### Video Transcript

Work out π£ hat cross π’ hat.

We can see that this is a question
about vector products. And weβre asked to calculate the
vector product between the unit vectors π£ hat and π’ hat. Recall that π£ hat is the unit
vector in the π¦-direction and π’ hat is the unit vector in the π₯-direction. Since weβre asked to work out a
vector product, letβs start by recalling the general expression for the vector
product between two vectors. Weβll call these general vectors π¨
and π© and assume that both lie in the π₯π¦-plane.

We can write these vectors in
component form as π¨ equals an π₯-component π΄ subscript π₯ multiplied by π’ hat
plus a π¦-component π΄ subscript π¦ multiplied by π£ hat, and similarly for vector
π©. Then, the vector product π¨ cross
π© is defined as 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 the
unit vector π€, which points in the π§-direction. So the vector product π¨ cross π©
produces a vector with this magnitude and with a direction that is perpendicular to
the direction of both π¨ and π©.

So thatβs the general case. Now letβs apply it to our two
vectors π£ hat and π’ hat. We can write these two unit vectors
in component form as weβve done for our general vectors π¨ and π©. We can write that π£ hat is equal
to zero times π’ hat plus one times π£ hat. This tells us that π£ hat, the unit
vector in the π¦-direction, has zero units in the π₯-direction and one unit in the
π¦-direction, which makes sense. Similarly, we can write the vector
π’ hat as equal to one multiplied by π’ hat plus zero multiplied by π£ hat. The question is asking us to
calculate the vector product π£ hat cross π’ hat. So, in this general expression π¨
cross π©, we want to replace the vector π¨ with the vector π£ hat and the vector π©
with the vector π’ hat. The first term in our expression
for the vector product is given by the π₯-component of π΄ multiplied by the
π¦-component of π΅.

Now, we said that to calculate our
specific vector product, when we see the vector π¨, we need to replace it with the
vector π£ hat. So, for the π₯-component of π΄, we
need to use the π₯-component of the vector π£ hat, which is zero. Likewise, when we see the vector
π©, we need to replace that with the vector π’ hat. So, for the π¦-component of π΅, we
need to use the π¦-component of the vector π’ hat, which is also zero. From this first term in our vector
product expression, we then subtract a second term. This second term is the
π¦-component of the vector π¨ multiplied by the π₯-component of the vector π©. And in our case, the vector π¨ is
replaced by the vector π£ hat.

So, for the π¦-component of π΄, we
need the π¦-component of π£ hat, which is one. And since the vector π© is replaced
by the vector π’ hat, then for the π₯-component of π΅, we need the π₯-component of
π’ hat, which is also one. Finally, this whole expression gets
multiplied by π€ hat, the unit vector in the π§-direction. Evaluating this expression, we have
that the first term is given by zero multiplied by zero, which gives us zero. And our second term, which we
subtract from this first one, is given by one multiplied by one, which gives us
one. And then of course this is all
multiplied by π€ hat, the unit vector in the π§-direction.

Since zero minus one is equal to
negative one, this gives us negative π€ hat. And so we have our answer to the
question that the vector product π£ hat cross π’ hat is equal to negative π€ hat, in
other words, the unit vector in the negative π§-direction.