### 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 dot π.

In this question, weβre given two vectors, π and π, in the form of arrows drawn on a diagram. We are then asked to work out the scalar product of these vectors, π dot π. Letβs begin by recalling the definition of the scalar product of two vectors. Weβll consider two general vectors, which weβll label π and π. And weβll 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 that π’ hat is the unit vector in the π₯-direction and π£ hat is the unit vector in the π¦-direction. Then, we have that the scalar product π dot π is equal to the π₯- component of π multiplied by the π₯-component of π plus the π¦-component of π multiplied by the π¦-component of π. So, in general, we can say that the scalar product of two vectors is equal to the product of the π₯-components of those two vectors plus the product of their π¦-components. This expression for the scalar product of two vectors tells us that if we want to calculate our scalar product π dot π, then weβre going to need to work out the π₯- and π¦-components of our vectors π and π.

Now, the vectors π and π are drawn as arrows on a diagram. And weβre told in the question that the grid squares in this diagram each have a side length of one. If we add a set of axes to our diagram with the origin positioned at the tail of the two vectors, then we can easily count the number of squares that each vector extends in the π₯-direction and the π¦-direction. And since we know that each square has a side length of one, then the number of squares directly gives the π₯- and π¦-components of the vectors. Letβs begin with vector π.

We see that π extends three units in the positive π₯-direction and four units in the positive π¦-direction. This means that the π₯-component of π is three and the π¦-component is four. So, we can write the vector π in component form as three multiplied by π’ hat plus four multiplied by π£ hat. Now, letβs do the same thing with vector π. We see that π extends two units in the negative π₯-direction and seven units in the positive π¦-direction. So, the π₯-component of π is negative two and the π¦-component is positive seven. And we can write π in component form as negative two π’ hat plus seven π£ hat.

Now that we have both of our vectors π and π written in component form, we are ready to calculate their scalar product. Looking at our general expression for the scalar product of two vectors, we see that the first term in this expression is the product of the π₯-components of the vectors. So, in our case, we need the π₯-component of π, which is three, multiplied by the π₯-component of π, which is negative two. Then, we need to add a second term which is equal to the product of the π¦-components of the vectors. So, for us, thatβs the π¦-component of π, which is four, multiplied by the π¦-component of π, which is seven.

The last step left to go is to evaluate this expression. The first term, three multiplied by negative two, gives us negative six, and the second term, four multiplied by seven, gives 28. Then adding negative six and 28, we get a result of 22. So, our answer to the question is that the scalar product π dot π is equal to 22.