### Video Transcript

The diagram shows two vectors π
and π in three-dimensional space. Both vectors lie in the
π₯π¦-plane. Each of the squares on the grid has
a side length of one. Calculate π cross π.

This is a question about vector
products. And specifically, we are asked to
calculate the vector product π cross π, where the vectors π and π are given to
us in the form of arrows drawn on a diagram. And we are told that both π and π
lie in the π₯π¦-plane.

Letβs begin by recalling the
definition of the vector product of two vectors. Letβs consider two general vectors
π and π and suppose that they both lie in the π₯π¦-plane. Then, we can write these vectors in
component form as an π₯-component labeled with a subscript π₯ multiplied by π’ hat
plus a π¦-component labeled with a subscript π¦ multiplied by π£ hat. Remember that π’ hat is the unit
vector in the π₯-direction and π£ hat is the unit vector in the π¦-direction. Then, the vector product π cross
π is the π₯-component of π multiplied by the π¦-component of π minus the
π¦-component of π multiplied by the π₯-component of π all multiplied by π€ hat,
which is the unit vector in the π§-direction.

What this expression for the vector
product is telling us is that in order to calculate our vector product π cross π,
weβre going to need to know the π₯- and π¦-components of the vectors π and π. The question tells us that the
squares on the grid in the diagram each have a side length of one. This means that to find the π₯- and
π¦-components of our vectors, we simply need to count the number of squares that
each vector extends in the π₯-direction and the π¦-direction.

Letβs begin by doing this for
vector π. We see that vector π extends one,
two, three, four, five squares into the negative π₯-direction and that it extends
one square into the negative π¦-direction. So we have that the π₯-component of
π is negative five and the π¦-component of π is negative one. This means we can write the vector
π in component form as negative five π’ hat minus one π£ hat.

Now, letβs do the same for vector
π. We see that vector π extends one,
two units in the positive π₯-direction and one, two, three units in the negative
π¦-direction. So the π₯-component of π is two,
and the π¦-component of π is negative three. And we can write π in component
form as two π’ hat minus three π£ hat.

Now that we have each of our
vectors π and π in component form, we are ready to calculate the vector product π
cross π. Looking at our general expression
for the vector product, we see that the first term is given by the π₯-component of
the first vector in the product multiplied by the π¦-component of the second vector
in the product. The first vector in our vector
product is π, and the second vector is π. So we need the π₯-component of
vector π, which is negative five, multiplied by the π¦-component of vector π,
which is negative three.

We then subtract a second term from
this first one. This second term is the
π¦-component of the first vector in the product multiplied by the π₯-component of
the second vector in the product. So in our case, thatβs the
π¦-component of π, which is negative one, multiplied by the π₯-component of π,
which is two. And then all of this gets
multiplied by the unit vector π€ hat.

Our last step is to evaluate this
expression here. This first term negative five
multiplied by negative three gives us positive 15. And this second term negative one
multiplied by two gives us negative two. So then we have 15 minus negative
two, which gives us 17. And so our final answer is that the
vector product π cross π is equal to 17π€ hat.