3.1: Vectors

Both momentum and angular momentum are vector quantities. Later, we’ll be talking about a more abstract form of vector used to represent the state of quantum systems. Here, we’re talking about a special kind of vector, a vector in regular old 3d-space. Distinguish these from more general vectors, I shall call them 3-vectors, in reference to 3-dimensional space. A 3-vector is anything that has both a magnitude (size) and direction. For example, consider speed: the speed an object is moving is just a number. (We would call that a scalar.) Likewise, the kinetic energy of an object is a scalar; it’s an amount of energy, and there is no direction associated with it. In contrast, the velocity of an object includes not just its speed, but also its direction. So, you could say that the speed of a car is 80 km/h. If you wanted to specify its velocity, you’d also have to give its direction. For example, you could say that the velocity of a car is 80 km/h due northwest.

3-vectors corresponding to different physical quantities will have different dimensionalites (and thus different units) associated with them. Displacement is a 3-vector form of distance. Distance just tells you how far apart two things are. Displacement tells you how far apart and in what direction. Just like distance, displacement comes in length units. So, you might say that one person is 1 meter due east of another person; in this case, the displacement from the other person to the first person is 1 m due east.. The magnitude of a 3-vector is just its size. Distance is the magnitude of displacement. If you consider that person whose displacement was 1 meter due east of the other person, you could also say that the distance between the two people was 1 meter. This is correct, even though it has less information.

A 3-vector can be visualized as an arrow in space. The length of the arrow represents the magnitude, and the direction the arrow points is the direction of the 3-vector. So, for example, let us consider a car going at 50 km/h due northwest.

The picture shows the \(\ x\) and \(\ y\) axes, representing East and North respectively. The \(\ z\) axis not drawn; it’s up, straight out of the page. We notate vectors by drawing a little arrow on top of them; you can see the \(\ \vec{v}\) in the diagram referring to the velocity of the car. Also shown are \(\ \vec{e}_{x}\) and \(\ \vec{e}_{y}\), the two basis vectors. In the case of 3-vectors, we can also call the basis vectors unit vectors, as they are 3-vectors whose length 1 (dimensionless), and that point right along the axes. The basis vectors define the coordinate system that we’re using; here, they just define \(\ x\) and \(\ y\) for us. The car’s velocity 3-vector \(\ \vec{v}\) is represented by the direction and length of the arrow sticking out of the front of its picture. We can see that it points partially in the negative-x direction, and partly in the positive-y direction.

If you have a complete set of basis vectors, you can construct any other vector out of them. The three unit 3-vectors \(\ \vec{e}_{x}, \vec{e}_{y}\), and \(\ \vec{e}_{z}\), all 3-vectors of length dimensionless 1 pointing (respectively) along the x, y, and z axes, form the most obvious and most generally useful set of basis 3-vectors. Any other 3-vector can be written as a sum of constants times those basis vectors. So, here, we could say that:

\(\ \vec{v}=v_{x} \vec{e}_{x}+v_{y} \vec{e}_{y}+v_{z} \vec{e}_{z}\)

From looking at the picture, we can see that in this case \(\ v_{x}\) is going to have to be negative, \(\ v_{y}\) is going to have to be positive, and \(\ v_{z}\) is going to be zero. How do we figure out where they are? Well, we know that the car is going due northwest, so we expect that the absolute value of \(\ v_{x}\) and \(\ v_{y}\) will be the same (it’s got just as much north velocity as west velocity). For the total speed, that is the total length of the 3-vector, we recognize that there’s a right triangle there, and use a generalization of the Pythagorean Theorem:

\(\ \begin{gathered}
v^{2}=|\vec{v}|^{2}=v_{x}{ }^{2}+v_{y}{ }^{2}+v_{z}{ }^{2} \\
v=\sqrt{v_{x}{ }^{2}+v_{y}{ }^{2}+v_{z}{ }^{2}}
\end{gathered}\)

In this example, we know that \(\ v=50 \mathrm{~km} / \mathrm{h}\). For this to work, we have to have \(\ v_{x}=-35 \mathrm{~km} / \mathrm{h}\) and \(\ v_{y}=35 \mathrm{~km} / \mathrm{h}\).