### 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.

The question is asking us about vector products, and in particular weβre asked to work out the vector product π cross π. The vectors π and π are given to us in the form of arrows on a diagram, and we are told that both vectors lie in the π₯π¦-plane. Weβll begin by recalling the definition of the vector product of two vectors. Weβll consider two general vectors, which weβll label π and π, and suppose that both of these vectors 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 π’ hat is the unit vector in the π₯-direction and π£ hat is the unit vector in the π¦-direction. Then we can write the vector product π cross π 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 π€ hat, where π€ hat is the unit vector in the π§-direction. Looking at this general expression for the vector product of two vectors that lie in the π₯π¦-plane, we see that if we want to calculate our vector product π cross π, then weβre going to need to know the π₯- and π¦-components of the vectors π and π.

Now, the vectors π and π are both given to us in the diagram shown in the question. And this question also tells us that the squares of the grid in this diagram have a side length of one. This means that to find the π₯- and π¦-components of our vectors π and π, all we need to do is count the number of squares that each vector extends in the π₯-direction and the π¦-direction. Since each square is one unit in length, the number of squares directly gives us the π₯- and π¦-components of the vectors.

Letβs start by doing this for vector π. By tracing down to the π₯-axis from the tip of vector π and then counting along the squares, we can see that vector π extends one, two, three, four, five squares in the positive π₯-direction. And tracing across from the tip of vector π to the π¦-axis, we can count that π extends one, two units in the positive π¦-direction. So we can write the vector π in component form as its π₯-component, five, multiplied by π’ hat plus its π¦-component, two, multiplied by π£ hat.

Now letβs do the same thing with vector π. Tracing down from the tip of vector π to the π₯-axis, we see that π extends one square in the negative π₯-direction. And tracing across to the π¦-axis, we can see that π extends one, two, three, four, five squares in the positive π¦-direction. So we can write π in component form as its π₯-component, negative one, multiplied by π’ hat plus its π¦-component, five, multiplied by π£ hat.

Now that we have both of our vectors π and π written in component form, we are ready to use this expression here 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. In our case, the first vector in our product is π and the second vector is π.

So for this first term, we need the π₯-component of vector π, which is five, multiplied by the π¦-component of vector π, which is also five. 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 for our specific case, thatβs the π¦-component of vector π, which is two, multiplied by the π₯-component of vector π, which is negative one. And as we said, we subtract this second term from the first one. This whole expression then gets multiplied by the unit vector π€ hat.

The last step left to go is to evaluate this expression here. The first term, which is five multiplied by five, gives us 25, while the second term, two multiplied by negative one, gives us negative two. Then when we calculate 25 minus negative two, we get 27. And so our answer for the vector product π cross π is 27π€ hat.