Dot

 Product  of  Vectors  

November  2017  
 Dot  Product  of  Vectors  
•  Defini'on:  
     If  u and  v  are  vectors  in  2-­‐space  or  3-­‐space  and  θ
is  the  angle  between  u and  v,  then  the  dot  
product  or  Scalar  Product    is  defined  by  
  % u v cosθ if u ≠ 0 and v ≠ 0
u ⋅ v =&
  '0 if u = 0 or v = 0
     By  the  angle  between  u  and  v,  we  mean  the  
angle  determined  by  u  and  v  that  saFsfies                              
       0      ≤      θ        ≤      π          (see  Figure  below).  
                                           
  
 Figure:The  
Figure 3.3.1 angle  θ between  u and  v saFsfies   0 ≤ θ ≤ π .
  The angle ! between u and v satisfies .
 
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INITION
          Component  Form  of  the  Dot  Product    
•  Let  u = (u1,u2, u3) and  v = (v1, v2, v3) be  two  
nonzero  vectors.  If  θ is  the  angle  between  u and  
v (as  shown  in  Figure  below),    
        then  the  law  of  cosines  yield  
2 2 2
PQ = u + v − 2 u v cos θ

•  Since,          PQ
             =     v − u
     we  can  rewrite  the  above  equaFon    as  
  2
v − u = u + v − 2 u v cosθ
2 2

 
€
1
  u v cos θ = ( 2
2 2
)
u + v − v −u
2

Or  
1
u⋅v= ( 2
2 2
)
u + v − v −u
2
•  SubsFtuFng  
2 2 2 2 2 2 2 2
      u = u1 + u2 + u3 , v = v1 + v 2 + v 3
       and  
2 2 2 2
                 v      −
       u              =
       (v
  1 − u1 ) + (v 2 − u2 ) + (v 3 − u3 )
we  obtain,  aRer   €simplifying  
  u ⋅ v = u1v1 + u2v 2 + u3v 3

•  If    u = (u1,u2) and  v = (v1, v2)    are  vectors  in  R2,

then  their  dot  product  is  defined  by  
u ⋅ v = u1v1 + u2v 2
 
•  If    u = (u1,u2,…, un) and  v = (v1, v2, …, vn)    are  
vectors  in  Rn, then  their  dot  product  is  defined  
by  
u ⋅ v = u1v1 + u2v 2 +! + un v n
   
•  If  u is a  vectors  in  Rn,  then  we  can  use  the  
definiFon  of  dot  product  in  Rn  to  write  
u = u⋅u

•  Or               u
2
=u ⋅u
€
 Theorem    (ProperFes  of  the  Dot  Product)  
       If  u, v,  and  w are  vectors  in  Rn  and  c is  a  scalar,  
then  
  (a) u ⋅ u > 0 for u ≠ 0;
  u ⋅ u = 0 if and only if u = 0.
  (b) u ⋅ v = v ⋅ u
  (c) (u + v) ⋅ w = u ⋅ w + v ⋅ w
  (d) (cu) ⋅ v = u ⋅ (cv) = c(u ⋅ v)
 
Proof:  Exercise  
 Angle  Between  Two  Vectors  
•  Let  θ be  the  angle  between  two  nonzero  
vectors  u = (u1, u2) and  v = (v1, v2), as shown
in the figure below.

   
•  Applying  the  law  of  cosines  to  the  triangle  in  that  
figure,  we  obtain  
2 2 2
u − v = u + v − 2 u v cosθ (*)

•  But  
2 2 2
u−v = (u1 − v1 ) + (u2 − v 2 )
2 2 2 2
= u + v + u + v − 2(u1v1 + u2v 2 )
1 1 2 2
2 2
= u + v − 2(u ⋅ v)

•  SubsFtuFng  this  expression  in (*), we  obtain    

2 2 2 2
u + v − 2 u v cosθ = u + v − 2(u ⋅ v)
 •  Or     u v cos θ = (u ⋅ v)
•  Recall  that  since  u and  v are  nonzero  vectors,  
then   u ≠ 0 and v ≠ 0
 
•  Therefore  
€
u⋅v
cosθ = (0 ≤ θ ≤ π ).
u v
•  Or          
% u⋅v (
−1
θ = cos ' *.
& u v )
Theorem  
•  If  the  vectors  u  and  v  are  nonzero  and  θ  is  the  
angle  between  them,  then  
(a) θ is acute (i.e.,0 o ≤ θ < 90 o ) if and only if u ⋅ v > 0
(b) θ is obtuse (i.e.,90 o < θ ≤ 180 o ) if and only if u ⋅ v < 0
   
(c) θ = 90 o if and only if u ⋅ v = 0
   
 Example    
•  If  u = (1, −2, 3), v = (− 3 ,4, 2),  and  w = (3, 6, 3),
then                        
 u ⋅ v = (1)(−3) + (−2)(4) + (3)(2) = −5
v ⋅ w = (−3)(3) + (4)(6) + (2)(3) = 21
u ⋅ w = (1)(3) + (−2)(6) + (3)(3) = 0

       Therefore,    
•  u  and  v  make  an  obtuse  angle,  
•  v  and  w  make  an  acute  angle,  and    
•  u  and  w  are  perpendicular.  
 Perpendicular  (Orthogonal)  Vectors  
       Defini'on:    
•  Two  nonzero  vectors  u  and  v  are  called  
perpendicular  or  orthogonal  if  the  angle  
between  them  is  90o,  denoted  as  u⊥v.
u ⊥ v iff u ⋅ v = 0

•  Example:  
       For  
€vectors  u = (3, 6, −4) and  v = (−2, k , 1),
determine  the  value  of  k  such  that  the  two  
vectors  are  perpendicular.
  Solu'on:  
               
u ⊥ v iff u ⋅ v = 0
u ⋅ v=0
⇒ (3, 6, − 4)⋅ (−2, k,1) = 0
€
3(−2) +€(6)(k) − (4)(1) = 0
−6 + 6k − 4 = 0
  6k = 10
5
k=
3
Example:  
a) Compute  the  distance  between  P1(3, -1, 2) and  
P2(-1, 1, 0)
b) Show  that  the  points  P(3, -1, 1), Q(4, 1, 4) and  
R(6, 0, 4) are  verFces  of  a  right  angled  triangle.  
 
         Solu'on:  
a)   The  distance  between  two  arbitrary  points                  
P1(x1, y1, z1) and  P2(x2, y2, z2) is  given  by  
                   
d = (x 2 − x1 ) 2 + (y 2 − y1 ) 2 + (z 2 − z1 ) 2
       
•  Therefore,  distance  between  P1(3, -1, 2) and        
P2(-1, 1, 0)  is  given  by  
2 2 2
d= (−1 − 3) + (1+1) + (0 − 2)
= 16 + 4 + 4
  = 24
 (b) A  right  angled  triangle  must  have  two  sides  
forming  a  right  angle,  and  this  happens  iff                            
two  of  its  sides  are  orthogonal  to  each  other,  
(i.e.,  iff  the  corresponding  vectors'  dot    product  is  
zero)  
 PQ = (4,1, 4) − (3, −1,1) = (1, 2, 3),
QR = (6, 0, 4) − (4,1, 4) = (2, −1, 0),
PR = (6,0,4) − (3, −1,1) = (3,1, 3).

•  Check    PQ ⋅ QR, PQ ⋅ PR and PR ⋅ QR

•  Thus,    
                         
PQ ⋅ QR = (1, 2, 3) ⋅ (2, −1, 0) = 2 − 2 + 0 = 0
€
  Hence PQ and QR are perpendicular.
•  Therefore  triangle  PQR  has  a  right  angle  at  Q.
 Theorem    (Cauchy-­‐Schwarz  Inequality)  
•   If  u and  v are  vectors  in  Rn,  then  
u⋅v≤ u v.
       (Note  that  |  |  on  the  leR  stands  for  the  absolute  
value  of  a  real  number;                    on  the  right  denotes  
the  length  of  a  vector.)  
€
Proof:    
•  If  either  u  or  v  is  the  zero  vector,  then    
€
     
       u   ⋅ v = 0 and u ⋅ v ≤ u v simply says that 0 ≤ 0,
       which  is  certainly  true.    
•  On  the  other  hand,  if    u  and  v  are  nonzero  vectors,  
we  have  

u ⋅ v = u v cosθ , so
u ⋅ v = u v cos θ , so
u ⋅ v ≤ u v since cos θ ≤ 1 for all values of θ .
 Theorem  (Triangle  Inequality)  
•   If  u and  v are  vectors  in  Rn,  then  
  u+ v ≤ u + v
•  This    triangle  inequality  in  R2  and  R3  merely  states  
that  the  length  of  a  side  of  a  triangle  does  not    
exceed  the  sum  of  the  lengths  of  the  other  two  
€
sides  (  see  the  figure  below)    
 
 Proof:  
•  We  have,  from  the  definiFon  of  the  length  of  a  
vector,  
2
u+ v = (u + v) ⋅ (u + v)
= u ⋅ u + 2(u ⋅ v) + v ⋅ v
2 2
= u + 2(u ⋅ v) + v .

•  So  since  a ≤ |a| for  any  real  number  a, we  have    

€   2 2 2 2
u + 2(u ⋅ v) + v ≤ u + 2 u ⋅ v + v
•  So  by  the  Cauchy-­‐Schwarz  inequality  we  have  
2 2 2 2
u + 2u ⋅ v + v ≤ u +2 u v + v
2
=( u + v )
•  Taking  square  roots,  we  obtain  
    u+ v ≤ u + v .

•  Note:  The  inequality  only  becomes  an  equality  

when  u  and  v  are  parallel.    
€
 Theorem  (Pythagorean  Theorem  )  
•  If  u and  v are  vectors  in  Rn,  then  
2 2 2
          u+ v = u + v
       if  and  only  if  u and  v are  orthogonal.  
•  Proof:  Exercise      
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