### Video Transcript

The diagram shows two vectors π and π. Work out π minus π in component form.

This question gives us two vectors in graphical form, and weβre asked to work out the vector subtraction π minus π in component form. So there are two steps needed to get to the answer. We need to subtract vector π from vector π. And we need to get the result into component form rather than a graphical form.

Now, there are actually two different ways we could go about solving this problem, because we could do these two steps in either order. We could opt to subtract the vectors using a graphical method and then convert the result to component form. Or we could convert our vectors π and π to component form and then subtract those two vectors algebraically. These two methods are entirely equivalent to each other, and therefore both methods will produce the same result.

In this video, weβll use method one. Weβll begin by subtracting the vector π from the vector π using a graphical method, and then weβll convert our result to component form. Weβre asked to find π minus π, which is just the same thing as adding the negative of vector π to vector π. We can write this as π plus negative π. Now, to get the negative of a vector, we simply take that vector and rotate it by 180 degrees about its starting position. So letβs draw the vector negative π onto our diagram. We see that the vector negative π is a vector with the same length as π and starting at the same point but just rotated around by 180 degrees so that itβs pointing in the opposite direction.

Now, we need to find the sum of π and negative π. To add together two vectors graphically, we can use the tip-to-tail method. In the tip-to-tail method, we slide one vector over so that its tail is on the tip of the other. Then, the sum of these two vectors is the vector that is drawn from the tail of the unmoved vector to the tip of the moved vector. So letβs apply this to our two vectors π and negative π. We begin by sliding the vector negative π over so that its tail is at the tip of vector π. Now, we can draw in our vector, which is equal to the sum of π and negative π. This starts at the tail of our unmoved vector, which in our case is vector π. So our vector starts here. It then extends to the tip of the moved vector, which in our case is vector negative π. So thatβs here. And so our vector sum π plus negative π is equal to this orange arrow here.

So we have a graphical representation of the vector π plus negative π. And we already know that this is equal to the vector π minus π, which is what we were asked to find. So we have successfully subtracted the vectors graphically, and we can check off our first step. All that remains is to convert this result into component form. To do this, we need to read off the number of units our vector extends into each of the π₯- and π¦-directions. If we look at our orange arrow in the diagram, we see that in the π₯-direction it extends one, two, three, four, five, six, seven units. And in the π¦-direction it extends one, two units in the negative π¦.

Letβs recall how we can write a general vector, which weβll label capital π for vector, in component form. We have that π is equal to an π₯-component π subscript π₯ multiplied by π’ hat, the unit vector in the π₯-direction, plus a π¦-component π subscript π¦ multiplied by π£ hat, the unit vector in the π¦-direction. Now, the π₯-component is simply the number of units that the vector extends in the π₯-direction. And the π¦-component is the number of units it extends in the π¦-direction.

Now, the vector weβre trying to write in component form is the vector π minus π. We have found that this vector extends seven units in the π₯-direction. So our π₯-component of this vector is seven. And this gets multiplied by the unit vector π’ hat. We also found that our vector π minus π extends negative two units in the π¦-direction. So our π¦-component is negative two, and this gets multiplied by the unit vector π£ hat.

So now we have successfully converted our result to component form. We have that the vector π minus π is equal to seven π’ hat plus negative two π£ hat, which we can write a little more simply as seven π’ hat minus two π£ hat. So we have our answer to the question that the vector subtraction π minus π is equal to seven π’ hat minus two π£ hat.