### Video Transcript

The diagram shows two vectors, π and π. Each of the grid squares in the diagram has a side length of one. Calculate π cross π.

This question is asking us about vector products, and specifically weβre asked to work out 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. Since weβre being asked to calculate a vector product, letβs begin by recalling the general definition of the vector product of two vectors. Weβll 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.

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 given by 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, where we should recall that π€ hat is the unit vector in the π§-direction. So the vector product π cross π produces a vector with this magnitude and that points in the π§-direction. Since we defined both of our vectors π and π to lie in the π₯π¦-plane, then this means that the vector product of two vectors produces a vector which is perpendicular to both of those vectors.

This general expression for the vector product of two vectors tells us that if we want to calculate the vector product π cross π, then weβre going to need to work out the components of the vectors π and π. Weβll assume that the plane of the diagram in which the vectors are drawn is the π₯π¦-plane. And then we can label our π₯- and π¦-directions. The question tells us that each of the grid squares in this diagram has a side length of one. This means that to work out the components of our vectors, we simply need to count the number of squares that each of our vectors extends in the π₯-direction and in the π¦-direction. Letβs begin by doing this for vector π.

Weβll start at the tail of vector π and then count the number of squares in the π₯-direction between the tail and the tip of this vector. So thatβs one, two, three, four, five, six squares in the positive π₯-direction. Weβll start back again at the tail of vector π, and this time weβll count the number of squares that π extends in the π¦-direction. So thatβs one, two squares in the positive π¦-direction. This means that the π₯-component of π is positive six and the π¦-component is positive two. And so we can write the vector π in component form as six π’ hat plus two π£ hat.

So now letβs do the same thing with vector π. If we start at the tail of vector π and count the number of squares it extends in the π₯-direction, we see that this is one square in the negative π₯-direction. And again starting back at the tail of vector π, this time counting squares in the π¦-direction, we see that π extends one, two, three squares in the positive π¦-direction. This means that the π₯-component of π is negative one and the π¦-component is positive three. So writing π in component form, we have that π equals negative one π’ hat plus three π£ hat. We now have both of our vectors π and π written in component form, which means that we are ready to calculate the vector product π cross π.

If we look 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 our product is the vector π and the second vector is the vector π. So this means that we need the π₯-component of π, which is six, multiplied by the π¦-component of π, which is 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 for us, thatβs the π¦-component of π, which is two, multiplied by the π₯-component of π, which is negative one. And then we multiply all of this by π€ hat.

The final step is to evaluate this expression here. The first term is six multiplied by three, which gives us 18. And the second term is two multiplied by negative one, which gives us negative two. Then, we have 18 minus negative two, which gives us 20. So our answer to the question is that the vector product π cross π is equal to 20π€ hat, in other words, a vector with a magnitude of 20 and which points in the positive π§-direction, which is perpendicular to the plane of the diagram.