In this video, we’re gonna find out about the notation for representing the magnitude and direction of a vector. We’re also gonna learn how to calculate the magnitude and direction of a vector.
Hopefully, you already know that a vector is a set of numbers that can be represented in a suitable space, by a line segment with a specific length and direction. We use them to represent things like velocity, forces, and translations that have magnitude and direction. As we said, in this video, we’re gonna use vector notation and will calculate the magnitude and direction of some vectors.
Let’s start with point 𝐴 at one, two and 𝐵 at five, five. The vector represented by the line segment from 𝐴 to 𝐵 denotes a translation of positive four in the 𝑥-direction and positive three in the 𝑦-direction. We can write this as 𝐴𝐵 is equal to four, three in angle brackets. We could also give the vector a name, let’s say 𝑉. So 𝑉 equals four, three in angle brackets. Now the 𝑉 would normally have a bold typeface or an italic typeface, depending on which particular textbook you happen to be looking at.
Now because the 𝑥- and 𝑦-axes are at right angles to each other, when we were calculating the 𝑥- and 𝑦-components, we made a right-angled triangle. So we can use some right-angled triangle maths, trigonometry and the Pythagorean theorem, to work out the vector’s magnitude and direction. Now the magnitude of 𝑉 could be written as a 𝑉 with vertical lines either side of it, that’s the notation for magnitude of 𝑉. Also, the Pythagorean theorem states that in a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other sides. In our case, the length of the hypotenuse is the length, or the magnitude, of the vector. So let’s use the Pythagorean theorem to work out the length of this. So the length of 𝑉 squared, of the magnitude squared, is equal to four squared plus three squared because they’re the other sides of the triangle. And four squared is sixteen, three squared is nine. Add those together, we get twenty-five. Then taking square roots on both sides, 𝑉 is equal to the square root of twenty-five. So the magnitude of 𝑉 is five. And if we haven’t called our vector 𝑉, and we just called it 𝐴𝐵, as we did here, this would be the way of writing that out. The magnitude of Vector 𝐴𝐵 means exactly the same thing, but obviously with a different name.
Now for the direction, we’re looking for the angle that the vector would make with the positive part of the 𝑥-axis, if it started — if it’s initial point was at zero, zero. And we work out what size counterclockwise rotation we’d need to make from the positive part of the 𝑥-axis to get to the vector. So we’re looking for, basically, this angle here. So thinking about this direction, what angle does that make with our vector in the counterclockwise direction. Now we can use trigonometry, in particular the tangent ratio, to work out the size of angle 𝜃. But to do that, I’m gonna draw a new diagram. So we know that vectors have length and direction, but we can pick them up and put them anywhere we like, provided we keep that same direction and that same length as well. So that’s just what we’ve done here. We’ve picked up vector 𝑣 and we’ve started it, its a- its initial point, at the origin here. So what we’re trying to measure is this angle here.
Now we know that the 𝑥-component in that vector is four. So this distance, on this side here, is four. And we know that the 𝑦-component is three. So this distance here, on this side, is three. So we’ll just label up our opposite, adjacent, and hypotenuse. And the adjacent side and the opposite are the side lengths that we know, in terms of working out this angle. So we’re gonna be using the tan ratio. So tan of our angle is the length of the opposite side divided by the length of the adjacent side. So that’s three over four. So if tan 𝜃 is equal to three quarters, that means the 𝜃 is tan to the minus one of three over four, of three quarters. So the three, remember, is the 𝑦-component and the four is the 𝑥-component on the bottom. And a little bit of calculator work tells us that 𝜃, in this case, is thirty-six point nine degrees. So for our vector 𝐴𝐵, or our vector 𝑉 in fact, the magnitude of 𝑉 was five and the direction, in-in a counterclockwise direction from what would be the 𝑥-axis, is thirty-six point nine degrees.
Okay. Let’s look at another example, which is just a little bit more tricky for a few reasons as we’ll find out. 𝐶 is at one, six and 𝐷 is at five, one. Find the magnitude of vector 𝐶𝐷 and the direction of vector 𝐶𝐷. So as we said, this is a bit trickier for a few reasons. One, you don’t have a diagram. Two, the Pythagorean theorem bit isn’t gonna work out quite so nicely. And three, the direction makes this question a little bit more awkward. So the thing I would recommend is that you, first of all, do a diagram. It doesn’t have to be an accurate dri- diagram, not a hundred percent accurate, but it has to be a sort of an attempted sketch at that situation.
So here’s our sketch. We’ve got points 𝐶 and 𝐷 sketched in. There, we haven’t bothered drawing the 𝑥- and 𝑦- axes in their coordinate system. They’re probably not a hundred percent accurate, but we’ve got this idea that 𝐶 — at one, six — is slightly to the left of and above point 𝐷. I’ve also just start to sketch in the right-angled triangle that that would make. So to work out the 𝑥-component of vector 𝐶𝐷, we’re going from an 𝑥-coordinate of one up to an 𝑥-coordinate of five. So the difference between those, five take away one, is positive four. That is positive four in the 𝑥-direction. And for the 𝑦-coordinates, we’ve gone from six down to one. So six taken away from one leaves us with negative five. We’ve gone negative five in the 𝑦-direction. So we can write out vector 𝐶𝐷 as — with the angle brackets — four, negative five.
And now we can start to apply the Pythagorean theorem to find the magnitude of that vector. So looking at our right-angled triangle, this is the length we’re trying to find out, the length of the hypotenuse. So 𝐶𝐷 squared is equal to four squared plus negative five squared. And four squared is sixteen and negative five squared is twenty-five. Add those two together, you get forty-one. We’re gonna take square roots of both sides now. And well, the square root of forty-one is a — it’s not a nice easy number to work out. So we just leave that as the square root of forty-one. But in some situations, you might do that in a calculator, and the answer would be six point four, correct to one decimal place.
Now remember that direction is measured counterclockwise from the positive part of the 𝑥-axis to the vector, assuming that it started at zero, zero. So that’s gonna be a little bit trickier in this situation. So I’m just gonna sketch out another diagram. So remembering that out vector 𝐶𝐷 was four, negative five, we’re gonna pick the vector up and make the starting point at zero, zero over here, the origin. So when we do that, this is what the vector looks like. We have an 𝑥-component of four and a 𝑦-component of negative five. So this is gonna make a nice little right-angled triangle again.
Now if you remember from the first example, we said that the direction is equal to the tan to the minus one at the 𝑦-component divided by the 𝑥-component of the vector that we were looking at. So that’s gonna be tan to the minus one of negative five over four. And that, in turn, gives us an angle of negative fifty-one point three degrees, correct one decimal place. Now what that’s done, is that’s given us this angle here. So if we were starting along the 𝑥-axis, travelling in a clockwise direction, counts as negative turn so that would be negative fifty-one point three degrees. So this angle here is fifty-one point three degrees. But what we need is the counterclockwise angle starting from the 𝑥-axis going all the way round here to the vector. It’s that angle there round the outside, that reflex angle.
So in the-the immortal words of Star Wars, this is not the angle we’re looking for. So what this is telling us, is that we are fifty-one point three degrees shy of a full turn of three hundred and sixty degrees in the counterclockwise direction. So the direction we are looking for is three hundred and sixty degrees take away that fifty-one point three degrees. You might say it’s negative fifty-one point three plus three hundred and sixty degrees, but that’s just another way of saying the same thing. So the direction we wanted was three zero eight point seven degrees, correct to one decimal place.
So let’s just summarize that whole process then. To work out the direction and the magnitude of a vector 𝐴𝐵, first of all, we pick up the vector and move it to the origin of the 𝑥𝑦 axes. We can then use the Pythagorean theorem to work out the length of that vector, or the magnitude of that vector. So if 𝐴 was our initial point and 𝐵 was our terminal point, and 𝐴 had coordinates 𝑥 one, 𝑦 one and 𝐵 had coordinates 𝑥 two, 𝑦 two, the magnitude of vector 𝐴𝐵 is simply the difference in the 𝑥-coordinates squared plus the difference in the 𝑦-coordinates all squared. And then take the square root of all of that.
And to work out the direction, it’s the angle here from the 𝑥-axis to the vector in a counterclockwise direction. So for our vector 𝐴𝐵, the 𝑥-component was 𝑥 two minus 𝑥 one, the terminal point 𝑥 component minus the initial point 𝑥 component. And likewise, for the 𝑦-component. And the direction 𝜃 is the inverse tan of the 𝑦-component divided by the 𝑥-component of that vector 𝐴𝐵. But we do need to be careful because if that angle comes out negative, we need to make sure that we end up with this positive turn here in the counterclockwise direction from the 𝑥-axis.
So whether you just remember these formulae or whether you remember the methods that we went through, as we described them in this video, either way, hopefully will help you to work out magnitudes and directions of vectors in the future.