1-page summary sheet:

Review of Vectors

$\bullet$ Vector Notation: a vector quantity is one that has both magnitude and direction. Another (equivalent) way of putting it is that a vector quantity has several components in orthogonal (perpendicular) directions. The idea of a vector is very abstract and general; one can define useful vector spaces of many sorts, some with an infinite number of orthogonal basis vectors, but the most familiar types are simple 3-dimensional quantities like position, speed, momentum and so on. The conventional notation for a vector is $\Vec{A}$, sometimes written $\vec{\bf A}$ or $\vec{A}$ or A but most clearly recognizable when in boldface with a little arrow over the top. On the blackboard a vector may be written with a tilde underneath, which is hard to generate in L^ATEX.

 

$\bullet$ Unit Vectors: In Cartesian coordinates (x,y,z) a vector $\Vec{A}$ can be expressed in terms of its three scalar components A_x,A_y,A_z and the corresponding unit vectors $\iH, \jH, \kH$ (sometimes written as $\xH, \yH, \zH$ or occasionally as $\xH_1, \xH_2, \xH_3$) thus:

$$\Vec{A} = \iH A_x + \jH A_y + \kH A_z$$ (1)

where the little "hat" over a symbol means (in this context) that it has unit magnitude and thus imparts only direction to a scalar like A_x. ¹

A unit vector $\Hat{a}$ can be formed from any vector $\Vec{a}$ by dividing it by its own magnitude a:

$$\Hat{a} = {\Vec{a} \over a} \qquad \hbox{\rm where ~ } a = \vert\Vec{a}\vert = \sqrt{a_x^2 + a_y^2 + a_z^2} \; .$$ (2)

Already we have used a bunch of concepts before defining them properly, the usual chicken-egg problem with mathematics. Let's try to catch up:

 

$\bullet$ Multiplying or Dividing a Vector by a Scalar: Multiplying a vector $\Vec{A}$ by a scalar b has no effect on the direction of the result (unless b=0) but only on its magnitude and/or the units in which it is measured - if b is a pure number, the units stay the same; but multiplying a velocity by a mass (for instance) produces an entirely new quantity, in that case the momentum.

Dividing a vector by a scalar c is the same as multiplying it by 1/c.

This type of product always commutes: $\Vec{A} b = b \Vec{A}$.

 

$\bullet$ Adding or Subtracting Vectors: In two dimensions one can draw simple diagrams depicting "tip-to-tail" or "parallelogram law" vector addition (or subtraction); this is not so easy in 3 dimensions,

so we fall back on the algebraic method of adding components. Given $\Vec{A}$ from Eq. (1) and

$$\Vec{B} = \iH B_x + \jH B_y + \kH B_z$$ (3)

we write

$$\Vec{A} + \Vec{B} = \iH (A_x + B_x) + \jH (A_y + B_y) + \kH (A_z + B_z) \; .$$ (4)

Subtracting $\Vec{B}$ from $\Vec{A}$ is the same thing as adding $-\Vec{B}$.

 

$\bullet$ Multiplying Two Vectors . . .

. . .

to get a Scalar: we just add together the products of the components,

$$\Vec{A} \cdot \Vec{B} = A_x B_x + A_y B_y + A_z B_z$$ (5)

also known as the "dot product", which commutes: $\Vec{A} \cdot \Vec{B} = \Vec{B} \cdot \Vec{A}$.

. . .

to get a Pseudovector:

 

$$\Vec{A} \times \Vec{B}$$ = i (A_yB_z - A_zB_y) (6)

+ j (A_zB_x - A_xB_z)

+ k (A_xB_y - A_yB_x)

This "cross product" is actually a pseudovector (or, more generally, a tensor), because (unlike the nice dot product) it has the unsettling property of not commuting ( $\Vec{A} \times \Vec{B} = - \Vec{B} \times \Vec{A}$) but we often treat it like just another vector.

We are going to be using these things continually throughout the rest of the course, so make sure you are adept with them.