### Video Transcript

Consider two vectors π equals 15π’
plus seven π£ and π equals four π’ plus nine π£. Calculate π cross π. Calculate π cross π.

We are tasked with calculating two
cross products, also called vector products, using the vectors π and π defined at
the top. Whenever we need to perform
multiple calculations, itβs worth taking a moment to see if those calculations are
related. In this particular case, the
calculations are directly related. It is a property of the vector
product that changing the order of the two vectors only changes the sign of the
final answer. Algebraically, we write π cross π
is equal to negative π cross π. So, we actually only need to
compute one of these cross products, either π cross π or π cross π, because we
can then find the other one immediately by flipping the sign.

Alright, so instead of two vector
products, we can actually answer both parts of this question after only computing
one vector product. So, letβs see how to form the
vector product of two vectors. We can write a simple formula for
this vector product by noting that π and π are both written in terms of the π’ and
π£ unit vectors, and π’ and π£ point, respectively, in the positive π₯- and positive
π¦-directions of the Cartesian set of axes. To understand why this is
important, we have drawn a set of three-dimensional Cartesian axes, where π₯
increases to the bottom left, π¦ increases to the right, and π§ increases
upward. Note, of course, that this is only
a two-dimensional picture of what is in reality a three-dimensional object.

Anyway, the vector π’ is defined as
pointing in the positive π₯-direction and having a magnitude of one. The vector π£ is defined as
pointing in the positive π¦-direction and having a magnitude of one. And as we can see, this leaves
space for a third unit vector pointing in the positive π§-direction also with a
magnitude of one, and we call this vector π€. Now, the vector product of two
nonparallel vectors is always perpendicular to the plane that contains those two
vectors. Since the vectors π and π are
both expressed in terms of the π’ and π£ unit vectors only, they both exist in the
π₯π¦-plane. And so their vector product must be
perpendicular to the π₯π¦-plane, that is, in the positive π€-direction or possibly
in the negative π€-direction.

Now, remember that π cross π and
π cross π only differ by a negative sign. So, one of these will be in the
positive π€-direction and one of these will be in the negative π€-direction. Now, because these vector products
will be in the π€- or negative π€-directions, we can write the simple formula π
cross π is the π₯-component of π times the π¦-component of π minus the
π¦-component of π times the π₯-component of π in the π€-direction. If the quantity in the parentheses
is positive, the direction will be positive π€. And if the quantity in the
parentheses is negative, the direction will be negative π€.

We can also see from this formula
why changing the order of the vectors changes the sign of the vector product. When we exchange π and π on the
left side of this equation, we also need to exchange the two terms in the
parentheses. But exchanging those two terms is
equivalent to multiplying the whole thing by negative one. We should also mention that when we
write a vector as a sum in terms of π’ and π£ unit vectors, the π₯-component is
whatever number is multiplying the π’ unit vector, and the π¦-component is whatever
number is multiplying the π£ unit vector. This is perfectly logical. As we can see from the bottom-left
diagram, π’ is along the π₯-axis and π£ is along the π¦-axis.

Now, all we need to do is
substitute numbers and calculate. Starting with π cross π, we have
15 times nine, which is πΆ π₯ times π· π¦, minus seven times four, which is πΆ π¦
times π· π₯, in the π€-direction. 15 times nine is 135, and seven
times four is 28. So, we have 135 minus 28, which is
107, in the π€-direction. This gives us π cross π, and all
we need to do to find π cross π is multiply by negative one. So, π cross π is 107π€, and π
cross π is negative 107π€. Or equivalently, π cross π is 107
in the π€-direction, and π cross π is 107 in the negative π€-direction.