Elementary Vector Analysis

In order to measure many physical quantities, such as force or velocity, we need to determine both a magnitude and a direction. Such quantities are conveniently represented as vectors.

The direction of a vector v in 3-space is specified by its components in the x, y, and z directions, respectively:

(x,y,z) or xi + yj + zk,
i = (1,0,0)
j = (0,1,0)
k = (0,0,1)

where i, j, and k are the coordinate vectors along the x-, y-, and z-axes.

The magnitude of a vector v = (x,y,z), also called its length or norm, is given by


|| v || =
x2+y2+z2
 

Notes

  • Vectors can be defined in any number of dimensions, though we focus here only on 3-space.

  • When drawing a vector in 3-space, where you position the vector is unimportant; the vector's essential properties are just its magnitude and its direction. Two vectors are equal if and only if corresponding components are equal.

  • A vector of norm 1 is called a unit vector. The coordinate vectors are examples of unit vectors.

  • The zero vector, 0 = (0,0,0), is the only vector with magnitude 0.

Basic Operations on Vectors

To add or subtract vectors u = (u1,u2,u3) and v = (v1,v2,v3),

add or subtract the corresponding coordinates:


u + v = (u1+v1,u2+v2,u3+v3)

u - v = (u1-v1,u2-v2,u3-v3)



To mulitply vector u by a scalar k, multiply each coordinate of u by k:


k u = (ku1,ku2,ku3)

Example

The vector v = (2,1,-2) = 2i + j - 2k has magnitude

|| v || =   ___________
22 +12 -(-2)2
 
= 3.

Thus, the vector 1/3v = (2/3,1/3,-2/3) is a unit vector in the same direction as v.

In general, for v0, we can scale (or normalize) v to the unit vector v/ ||v|| pointing in the same direction as v.

Le componenti del versore (vettore unitario) u ottenuto normalizzando v = (v1,v2,v3) vengono dette anche coseni direttori (direction cosines) di v in quanto sono dati dal coseno degli angoli che v forma con i tre assi coordinati:  ui = cos(∠ v xi) (avendo indicato gli assi x, y e z con x1, x2 e x3).
La figura a lato illustra il caso di u3. Con un'analoga costruzione si possono illustrare gli altri due casi.
Si noti che posso scrivere sia cos(∠ v xi) che cos(∠ xi v) in quanto cos(α) = cos(–α).
 
Una somma del tipo  a1v1+a2v2+…+anvn  (con ai numeri reali e vi vettori) viene detta combinazione lineare dei vettori v1, v2, …, vn.  Ogni vettore può essere espresso (in modo unico) come combinazione lineare di i, j e k.

 

Dot Product   (prodotto scalare)

Let u = (u1,u2,u3) and v = (v1,v2,v3). The dot product u · v (also called the scalar procuct or Euclidean inner product) of u and v is defined in two distinct (though equivalent) ways:


u · v
=
u1v1+u2v2+u3v3
u · v is a number
=





|| u || || v || cosθ
if u0, v0
0
if u = 0 or v = 0
  where 0 ≤ θ ≤ π is the angle between u and v

Why are the two definitions equivalent?

Properties of the Dot Product

  • u · v = v · u

  • u · (v + w) = (u · v) + (u · w)

  • u · u = || u ||2

See if you can verify each of these!

Example

If u = (1,-2,2) and v = (-4,0,2), then
u · v
=
(1)(-4)+(-2)(0)+(2)(2)
=
-4+0+4
=
0

Using the second definition of the dot product with || u || = 3 and || v || = 2√5,

u · v = 0 = 6√5cosθ
so cosθ = 0, yielding θ = π/2.

Though we might not have guessed it, u and v are perpendicular to each other!

In general,


Two non-zero vectors u and v are perpendicular (or orthogonal) if and only if u ·v = 0.

Proof

Projection of a Vector

It is often useful to resolve a vector v into the sum of vector components parallel and perpendicular to a vector u.

Consider first the parallel component, which is called the projection of v onto u. This projection should be in the direction of u and should have magnitude ||v||cosθ, where 0 ≤ θ ≤ π is the angle between u and v. Let's normalize u to u/|| u || and then scale this by the magnitude || v ||cosθ:

Projection of v onto u
= (|| v || cosθ) u
||u||
 
= ||v|| ||u|| cosθ
||u||2
u
= v · u
||u||2
u

The perpendicular vector component of v is then just the difference between v and the projection of v onto u.

In summary,

projection of v onto u:
= v · u
||u||2
u
vector component of v
perpendicular to u:
= v - v · u
||u||2
u

Cross Product   (prodotto vettoriale)

Let u = (u1,u2,u3) and v = (v1,v2,v3). The cross product u × v yields a vector perpendicular to both u and v with direction determined by the right-hand rule. Specifically,


u × v is a vector
u × v = (u2v3-u3v2)i - (u1v3-u3v1)j + (u1v2-u2v1)k
or:   u × v =  |
|
|
|
|
  i   j   k  |
|
|
|
|
uxuyuz
vxvyvz

It can also be shown that


|| u × v || = || u || || v || sinθ for u0, for v0

where 0 ≤ θ ≤ π is the angle between u and v.

Proof

Thus, the magnitude || u ×v || gives the area of the parallelogram formed by u and v.

As implied by the geometric interpretation,


Non zero vectors u and v are parallel if and only if u × v = 0.

Proof

Properties of the Cross Product

  • u × v = - ( v × u)

  • u × ( v + w ) = (u × v ) + ( u × w )

  • u × u = 0

Again, see if you can verify each of these.

Connections between the dot product and cross product (Lagrange's Identity)


Key Concepts [index]

Let u = (u1,u2,u3) and v = (v1,v2,v3).

  • Basic Operations, Norm of a vector

    u + v
    =
    (u1+v1,u2+v2,u3+v3)
    u - v
    =
    (u1-v1,u2-v2,u3-v3)
    k u
    =
    (ku1,ku2,ku3)
    || v ||
    =

    x2+y2+z2
     

  • Dot Product

    u · v
    =
    u1v1+u2v2+u3v3
    =





    || u || || v || cosθ
    if u0, v0
    0
    u = 0 or v = 0
          where 0 ≤ θ ≤ π is the angle between u and v

    For u0, v0, u · v = 0 if and only if u is orthogonal to v.

  • Projection of a Vector

    projection of v onto u:
    = v · u
    ||u||2
    u
    vector component of v
    perpendicular to u:
    = v - v · u
    ||u||2
    u

  • Cross Product

    u × v = (u2v3-u3v2)i - (u1v3-u3v1)j + (u1v2-u2v1)k

    || u × v || = || u || || v || sinθ for u0, for v0

    where 0 ≤ θ ≤ π is the angle between u and v.

    For u0, v0, u × v = 0 if and only if u is parallel to v.