### 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 of the grid has a side length of one. Calculate π cross π.

So 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 on a diagram. Letβs begin by recalling the definition of the vector product. Weβll consider two general vectors π and π and suppose that they lie in the π₯π¦-plane. Then we can write these vectors in component form as an π₯-component, which weβve labeled with a subscript π₯, multiplied by π’ hat plus a π¦-component, labeled with a subscript π¦, multiplied by 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. 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. And this is all multiplied by π€ hat, which is a unit vector in the π§-direction. What this expression tells us is that in order to calculate the vector product π cross π, then weβre going to need to work out the π₯- and the π¦-components of the vectors π and π.

Now, weβre told in the question that each of the squares of the grid in the diagram has a side length of one. So to get the π₯- and π¦-components of each of our vectors, we simply need to count the number of squares that they extend in each of the π₯- and π¦-directions. Weβll begin with vector π. Vector π extends one, two, three, four units in the π₯-direction and one, two, three, four, five units in the π¦-direction. So we can write the vector π as an π₯-component of four multiplied by π’ hat plus a π¦-component of five multiplied by π£ hat.

Next, letβs look at the vector π. We see that π extends one, two, three, four units in the negative π₯-direction and one, two, three, four, five units in the negative π¦-direction. So we can write the vector π as negative four π’ hat minus five π£ hat.

Now that we have each of the 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 the π₯-component of the first vector in the product multiplied by the π¦-component of the second vector in the product. In our case, the first vector in the product is π and the second vector is π. So this means that we need the π₯-component of π, which is four, multiplied by the π¦-component of π, which is negative five.

Then we subtract a second term from this. The 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 five, multiplied by the π₯-component of π, which is negative four. And then, finally, this whole thing gets multiplied by the unit vector π€ hat.

All thatβs left to do now is to evaluate this expression here. When we do this, we find that the first term, four multiplied by negative five, gives us negative 20. And the second term, five multiplied by negative four, also gives us negative 20. And so we have that the vector product π cross π is equal to negative 20 minus negative 20 multiplied by π€ hat. When we subtract negative 20 from negative 20, we get zero. And so our final answer is that the vector product π cross π is equal to zero π€ hat.