The Prodigal Physicist

A vector is just a set of numbers used to describe a single quantity (direction and magnitude) and from here on in, I’ll indicate a variable that is a vector by making it boldface. So the components of a vector a are:

$\textbf{a}_x=\textbf{a}\cos\theta\hspace{10 mm}and\hspace{10 mm}\textbf{a}_y=\textbf{a}\sin\theta$

This means that each component is equal to the product of the vector’s magnitude and the cosine of the angle between the positive axis and the vector.

So the magnitude of the vector and the angle between the vector and the positive axis are, respectively,

$|a|=\sqrt{{a_x}^2+{a_y}^2}$

$\theta=tan^{-1}\frac{a_y}{a_x}$

Vector Addition and Subtraction

To add or subtract vectors, just add or subtract the components:

$\textbf{a}\pm\textbf{b}=(a_x,a_y,a_z)\pm(b_x,b_y,b_z)=(a_x{\pm}b_x,a_y{\pm}b_y,a_z{\pm}b_z)$

The geometric interpretation is similarly simple. If you start b where a ends, the sum a + b is given by the hypotenuse that makes triangle abc. And if you start a and b at the same point, the hypotenuse gives a — b. Easy!

Vector Multiplication

There are two ways to multiply vectors and each has a nifty geometric interpretation as well.

The dot product of two vectors is a scalar and the rules are the same as vector addition (a • b = a_xb_x + a_yb_y + a_zb_z) but there’s a shortcut!

Since a and b both start at (0, 0, 0), it is easier to calculate the angle θ between them rather than struggle individually with the x, y and z components. We can rotate our axes so that a runs along the x axis and b lies in the xy-plane. This would make the components (a, 0, 0) and (b cos θ, b sin θ, 0). So,

$\textbf{a}\cdot\textbf{b}=a_xb_x=|\textbf{a}||\textbf{b}|\cos\theta_{ab}$

That is, a • b, a scalar, is b times the portion of a that extends in the b direction.

Now’s a good a time as any to introduce the unit coordinate vectors. These are just vectors whose magnitudes are unity and whose directions run along the Cartesian axes.

The unit coordinate vectors are:

$\hat{\textbf{i}}=(1,0,0)\hspace{10 mm}\hat{\textbf{j}}=(0,1,0)\hspace{10 mm}\hat{\textbf{k}}=(0,0,1)$

(Pronounced “eye-hat,” “jay-hat” and “kay-hat”) This lets us separate the direction of the vector from the magnitude. Now we can express any vector like so:

$\textbf{a}=(a_x,a_y,a_z)=(a_x,0,0)+(0,a_y,0)+(0,0,a_z)\\=a_x(1,0,0)+a_y(0,1,0)+a_z(0,0,1)\\=a_x\hat{\textbf{i}}+a_y\hat{\textbf{j}}+a_z\hat{\textbf{k}}$

From the definitions of the unit coordinate vectors, we get the following relations:

$\hat{\textbf{i}}\cdot\hat{\textbf{i}}=\hat{\textbf{j}}\cdot\hat{\textbf{j}}=\hat{\textbf{k}}\cdot\hat{\textbf{k}}=|\hat{\textbf{i}}||\hat{\textbf{i}}|\cos0=(1,0,0)(1,0,0)(1)=1 \\\hat{\textbf{i}}\cdot\hat{\textbf{j}}=\hat{\textbf{i}}\cdot\hat{\textbf{k}}=\hat{\textbf{j}}\cdot\hat{\textbf{k}}=|\hat{\textbf{i}}||\hat{\textbf{j}}|\cos90=(1,0,0)(0,1,0)(0)=0$

The cross product of two vectors is a vector. Written as a matrix, we have:

$\textbf{a}\times\textbf{b}=\left(\begin{array}{ccc}\hat{\textbf{i}}&\hat{\textbf{j}}&\hat{\textbf{k}}\\a_x&a_y&a_z\\b_x&b_y&b_z\end{array}\right)$

To put this (or any) matrix into algebraic form, do the following to get your x, y, and z components. For each component in the top row (e.g. i) multiply diagonally to the right (e.g. ia_yb_z) and then to the left (e.g. ia_zb_y) and subtract the second from the first (e.g. ia_yb_z — ia_zb_y). Do this for all the components and evaluate. The above matrix then gives us:

$\textbf{a}\times\textbf{b}=\textbf{c}=\hat{\textbf{i}}(a_yb_z-a_zb_y)+\hat{\textbf{j}}(a_zb_x-a_xb_z)+\hat{\textbf{k}}(a_xb_y-a_yb_x)$

Geometrically speaking, the vector c that results from the cross product of vectors a and b has a magnitude given by the product of |a| and |b| times the sine of the angle from a to b, or |c| = ab sin θ.

The direction is given by the “right hand rule.” If you lay your right hand palm-up along a, then curl your fingers toward b, the resulting vector c is in the direction of your thumb. Take a look:

Properties of the cross product:

$\textbf{a}\times\textbf{b}=-\textbf{b}\times\textbf{a}$
$\textbf{a}\times(\textbf{b}+\textbf{c})=(\textbf{a}\times\textbf{b})+(\textbf{a}\times\textbf{c})$
$c\hspace{1mm}(\textbf{a}\times\textbf{b})=(c\hspace{1 mm}\textbf{a}) \times\textbf{b}=\textbf{a}\times(c\hspace{1 mm}\textbf{b})$
$|\textbf{a}\times\textbf{b}|^2=a^2b^2-(\textbf{a}\cdot\textbf{b})^2$

From this we get some more unit coordinate vector relations:

$\hat{\textbf{i}}\times\hat{\textbf{i}}=\hat{\textbf{j}}\times\hat{\textbf{j}}=\hat{\textbf{k}}\times\hat{\textbf{k}}=\hat{\textbf{i}}(0-0)+\hat{\textbf{j}}(0-0)+\hat{\textbf{k}}(0-0)=0$

$\hat{\textbf{i}}=\hat{\textbf{j}}\times\hat{\textbf{k}}=-\hat{\textbf{k}}\times\hat{\textbf{j}}\\\hat{\textbf{j}}=\hat{\textbf{k}}\times\hat{\textbf{i}}=-\hat{\textbf{i}}\times\hat{\textbf{k}}\\\hat{\textbf{k}}=\hat{\textbf{i}}\times\hat{\textbf{j}}=-\hat{\textbf{j}}\times\hat{\textbf{i}}$

Triple Products

The scalar triple product is the dot product of two vectors (since a vector “crossed” with a vector is a vector), which is a scalar:

$\textbf{a}\cdot(\textbf{b}\times\textbf{c})=\left(\begin{array}{ccc}a_x&a_y&a_z\\b_x&b_y&b_z\\c_x&c_y&c_z\end{array}\right)=(\textbf{a}\times\textbf{b})\cdot\textbf{c}$

Now we know that the dot and cross products may be interchanged in the scalar triple product.

The vector triple product comes in handy when evaluating rotating coordinate systems or the rotation of rigid bodies. But more on that later.

It can be shown that a × (b × c) = b(a • c) — c(a • b) but that’s a hell of a lot of LaTeX code. Try it out!

By the way, this is cordially referred to as the “back minus cab” rule for obvious reasons.

Classical Mechanics

Vectors

Vector Addition and Subtraction

Vector Multiplication

Triple Products

Problem Set 1 » Vectors