### Video Transcript

In this video, weβre going to learn
about products of vectors, what they are, why they are useful, and how to calculate
them. To give an idea of how useful
products of vectors are in describing physical phenomena, imagine a situation where
youβre moving a large heavy box across the floor. You push up on the box a bit such
that the force vector points above the horizontal. Over time, youβre able to push the
box along the floor such that it ends up with the horizontal displacement vector
π. If we want to calculate how much
work is done in moving the box that distance, weβll need to make use of a product of
vectors. Or imagine a different
scenario. In this one, weβre sending a screw
into a block of wood. We use a screwdriver to apply a
force πΉ, a displacement of π away from the screwβs axis of rotation. By using a particular vector
product, weβre able to tell which direction, whether into or out of the board, the
screw goes.

As we start out, it can be helpful
to contrast vector products with products we may be more familiar with. And thatβs the products of
scalars. Recall that a scalar is a quantity
that has magnitude but no direction to it. Examples of scalars would be
numbers such as five kilograms or 10 seconds or even 65 degrees. Each of these values has a
magnitude without an associated direction. If we wanna find the product of two
scalars, say five kilograms and 10 seconds. Then we simply multiply them as
weβre used to multiplying scalar values. The result of this multiplication
would be 50 kilogram-seconds. Vectors, of course, are quantities
that have both magnitude and direction. Examples of vectors might be 10
meters to the east or two kilometers down.

Now if someone said what is the
product of these two vectors, we might not be quite so confident as how to combine
them together as we were with scalars. After all, how would we go about
combining the different directions of east and down? It turns out that vector products
come in two particular types. That let us navigate this tricky
question of how to combine unlike things.

The first approach we have to
finding the product of two vectors, we can call them π΄ and π΅, is to take whatβs
called the dot product. It looks like a dot placed between
those two vectors. If we had vector π΄ and vector π΅
with their tails at the same location. When we take their dot product,
what weβre doing is finding out how much of the vectors lie along one another. Thatβs the geometrical meaning of
the dot product. And if we give the name π to the
angle between these two vectors π΄ and π΅, we can write that the dot product of π΄
and π΅ is equal to the product of their magnitudes multiplied by the cos of that
angle π. So the dot product is useful to us
in finding the angle between vectors. With this mathematical relationship
in front of us, we can try out a few simple cases to explore this equation.

Imagine we had two vectors, πΆ and
π·, that were perpendicular to one another. When we consider that the dot
product means the overlap of two vectors β that is, how much of one lies along the
other. By looking at this diagram, we
might naturally guess that the dot product of πΆ and π· is zero since theyβre
perpendicular. This intuition is confirmed when we
consider the equation for dot product. We know that the cos of 90 degrees,
which is the angle between πΆ and π·, is zero. So weβre correct in saying that
this dot product itself is zero.

Going just a bit further, imagine
that we are given π΄ and π΅ in their component forms, not graphically but
numerically. And in this case, both vectors have
three dimensions to them: π, π, and π. If we wanted to take the dot
product of these two vectors, how would we do that mathematically? This gets a bit of the question we
talked of earlier, of how do we combine or sort out unlike types as we have with
different unit vectors in this example. The answer to this is that we
multiply our vectors according to their type. π΄ π₯ times π΅ π₯ plus π΄ π¦ times
π΅ π¦ plus π΄ π§ times π΅ π§, keeping the different vector components separate from
one another. Then finally add those different
component products together to find the dot product, which will ultimately be a
scalar. Both these relationships are
helpful to keep in mind when weβre working with dot products.

The second type of vector product
is called a cross product. And it is symbolized with an x or a
cross symbol in between our test vectors, π΄ and π΅. While the dot product creates a
scalar, taking a cross product of two vectors creates a vector. Another difference with the dot
product is that the dot product is commutative. Meaning that π΄ dot π΅ is equal to
π΅ dot π΄. But the cross product is not like
that. That is, in general, π΄ cross π΅ is
not equal to π΅ cross π΄. So the order is very important when
we calculate cross products. If we again had two vectors, vector
π΄ and vector π΅, then we can understand the physical meaning of the cross product
this way. When we cross π΄ into π΅, we are
solving for the components that are perpendicular to the plane in which vector π΄
and vector π΅ lie. For example, if our two vectors π΄
and π΅ were in a plane with π- and π-components, then their cross product would
have no components in that direction. It would be entirely either into
the page or out of the page.

So when we think of cross products,
think of results that are perpendicular to the input vectors we use. There is a very helpful
mathematical tool for calculating cross products. The tool involves a three-by-three
matrix of values, where we take the determinant of that matrix in order to solve for
the cross product. The three common unit vectors π,
π, and π are the head of each column. And the two vectors weβre crossing
together, π΄ and π΅, have their components listed as the last two rows of this
matrix. If we were to calculate the
determinant of this matrix, weβd find a result with an π-, π-, and
π-component. In other words, our result would be
a vector. The particular value for each of
those three components depends on the values of π΄ and π΅. But notice as we look at the three
different component results, the components of π΄ and π΅ that go into each one are
perpendicular to that component direction.

For example, with π, the only
components of π΄ and π΅ that affect π are in the π- and π-directions. In a similar way, itβs the
components in the π- and π- directions that determine the magnitude of the
π-component. And itβs the components of π΄ and
π΅ in the π- and π-direction that determine the magnitude of the π-component. This confirms what we spoke of
earlier where the cross product has to do with results perpendicular to the
direction of the input vectors. With all that as background, letβs
try a couple of example problems having to do with vector products.

Find the angle between the two
vectors π² equals 2.0π’ plus 4.0π£ plus 8.0π€ and π³ equals 6.0π’ plus 4.0π£ plus
6.0π€.

If we call the angle between these
two vectors π, itβs that we want to solve for. To start out, we can recall that
the dot product between two vectors, π and π, is equal to the product of their
magnitudes times the cosine of the angle between them. And we can recall further that the
dot product is also equal to the product of the π₯-components plus the product of
the π¦-components plus the product of the π§-components of our two vectors. Bringing those two relationships
together, we can write that, in our case, the product of the π₯-components of our
vectors plus the product of the π¦-components plus the product of the
π§-components. Is equal to the product of their
magnitudes multiplied by the cos of π, the angle between them. If we divide both sides of this
equation by the magnitude of π² times the magnitude of π³ and then take the inverse
cosine of both sides. So π, the angle weβre interested
in, is equal to the inverse cos of π¦ π₯, π§ π₯ plus π¦ π¦, π§ π¦ plus π¦ π§, π§ π§
divided by the product of the magnitudes of our two vectors.

We can recall that the magnitude of
a vector, when it has three dimensions, is equal to the square root of the
π₯-dimension squared plus the π¦-dimension squared plus the π§-dimension
squared. If we insert the magnitude
expansion for both π¦ and π§, then we now have an expression for π entirely in
terms of the components of our two vectors. In the case of our vector π², 2.0
is π¦ π₯, 4.0 is π¦ π¦, and 8.0 is π¦ π§. And for π³, 6.0 is π§ π₯, 4.0 is π§
π¦, and 6.0 is π§ π§. When we plug each of these values
in where it fits in our equation. With all these values plugged in,
when we enter this expression on our calculator, we find π is 28 degrees. Thatβs the angle between our two
vectors π² and π³.

Thatβs an example using the dot
product combination of vectors. Now letβs try an example using the
cross product.

Calculate the cross product of π²
equals two π’ plus four π£ plus eight π€ and π³ equals six π’ plus four π£ plus two
π€.

In this example, we want to solve
for π² cross π³, where π² and π³ are both three-dimensional vectors. We can begin our solution by
recalling the mathematical definition for a cross product. The cross product of two vectors,
we can call them π and π, is equal to the determinant of a three-by-three matrix ,
where the columns are headed by the three unit vectors π’, π£, and π€. And the last two rows are populated
by the respective components of vectors π and π. We can apply this formula to our
vectors π² and π³. As we set up our matrix, we know
the top row will be the π’, π£, and π€ unit vectors. The next row will be the respective
components of the vector π². Looking at π², we see that itβs
π₯-component is two, its π¦-component is four, and its π§-component is eight. So we write those values in to our
matrix. In the next row, weβll write the
components of π³. The π₯-component of π³ is six. Its π¦-component is four. And its π§-component is two.

Now weβre ready to calculate the
determinant of this matrix and solve for the cross product of π² cross π³. When we compute this determinant,
we find that the π’-component is eight minus 32, or negative 24. The π£-component is four minus 48,
or negative 44. And the π€-component is eight minus
24, which equals negative 16. Our overall cross product then is
negative 24π’ minus 44π£ minus 16π€. Thatβs the cross product of the
vectors π² and π³.

To summarize, products of vectors
are useful because they help us to understand and themselves reveal physical
phenomena. These could include the direction
to point a windmill or the best way to get power out of an engine. There are two types of products of
vectors that weβve learned: the dot product, or the scalar product, and the cross
product, which produces a vector. And finally, when calculating
vector products, matrices can be helpful. For example, the dot product of π
and π can be expressed as a combination of two matrices. And the cross product of π and π
can be expressed as the determinant of a three-by-three matrix.