product of two vectors. As weβll see, this is a method for
taking two vectors that lie in the π₯- and π¦-plane and combining them to yield a
third vector in the π§-dimension.

To get started, we can recall that
there are a number of ways of combining vectors. Itβs possible to add vectors
together or to subtract one from another. And we can also combine vectors
using whatβs called a scalar or dot product. In this lesson, weβre talking about
vector products which are unique in that a vector product is a way of multiplying
two vectors together so that a vector, a quantity with magnitude and direction,
results. In other words, a vector product
creates another vector.

To see how this works
mathematically, letβs give these two vectors names. Letβs say the vector in blue is
vector π and that in red is vector π. Without knowing anything more about
π and π, we can write down a general rule for their vector product. We can represent the vector product
of π with π this way, using an Γ or a cross in between the two vector symbols.

Now, the equation weβre going to
write applies specifically to vectors that lie in the π₯π¦-plane. This means that we can write vector
π and vector π mathematically like this. We can represent π as the sum of a
component in the π₯-direction and a component in the π¦-direction. And we can represent vector π in a
similar way. And note these vector symbols that
apply in our definitions for vectors π and π. In both cases, we have a vector
called π’ and a vector called π£.

We can recall that these are unit
vectors for the π₯- and π¦-axes, respectively. Note, by the way, that when we
include a third dimension and call it π§, the unit vector for that dimension is
typically represented by the letter π€. Anyway, when we write π and π in
terms of their components this way, weβre then able to take the next step and fill
out the right-hand side of this vector product equation. We do that using the π₯- and
π¦-components of π and π.

If we take the π₯-component of π,
thatβs this component here, and multiply it by the π¦-component of π, thatβs this
component here. And then, subtract from that the
π¦-component of π here times the π₯-component of π here. Then we will have the correct
expression for the magnitude of our vector product. But remember we said that a vector
product results in another vector. And thatβs where this π€ unit
vector comes in. This gives the magnitude of π π₯
times π π¦ minus π π¦ times π π₯ a direction.

So indeed, we are calculating a
vector. And notice something about that
direction. While π and π were both entirely
in the π₯π¦-plane and, therefore, had only π’- and π£-components, the vector product
of π and π is entirely in a different dimension. Itβs in the π€-direction. Itβs always the case that a vector
product of two vectors is perpendicular to those two original vectors. Now, letβs get some practice
applying this equation using these two vectors here.

Say that we locate π and π on a
set of coordinate axes. And we label the π₯- and π¦-axes
using these numbers, but we donβt specify the units they represent. For our purposes, those units
arenβt important, so weβll leave them undefined. Studying these axes, we can see
that vector π has an extension of one unit in the π₯-direction. And then, it extends five units in
the positive π¦-direction. This means we can write π as π’
plus five times π£. Likewise, with vector π, we see
that this extends four units in the positive π₯-direction and two units in the
positive π¦. So, π is equal to four times π’
plus two times π£.

Now that we know the π₯- and
π¦-components of π and π, we can calculate their vector product. Specifically, letβs calculate π
cross π. We see, based on our vector product
equation, that this is equal to the π₯-component of π, thatβs one in our case,
times the π¦-component of π, thatβs equal to two based on our definition of vector
π. And then, from this, we subtract
the π¦-component of π, thatβs equal to five, multiplied by the π₯-component of π,
and that is equal to four.

And just as we saw for our general
vectors that lay in the π₯π¦-plane, so these specific vectors π and π whose
components we know, according to this set of coordinate axes, will yield a vector
product that points either in the positive or the negative π€-direction. In other words, this vector product
will point somewhere along the π§-dimension. And it will have no component in
the π₯- or π¦-dimensions.

So then, looking at this equation,
we can now calculate whatβs inside the parentheses in blue. One times two is two. And five times four is 20. So, we have two minus 20 or
negative 18π€ as a resulting vector. On our coordinate axes then, if we
represented the negative π§-direction using this dashed line, then the vector
product of π and π would point 18 units along that line.

Now, thereβs something very
important to remember whenever weβre computing a vector product. And that is that the order of the
vectors we use is very important. Notice in this equation that when
we cross π with π, π coming first, then itβs the π₯-component of that first
vector which multiplies the π¦-component of the second one. And then, in the next term, the
π¦-component of that first vector times the π₯-component of the second. If we were to reverse these two
vectors at the outset β instead of taking π cross π, we take π cross π β then we
would calculate a different result on the right-hand side.

To see that thatβs so, letβs do
this. Weβve already calculated π cross
π using these specific π and π vector examples. Letβs now reverse the order. Letβs take the vector product of π
and π. As we do this, it will be a bit
confusing to refer to our reference equation over here because this equation has a
vector called π coming first and a vector called π coming second. But if weβre careful about it,
weβll be able to work it out and calculate π cross π.

Okay, the first term in a
vector-product calculation is the π₯-component of the first vector. In the case of our specific
calculation, that first vector is π and its π₯-component is four. That value is then multiplied by
the π¦-component of the second vector. Again, in our case, that second
vector is π and its π¦-component is five. So, we have four times five.

And from this, we subtract the
π¦-component of our first vector. The π¦-component of vector π over
here is two times the π₯-component of our second vector. Our second vector in this
calculation is vector π and its π₯-component is one. And now, when we calculate the
value inside the blue parentheses, we have four times five, or 20, minus two times
one, or two. And 20 minus two is 18.

As we compare this result with our
result for calculating π cross π, we can notice two things. First, these results are not the
same. And as it turns out, this is
generally true. That is, itβs generally the case
that π cross π is not equal to π cross π. But notice something else about
these results. These two vector products have the
same magnitude, 18 units, and they point in opposite directions. This also is true in general.

And we can write that this way. We can say that the vector product
of π and π is equal to negative the vector product of π and π. As a corollary of this, by the way,
we can write that the magnitude of π cross π is equal to the magnitude of π cross
π. So, there are a lot of equations
here, but weβre discovering important relations for vector products. Now that weβve calculated the
vector products of our two vectors π and π on this graph, letβs clear some space
for a new calculation.

What would happen if we were to
compute the vector product of two unit vectors? That is, what if we were to take
the unit vector π’ and then cross π’ with the unit vector π£? We know that technically both of
these are vectors, so this should work. And in fact, just like vectors π
and π, we can write out the π₯- and π¦-components of these unit vectors π’ and
π£.

Looking down at our diagram at the
bottom left, we can write the unit vector π’ this way. The unit vector π’ has one unit of
magnitude in the π’-direction, the π₯-dimension. And then, it doesnβt point at all
in the π£-direction along the π¦-axis. We could say then that π’ is equal
to one π’ plus zero π£, and similarly with the unit vector π£. This vector has zero magnitude in
the π’-direction or the π₯-dimension. And it has a magnitude of one in
the π£-direction.

With π’ and π£ written out
according to their π₯- and π¦-components, we can now calculate π’ cross π£. Itβs equal to the π₯-component of
the first vector, thatβs π’. And that π₯-component is one
multiplied by the π¦-component of the second vector. That second vector is π£, and its
π¦-component is one as well. And then, from that product, we
subtract the π¦-component of the first vector. Thatβs vector π’. We see that π¦-component is
zero. And we multiply that by the
π₯-component of the second vector. That second vector is π£. And that π₯-component is zero as
well.

So now, calculating the result of
this, we have one times one, or one, minus zero times zero, which is zero. And when this whole expression
simplifies, we find that π’ cross π£ is equal to the π€ unit vector. And we can write this equation in
green because itβs generally true. Knowing this, what if we wanted to
reverse the order and calculate the vector product of π£ and π’?

To do that, we could once more go
through this calculation. But letβs not forget about the
vector product relationships we discovered, in particular this one, which says the
vector product of π and π equals the negative of the vector product of π and
π. This saves us some calculation
work. Knowing that π’ cross π£ is equal
to π€, we know that π£ cross π’ is equal to the negative of that, negative π€. So, as strange as it may seem,
indeed we can calculate a vector product of unit vectors, and the result is a vector
thatβs perpendicular to both π’ and π£.

Now, thereβs one more thing weβll
want to point out about vector products. And it has to do with the magnitude
of those products. Say that we return to our graph of
vectors π and π and that, for now, we ignore the π§-dimension. So, we have these two vectors in
the π₯π¦-plane, π and π. And say that we want to calculate
the magnitude of their vector product. It turns out that weβre able to do
that given three pieces of information. First, we need to know the
magnitude of π. Second, we need to know the
magnitude of π. And third, weβll need to know the
angle between these two vectors. And we can call this angle π.

So, hereβs how this works. The magnitude of the vector product
of π and π is equal to the magnitude of π times the magnitude of π multiplied by
the sin of the angle between these two vectors. This relationship can come in very
useful when we know the magnitudes of our two vectors but donβt know their component
parts. We wonβt need to know those
component parts to calculate the magnitude of the vector product so long as we also
know the angle between the two vectors. We can understand a bit more about
this equation by considering this sin π factor here.

We know that if the angle π is
zero degrees and that would indicate that π and π are parallel vectors, then since
the sin of zero is zero, we know that would result in this whole vector product
being zero. In other words, for parallel
vectors, their vector product is zero. But then, what if instead of being
a zero-degree angle, π is a 90-degree angle? Well, we know that the sin of 90
degrees is equal to one. And that would lead to a maximum
possible value for the magnitude of π times the magnitude of π times the sin of
the angle between them.

This tells us that vectors that are
perpendicular to one another at 90 degrees have the largest possible vector product
magnitude, given their individual magnitudes. So then, just by knowing how the
sin of π behaves, we can understand a bit more about this equation. Knowing all this about vector
products, letβs try out an example exercise.

Consider the two vectors π equals
three π’ plus two π£ and π equals five π’ plus eight π£. Calculate π cross π.

Well, we see this exercise is about
computing a vector product between these two vectors π and π. Weβve been given those two vectors
in component form. And we see that each one has an
π’-component, that is a component along the π₯-axis, as well as a π£-component, some
component along the π¦-dimension. We see then that both of these
vectors, π and π, lie in the π₯π¦-plane. In fact, if we were to sketch out
π₯- and π¦-axes, then we can draw in vectors π and π on this graph.

Vector π extends three units in
the positive π₯-direction and two units in the positive π¦, giving a vector like
this. While vector π extends five units
in the positive π₯-direction and then eight units in the positive π¦-direction. And that vector looks like
this. Now, to calculate this vector
product, π across π, weβll want to recall the mathematical form of the vector
product of two vectors that lie entirely in the π₯π¦-plane like π and π do.

If we call two general vectors that
both are constrained to lie in the π₯π¦-plane π and π. Then the vector product of π and
π, also called the cross product of π and π, is equal to the π₯-component of π
times the π¦-component of π minus the π¦-component of π times the π₯-component of
π. Note that a vector product of two
vectors results in a vector. That is, it has both magnitude as
well as direction. And since our vectors π and π
both lie in the π₯π¦-plane, their vector product points perpendicularly to that, in
the π€-direction.

Letβs now use this relationship
along with π and π written in their component form to calculate π cross π. This vector product is equal to the
π₯-component of π times the π¦-component of π minus the π¦-component of π times
the π₯-component of π. And this will also point in the
π€-direction, either positive or negative. Looking at these different terms in
our parentheses, we can see that the π₯-component of π is three, that the
π¦-component of π is eight and then the π¦-component of π is two and the
π₯-component of π is five.

When we substitute in these values,
our next task is to calculate their result. Three times eight is 24, and two
times five is 10. And then, 24 minus 10 is equal to
14. So, our vector product points 14
units in the positive π€-direction. So, if we drew in a π§-axis on our
graph, where that axis pointed out of the screen at us, then this vector product
would point 14 units toward us along that direction. This is the vector product of π
and π.

Letβs summarize now what weβve
learned about the vector product of two vectors. In this lesson, we learned that the
vector product of two vectors is another vector. Along with this, we saw that if we
have two vectors π and π that lie entirely in the π₯π¦-plane. Then the vector product of π and
π is equal to the π₯-component of π times the π¦-component of π minus the
π¦-component of π times the π₯-component of π. And that this vector product points
in the π€-unit vector direction.

We also learned that the order in
which two vectors appear in a vector product makes a difference. The vector product, also called the
cross product, of π and π is equal to negative the vector product of π and
π. Along with this, we discovered
relationships among the unit vectors themselves. That the vector product of π’ and
π£ is equal to the π€ unit vector, while the vector product of π£ and π’ is equal to
negative that vector.

And lastly, we saw that if we have
two vectors π and π in the π₯π¦-plane that are separated by an angle π, then the
magnitude of their vector product is equal to the magnitude of the individual
vectors times the sin of π.