### 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 in particular, 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 this question is asking us to find a vector product, letβs begin by recalling the general definition of the vector product of two vectors. Weβll define two general vectors which weβll label π and π. Letβs suppose that both of these vectors lie in the π₯π¦-plane. Then we can write them 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 given by the π₯-component of πΆ multiplied by the π¦-component of π· minus the π¦-component of πΆ multiplied by the π₯-component of π· all multiplied by π€ hat, where 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 define both our vectors π and π to lie in the π₯π¦-plane, this means that the vector product of two vectors produces a vector that is perpendicular to the direction of both of the vectors involved in the product.

Our 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 know the components of the vectors π¨ and π©. Weβll assume that the plane in 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 the diagram has a side length of one. This means that in order to find the π₯- and π¦-components of our two vectors, we simply need to count the number of squares that each vector extends in the π₯-direction and the π¦-direction.

Starting with vector π¨, weβll begin at the tail of the vector and count across squares in the π₯-direction until we reach the π₯-position of the tip of the vector. So in this case, thatβs one, two, three squares. If we do the same in the π¦-direction, we count one, two, three squares. But notice that the tip of vector π¨ is displaced downwards relative to the tail. In other words, weβve been counting squares in the negative π¦-direction, so this gives us a value of negative three squares. Since the vector π¨ extends positive three units in the π₯-direction and negative three units in the π¦-direction, we have that the π₯-component of π¨ is three and the π¦-component is negative three. So we can write this vector in component form as three multiplied by π’ hat minus three multiplied by π£ hat.

Weβll now do the same thing for vector π©. If we start at the tail of vector π©, then we see that we have to count across one, two squares in the positive π₯-direction to reach the π₯-position of the tip of the vector. And in the π¦-direction again starting at the tail of π©, we count one, two, three, four squares in the positive π¦-direction until we reach the π¦-position of the tip. So the π₯-component of the vector π© is two and the π¦-component is four. Then in component form, we have that the vector π© is equal to two π’ hat plus four π£ hat. Now that we have both 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 of two vectors, 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 vector π¨ and the second vector is vector π©. So we need the π₯-component of π¨, which is three, multiplied by the π¦-component of π©, which is four. Then we subtract a second term from this first one. 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 vector π¨, which is negative three, multiplied by the π₯-component of vector π©, which is two. Then all of this gets multiplied by π€ hat.

Our final job is to evaluate this expression here. Our first term is three multiplied by four, which gives us 12. And our second term is negative three multiplied by two, which gives us negative six. Then 12 minus negative six gives us 18. And so our answer to the question is that the vector product π¨ cross π© is equal to 18π€ hat, in other words, a vector with a magnitude of 18 and which points in the π§-direction. So thatβs the direction perpendicular to the plane of the diagram.