Part 1: Elementary Properties

In elementary physics and mathematics courses, vectors are often defined to be quantities that have magnitude and direction. In reality, however, the magnitude and direction of a vector depend on the choice of an inner product. In this section, we introduce the standard inner product and how it is used to determine the magnitude and direction of a vector.

To begin with, if u = á u1,u2,u3 ñ and v = á v1,v2,v3 ñ , then their inner product is defined
u·v = u1v1+u2v2+u3v3
For example, if u = á 1,2,3 ñ and v = á 5,6,7 ñ , then
u·v = 1·5+2·6+3·7 = 38
Likewise, if u = á -2,3 ñ and v = á 9,6 ñ , then
u·v = -2·9+3·6 = 0

The result of an inner product, which is also known as a dot product, is a scalar-i.e., the result is a number. Other properties of the inner product which follow immediately from the definition include the following:      

Theorem 1.1: If u,v and w are vectors and k is a scalar, then
i)
u·v = v·u
ii)
( ku) ·v = k( u·v) = u·( kv)
iii)
u·( v+w) = u·v+u·w
iv)
v·v ³ 0 and v·v = 0 only if v = 0

If v = á a,b ñ , then v·v = a2+b2 , so that in accordance with the Pythagorean theorem, the length ||v|| of the vector v is defined to be
|| v|| =  

a2 + b2
  = 

v · v

That is, the length || v|| of a 2 dimensional vector satisfies the identity
||v||2  =  v · v.

EXAMPLE 2    Find the length of v = á3,4 ñ       

Solution: Our definition implies that the length of v satisfies
|| v|| 2 = v·v = 32+42 = 25
Thus, the length of vector v = á 3,4 ñ  is || v|| = 5.


(Vector in applet can be dragged about screen and resized by click/drag on endpoints).

Likewise, we define the length of a 3 dimensional vector v = á a,b,c ñ to be

|| v|| =   

a2+b2+c2
 = 

v·v

It then follows that if u,v are vectors and k is a scalar, then the following properties hold:
i.
|| u+v|| £ || u||+|| v||         (triangle inequality)
ii.
|| kv|| = | k|   || v||
iii.
|| v|| = 0 if and only if v = 0.
We will prove these in the exercises.           

EXAMPLE 2    The distance between two points P1(x1,y1,z1) and P2( x2,y2,z2) is defined to be the length of the vector between them. What is the distance from P1 to P2?    

Solution: The vector between P1 and P2 is

v =

P1P2

  =  áx2-x1, y2-y1, z2-z1 ñ
 
Consequently, the distance between P1 and P2 is
Dist( P1,P2) =  
v·v
  =  
(x2-x1)2 + ( y2-y1)2 + (z2-z1)2

 

   

Since a sphere of radius R centered at ( a,b,c) is the set of all point ( x,y,z) that are a distance R from ( a,b,c) , the distance formula in example 2 implies that the equation of the sphere of radius R centered at ( a,b,c) is
( x-a)2 + ( y-b)2 + ( z-c)2 = R2
The unit sphere is the sphere with equation x2+y2+z2 = 1.

Check your Reading: How long is the vector v = á 2,2,1 ñ ?