Elementary Vector Analysis

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

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 ì
í
î || u || || v || cosq if u ¹ 0, v ¹ 0 0 if u = 0 or v = 0 where 0 £ q £ p is the angle between u and v

 

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 || = 2Ö5,

u · v = 0 = 6Ö5cosq

so cosq = 0, yielding q = p/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.

 

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||cosq, where 0 £ q £ p is the angle between u and v. Let's normalize u to u/|| u || and then scale this by the magnitude || v ||cosq:

 

Projection of v onto u	= (|| v || cosq) u ||u||
	= ||v|| ||u|| cosq ||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

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

It can also be shown that

 

|| u × v || = || u || || v || sinq for u ¹ 0, for v ¹ 0

where 0 £ q £ p is the angle between u and v.

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.

 

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.

In the following Exploration, select values for the components of u and v. You will see u · v and u × v computed and u, v, and u × v displayed on a coordinate system.

 

Exploration

 

Key Concepts

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 || =   Ö x2+y2+z2

   

• Dot Product

  u · v

  nowrap="nowrap">u₁v₁+u₂v₂+u₃v₃ = ì
  í
  î || u|| ||vvcosqif u ¹ 0, v ¹ 0 0u = 0 or v = 0

  where 0 £ q £ p is the angle between u and v

  For u ¹ 0, v ¹ 0, 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 || sinq for u ¹ 0, for v ¹ 0

  where 0 £ q £ p is the angle between u and v.

  For u ¹ 0, v ¹ 0, u × v = 0 if and only if u is parallel to v.