### Video Transcript

In this video, weβre talking about
the cross product in 3D. This is a mathematical operation
that we apply to two vectors in three-dimensional space. Weβre going to learn how to do
this, why itβs useful, as well as the geometric meaning behind the cross
product.

As we get started, the first thing
weβll say about this operation is that it applies to two vectors that may have as
many as three dimensions each. For example, imagine that we have
two vectors π and π, each with an π’-, π£-, and π€-component. The three-dimensional cross product
combines vectors like these, and what results is itself a vector. And not only that, this vector is
perpendicular to the original too.

So if we took the cross product of
vectors π and π, and thatβs represented by this symbol Γ like this, then this
operation would give us a third vector, weβve called it π, thatβs perpendicular to
π and π. So say that this here is vector π
and this is vector π. And even though both are
three-dimensional, we can imagine viewing them from a perspective perpendicular to
both vectors. Viewed this way, the vector π
resulting from π cross π would point out of the screen at us in this
direction.

And as far as the magnitude of
vector π goes, if we were to make a parallelogram where two adjacent sides are our
two input vectors π and π, then the area of that parallelogram is equal to the
magnitude of π cross π. And itβs this area that tells us
the magnitude of π. And that tells us one more thing
about the three-dimensional cross product, that when our two input vectors define
adjacent sides of a parallelogram, that shapeβs area is equal to the magnitude of
the cross product of those vectors. Carrying out a cross product in 3D
allows us to compute a number of useful physical quantities, for example, torque or
angular momentum.

Now that weβve looked into what the
cross product means, letβs see how we actually go about calculating it. The way we mathematically represent
a three-dimensional cross product is using a three-by-three matrix. In the first row of this matrix, we
have the three orthogonal unit vectors π’, π£, and π€. In the next row, we have the
corresponding components of the first vector listed in our cross product, in this
case vector π. And in the last row, we have those
same components of our second vector.

Notice then that when we take a
cross product, the order of the vectors we input matters. In general, π cross π is not the
same as π cross π. Now, note that weβve explicitly
indicated taking the determinant of our three-by-three matrix. Itβs common though to represent the
same mathematical operation this way. And it means the same thing. When we calculate this cross
product, weβll end up with a three-dimensional vector. The way we compute this result is
component by component.

If we first calculate the
π’-component, this involves multiplying that unit vector by the determinant of the
two-by-two matrix that remains if we cross out the row and column that contain the
π’ hat vector. And we can see that that
determinant is equal to ππ¦ times ππ§ minus ππ§ times ππ¦. Whatever that works out to you,
that is the π’-component of our resulting vector.

We next move on to calculating the
π£-component. This involves a similar process,
but note that we precede this component with a negative sign. Finding this componentβs magnitude
involves calculating the determinant once again of a two-by-two matrix, this time
with these four elements. We see thatβs equal to ππ₯ times
ππ§ minus ππ§ times ππ₯. And lastly, we calculate the π€ hat
component.

Once more, crossing out the row and
column in which this element appears, we find this component as the determinant of
this two-by-two matrix, ππ₯ times ππ¦ minus ππ¦ times ππ₯. And with that, weβve calculated the
cross product π cross π.

And recall that earlier we claimed
that this result is perpendicular to both of our input vectors. To see if that holds true, letβs
try this operation out on a test case. Say that we have two vectors π one
and π two, where π one is the π’ hat unit vector and π two is the π£ hat unit
vector. If we take the cross product π one
crossed with π two, then according to our formula, thatβs the determinant of a
three-by-three matrix, where the second row and the third row, which weβll fill in,
are the respective components of π one and π two.

Since π one is simply the π’ unit
vector, its magnitude in that direction is one and its magnitude in the other two
directions is zero. Similarly, for π two, its
magnitude in the π’ hat direction is zero. Its magnitude in the π£ hat
direction is one and zero in the π€ hat. Calculating the resulting
π’-component of this cross product, itβs equal to the determinant of this matrix
here. We see that equals zero times zero,
which is zero. And from that, we subtract zero
times one, also zero. This tells us that the cross
product has no π’-component itself. If we then calculate the
π£-component, thatβs equal to negative one times one times zero, which is zero,
minus zero times zero. So this cross product also has no
π£-component to it.

Lastly, we look at the
π€-component, equal to the determinant of this two-by-two matrix. One multiplied by one is one. And from that, we subtract zero
times zero. Finally, then, we have a nonzero
vector component. This whole cross product is simply
equal to the π€ hat unit vector. And recalling that π one equals
the π’ hat vector and π two equals the π£ hat vector, we see that the resultant
really is perpendicular to both of these input vectors.

Letβs continue to get practice with
a cross product in 3D by working through a few examples.

Let π equal the π’ hat unit vector
and π equal three π’ plus two π£ plus four π€. Calculate π cross π.

So here we have these two vectors
π and π, and one of them is three-dimensional. To calculate their cross product
then, weβll want to recall the mathematical rule for calculating a three-dimensional
cross product. Given two vectors π and π with as
many as three dimensions each, π cross π is equal to the determinant of this
three-by-three matrix.

Notice that the first row is
populated by the three orthogonal unit vectors and the second and third rows by the
corresponding components of our two input vectors π and π. Remembering this rule, we can now
apply it to our two particular vectors π and π. Weβll set up π cross π like
this. And with the unit vectors at the
top of each column, we know that the components of π are one, zero, zero in these
terms, while the π-vector has a magnitude of three in the π’-direction, two in the
π£-direction, and four in the π€-direction.

Now weβre ready to calculate this
cross product, and weβll do it component by component, starting with π’. That component will equal the
determinant of this two-by-two matrix. Thatβs equal to zero times four,
which is zero, minus zero times two, which is also zero. So the π’-component of our cross
product is zero. And now letβs look at the
π£-component. Its value is equal to negative one
times the determinant of this two-by-two matrix. Thatβs one times four or four minus
zero times three, which is zero. The π£-component then has a nonzero
value of negative four. And then, lastly, the π€-component
equal to the determinant of this two-by-two matrix. Thatβs one times two or two minus
zero times three, or zero. So here is our resulting
vector. This is the cross product of π and
π.

We could leave our answer in this
form, or we can write it in vector form. Doing that, we use this angled
bracket notation and we just write down the magnitude of each component. This is π cross π.

Now letβs look at an example where
all of the vectors involved are three-dimensional.

If vector π equals three, four,
negative four; vector π equals two, five, negative four; and vector π equals
negative four, negative four, two, find π minus π cross π minus π.

Okay, so here we have these three
vectors π, π, and π written out by their components. So, for example, considering vector
π, this is its π’-component, this is its π£-component, and this is its
π€-component. Our goal is to calculate this cross
product, where the two vectors involved are differences between two of our π, π,
and π vectors.

What we can do here is give these
differences their own names. Weβll call π minus π vector π,
and π minus π weβll call vector π. What weβll eventually do then is
cross π and π. But before that, weβll need to
figure out what each of these vectors are. Starting with vector π, we know
that thatβs equal to vector π, which is three, four, negative four, minus vector
π, which is two, five, negative four. Working component by component,
three minus two is one, four minus five is negative one, and negative four minus
negative four is zero. So thatβs vector π.

And vector π is equal to vector π
here minus vector π here. Negative four minus three is
negative seven, negative four minus four is negative eight, and then two minus
negative four is positive six. So we now have the two vectors π
and π. And weβll write them off to the
side that will combine in our cross product. To do that, letβs recall this rule
for the three-dimensional cross product of two vectors. Weβve called them π and π
here. This cross product equals the
determinant of a three-by-three matrix.

In the first row, we have our unit
vectors, in the second the corresponding components of vector π, and in the third
those of vector π. So when we set up the cross product
of π and π, weβll once again have π’, π£, and π€ unit vectors at the top of each
column. And in the next row, weβll write in
the components of vector π. Those are one, negative one, and
zero. Lastly, the components of vector
π: negative seven, negative eight, positive six.

Computing this cross product is now
a matter of calculating the determinant of this matrix. Beginning with the π’-component, if
we cross out the row and column that contain this element, then this component is
equal to the determinant of the two-by-two matrix remaining. This is equal to negative one times
six or negative six minus zero times negative eight, or zero. So the π’-component of our
resulting vector is negative six.

And then we move on to calculate
the π£-component. This has a value of negative one
times one times six, or six, minus zero times negative seven, or zero. This comes out to negative six
π£. And lastly, for our π€-component,
this is equal to one times negative eight, or negative eight, minus negative one
times negative seven, which comes out to negative seven. Our π€-component then simplifies to
negative 15.

And now weβve computed all three
components of a resulting cross product. The vector we end up with is
negative six π’ minus six π£ minus 15π€. Going back to our original three
vectors π, π, and π, this equals π minus π cross π minus π.

Letβs now look at an example where
we consider the geometric meaning of the cross product.

π΄π΅πΆπ· is a parallelogram with
vector ππ equals negative one, one, three and vector ππ equals three, four,
one. Find the area of π΄π΅πΆπ·. Give your answer to one decimal
place.

Okay, so here π΄π΅πΆπ· is a
parallelogram. And two of its sides are made of
these two vectors weβre given, ππ and ππ. So letβs say that this point here
is point π΄. And weβll say that it has the
coordinates zero, zero, zero. In other words, π΄ is the origin in
a three-dimensional set of axes.

From this point, we can draw
vectors ππ and ππ. If we say that they both lie in the
plane of our screen, then vector ππ could look like this and vector ππ like
this. We know that each of these two
vectors connects two corners on our parallelogram. This tells us that point π΅ is
right here at the end of vector ππ and point π· is at the end of vector ππ. And because weβre working with a
parallelogram, we can sketch in the remaining sides. They would look like this. And point πΆ in our parallelogram
is here.

In our question, weβre asked to
find this area, the area of π΄π΅πΆπ·. To do this, we can recall the fact
that the magnitude of the cross product of two vectors making up adjacent sides of a
parallelogram is equal to that shapeβs area. In other words, if we take vector
ππ and we cross it with vector ππ, then the magnitude of that cross product is
equal to the area weβve shaded in in blue.

Our first step then is to figure
out how to calculate this cross product. We can recall that, given two
three-dimensional vectors β weβll call them π and π β π cross π is given by this
expression. Here we have a three-by-three
matrix where in the first row we have the unit vectors, in the second the
corresponding components of π, and in the third those of π. So letβs now apply this general
rule to the vectors given to us here, vectors ππ and ππ.

Having started our three-by-three
matrix, weβll fill in the second row with the components of vector ππ. And we see that those are negative
one, positive one, and three. And then the components of ππ are
positive three, positive four, and positive one. To calculate this determinant,
letβs clear some space on screen and then calculate this cross product, starting
with the π’-component. We cross out the row and column
containing this element. And the value of the π’-component
then is equal to the determinant of this two-by-two matrix. One times one is one. And from that, we subtract three
times four, or 12.

Then, we move on to the
π£-component. This value equals negative negative
one times one, or negative one, minus three times three, or nine. And lastly, thereβs the
π€-component, equal to the determinant of this matrix. Negative one times four is negative
four, minus one times three or three. These components simplify to
negative 11π’, positive 10π£, and negative seven π€. This then is the cross product of
ππ and ππ.

But weβre not quite to our answer
because remember, we want to solve for the area of our parallelogram. To do that, weβll need to calculate
the magnitude of the vector we just computed. If we have a vector π with
components in the π’-, π£-, and π€-directions, then we can recall that the magnitude
of π is equal to the square root of the sum of the squares of its components. We can apply the same rule to solve
for the magnitude of ππ cross ππ. Here, the components are negative
11, positive 10, and negative seven. If we square all these, we get 121,
100, and 49. And adding the squares together
gives us 270. Entering the square root of 270 on
our calculator, if we round it to one decimal place, the answer we get is 16.4. This then, in whatever relevant
area units, is the area of π΄π΅πΆπ·.

Letβs finish up now by reviewing
some key points about the cross product in 3D. In this lesson, we learned that a
cross product in 3D combines two three-dimensional vectors and yields another vector
perpendicular to the original two. So if vectors π and π were lying
in the plane of our screen, then π cross π would point out of the screen at
us.

Along with this, we learned that
the magnitude of a cross product gives the area of the parallelogram defined by the
two vectors in the cross product. And lastly, we saw that if π and
π are two three-dimensional vectors, then π cross π is equal to the determinant
of this three-by-three matrix. Written in component form, that
looks like this.