# 8.2: What you don’t know about vectors may surprise you!

Remember that a scalar has only a magnitude while a vector has both a magnitude and a direction. The following video (12:33) makes this difference clear.

Scalars and Vectors

Typically the vectors used in meteorology and atmospheric science have two or three dimensions. Let’s think of two three-dimensional vectors of some variable (e.g., wind, force, momentum):

$\vec{A}=\vec{i} A_{x}+\vec{j} A_{y}+\vec{k} A_{z}$

$\vec{B}=\vec{i} B_{x}+\vec{j} B_{y}+\vec{k} B_{z}$

Sometimes we designate vectors with bold lettering, especially if the word processor does not allow for arrows in the text. When Equations [8.3] are written with vectors in bold, they are:

$\mathbf{A}=\mathbf{i} A_{x}+\mathbf{j} A_{y}+\mathbf{k} A_{z}$

$\mathbf{B}=\mathbf{i} B_{x}+\mathbf{j} B_{y}+\mathbf{k} B_{z}$

Be comfortable with both notations.

In the equations for vectors, Ax and Bx are the magnitudes of the two vectors in the x (east–west) direction, for which $$\vec{i}$$ or i is the unit vector; Ay and By are the magnitudes of the two vectors in the y (north–south) direction, for which $$\vec{j}$$ or is the unit vector; and Az and Bz are the magnitudes of the two vectors in the z(up–down) direction, for which $$\vec{k}$$ or k is the unit vector. Unit vectors are sometimes called direction vectors.

Sometimes we want to know the magnitude (length) of a vector. For example, we may want to know the wind speed but not the wind direction. The magnitude of $$\vec{A}$$ , or A, is given by:

$|\vec{A}|=\sqrt{\left(A_{x}^{2}+A_{y}^{2}+A_{z}^{2}\right)}$

We often need to know how two vectors relate to each other in atmospheric kinematics and dynamics. The two most common vector operations that allow us to find relationships between vectors are the dot product (also called the scalar product or inner product) and the cross product (also called the vector product).

The dot product of two vectors A and B that have an angle ββ between them is given by:

$\vec{A} \bullet \vec{B}=A_{x} B_{x}+A_{y} B_{y}+A_{z} B_{z}$

$=|\vec{A}||\vec{B}| \cos \beta$

$=|\vec{A}||\vec{B}|\) if $$\vec{A}$$ is parallel to $$\vec{B}$ $=0$$ if $$\vec{A} \perp \vec{B}$ The dot product is simply the magnitude of one of the vectors, for example A, multiplied by the projection of the other vector, B, onto A, which is just B cosβB cosβ A and B are parallel to each other, then their dot product is AB. If they are perpendicular to each other, then their dot product is 0. The dot product is a scalar and therefore has magnitude but no direction. Also note that the unit vectors (a.k.a., direction vectors) have the following properties: $\vec{i} \cdot \vec{i}=\vec{j} \cdot \vec{j}=\vec{k} \cdot \vec{k}=1$ $\vec{i} \cdot \vec{j}=\vec{i} \cdot \vec{k}=\vec{j} \cdot \vec{k}=\vec{j} \cdot \vec{i}=\vec{k} \cdot \vec{i}=\vec{k} \cdot \vec{j}=0$ $\vec{i} \cdot \vec{A}=A_{x}$ $\vec{B} \cdot \vec{A}=\vec{A} \cdot \vec{B}$ Note that the dot product of the unit vector with a vector simply selects the magnitude of the vector's component in that direction \(\left.\overrightarrow{(i} \cdot \vec{A}=A_{x}\right)$$ and that the dot product is commutative $$(\vec{A} \cdot \vec{B}=\vec{B} \cdot \vec{A})$$

Equation [8.4] can be rearranged to yield an expression for cosβcosβ in terms of the vector components and vector magnitudes:

$\cos \beta=\frac{A_{x} B_{x}+A_{y} B_{y}+A_{z} B_{z}}{|\vec{A}||\vec{B}|}$

The cross product of two vectors A and B that have an angle ββ between them is given by:

$\vec{A} \times \vec{B}=\left(\begin{array}{ccc}{\vec{i}} & {\vec{j}} & {\vec{k}} \\ {A_{x}} & {A_{y}} & {A_{z}} \\ {B_{x}} & {B_{y}} & {B_{z}}\end{array}\right)$

$\vec{A} \times \vec{B}=\left(A_{y} B_{z}-A_{z} B_{y}\right) \vec{i}-\left(A_{x} B_{z}-A_{z} B_{x}\right) \vec{j}+\left(A_{x} B_{y}-A_{y} B_{x}\right) \vec{k}$

The magnitude of the cross product is given by:

$|\vec{A} \times \vec{B}|=|\vec{A}||\vec{B}| \sin \beta$

$=0 \quad\) if $$\vec{A}$$ is parallel to $$\vec{B}$ $=|\vec{A}||\vec{B}|$$ if $$\vec{A} \perp \vec{B}$ where ββ is the angle between A and B, with ββ increasing from A to B. Note that the cross product is a vector. The direction of the cross product is at right angles to A and B, in the right hand sense. That is, use the right hand rule (have your hand open, curl it from A to B, and A x B will be in the direction of your right thumb). The magnitude of the cross product can be visualized as the area of the parallelogram formed from the two vectors. The direction is perpendicular to the plane formed by vectors A and B. Thus, if A and B are parallel to each other, the magnitude of their cross product is 0. If A and B are perpendicular to each other, the magnitude of their cross product is AB. The following video (2:06) reminds you about the right-hand rule for cross products. Right-hand Rule for Vector Cross Product Click here for transcript of the Right-hand Rule for Vector Cross product We're going to do a couple more examples of finding vector cross product. Suppose that I give you these two vectors a and B, which both lie in the plane of, look its my hands, which both lie in the plane of the page. Ok, so there are a and B. You want to find the direction of a cross B. To find the magnitude you do a times B times the sine of the angle between them, but we just want to find the direction right now, and to do this we're going to use the right hand rule, but first we can use a little bit of logic. So, first of all logic says this, whatever the direction of a cross B is which let's call that c, a cross b the we'll call that c. It has to be perpendicular to both a and B or perpendicular to the plane of the page. Well there are only two directions that that could be, right. What that means is that c either must point straight out of the page or it must point straight into the page. And, to figure out which one of those two directions it is, what we're going to have to do is we're gonna have to put our fingers along a. So there are two ways to do that. You can either put your fingers along a this way, or you could put your fingers along a this way, and you have to do it in the way that will let you swing a down into b like it was a little hinge. So, if you try that notice if you do it this way, yeah it's the wrong way right. You'd have to swing all the way the long way around. If you want to just simply fold a into b the way to do that is to put your fingers this way then you can curl them down this way. Notice when you do that your thumb is pointing into the page, so therefore, the answer is that c is into the page... and actually I got marker on my wall. Actually, the way we represent that is that's represented into the page is represented by a little X with a circle around it. You're supposed to think of it like the tail feathers of an arrow that's pointing into the page. It follows that the cross products of the unit vectors are given by: $\vec{i} \times \vec{j}=\vec{k} \quad \vec{j} \times \vec{k}=\vec{i} \quad \vec{k} \times \vec{i}=\vec{j}$ $\vec{i} \times \vec{j}=-\vec{j} \times \vec{i}$ Note finally that $\vec{A} \times \vec{B}=-\vec{B} \times \vec{A}$ We sometimes need to take derivatives of vectors in all directions. For that we can use a special vector derivative called the Del operator, \(\vec{\nabla}$$

Del is a vector differential operator that tells us the change in a variable in all three directions. Suppose that we set out temperature sensors on a mountain so that we get the temperature, T, as a function of xy, and z. Then $$\vec{\nabla}$$ T would give us the change of T in the xy, and z directions.

$\vec{\nabla}=\vec{i} \frac{\partial}{\partial x}+\vec{j} \frac{\partial}{\partial y}+\vec{k} \frac{\partial}{\partial z}$

The Del operator can be used like a vector in dot products and cross products but not in sums and differences. It does not commute with vectors and must be the partial derivative of some variable, either a scalar or a vector. For example, we can have the following with del and a vector A:

$$\vec{\nabla} \cdot \vec{A}=\frac{\partial A_{x}}{\partial x}+\frac{\partial A_{y}}{\partial y}+\frac{\partial A_{z}}{\partial z},$$ which is a scalar

$$\vec{\nabla} T=\vec{i} \frac{\partial T}{\partial x}+\vec{j} \frac{\partial T}{\partial y}+\vec{k} \frac{\partial T}{\partial z},$$ which is a vector even though $$T$$ is a scalar

$$\vec{A} \cdot \vec{\nabla} T=A_{x} \frac{\partial T}{\partial x}+A_{y} \frac{\partial T}{\partial y}+A_{z} \frac{\partial T}{\partial z},$$ which is a scalar

#### Quiz 8-1: Partial derivatives and vector operations.

1. Find Practice Quiz 8-1 in Canvas. You may complete this practice quiz as many times as you want. It is not graded, but it allows you to check your level of preparedness before taking the graded quiz.
2. When you feel you are ready, take Quiz 8-1. You will be allowed to take this quiz only once. Good luck!