The Dot Product
Sections 6.7
Objectives
• Calculate the dot product of two
vectors.
• Calculate the angle between two
vectors.
• Use the dot product to determine
if two vectors are orthogonal,
parallel, or neither.
Vocabulary
• dot product
a product of two vectors formed by summing the
product of their vertical components and the product
of their horizontal components which produces a
scalar number (not a vector)
• orthogonal
two vectors are orthogonal if their dot product is 0
• parallel
two vectors are parallel if the angle between them is
either 0° or 180°
Formulas
• Dot Product of v  a1i  b1 j and w  a2i  b2 j
v  w  a1a2  b1b2
OR
v  w  v w cos 
• Angle between two vectors (θ is the
smallest non-negative angle between the
two vectors)
vw
cos  
v w
and
 vw 

  cos 

v
w


1
Compute the dot product of each pair
of vectors:
Our first step in find each of these dot products is to put
the vectors into standard rectangular coordinates. The
vector i in rectangular coordinates is i = 1i + 0j. The vector j
in rectangular coordinates is j = 0i + 1j.
ii
i j
j j
is
(1i  0 j)  (1i  0 j)
is
(1i  0 j)  (0i  1 j)
is
(0i  1 j)  (0i  1 j)
continued on next slide
Compute the dot product of each pair
of vectors:
ii
is
(1i  0 j)  (1i  0 j)
For this problem, the dot product formula to use is v  w  a1a2  b1b2
In our problem a1 is 1, a2 is 1, b1 is 0, and b2 is 0.
i  i  (1)(1)  (0)(0)
ii  1
continued on next slide
Compute the dot product of each pair
of vectors:
i j
is
(1i  0 j)  (0i  1 j)
For this problem, the dot product formula to use is v  w  a1a2  b1b2
In our problem a1 is 1, a2 is 0, b1 is 0, and b2 is 1.
i  j  (1)(0)  (0)(1)
i j  0
continued on next slide
Compute the dot product of each pair
of vectors:
j j
is
(0i  1 j)  (0i  1 j)
For this problem, the dot product formula to use is v  w  a1a2  b1b2
In our problem a1 is 0, a2 is 0, b1 is 1, and b2 is 1.
j  j  (0)(0)  (1)(1)
j j 1
Given the vectors u = 8i + 8j and
v = —10i + 11j find the following.
uv
is
(8i  8j)  (10i  11 j)
For this problem, the dot product formula to use is v  w  a1a2  b1b2
In our problem a1 is 8, a2 is -10, b1 is 8, and b2 is 11.
u  v  (8)(10)  (8)(11)
u  v  80  88
uv  8
continued on next slide
Given the vectors u = 8i + 8j and
v = —10i + 11j find the following.
vu
is
(10i  11 j)  (8i  8j)
For this problem, the dot product formula to use is v  w  a1a2  b1b2
In our problem a1 is -10, a2 is 8, b1 is 11, and b2 is 8.
v  u  ( 10)(8)  (11)(8)
v  u  80  88
vu 8
You should notice that when we computed the dot product
with the vectors u•v first we got the same answer as when
we switched the order of the vectors and calculated the
dot product as v•u. This means that the dot product is
commutative.
continued on next slide
Given the vectors u = 8i + 8j and
v = —10i + 11j find the following.
vv
is ( 10i  11 j)  ( 10i  11 j)
For this problem, the dot product formula to use is v  w  a1a2  b1b2
In our problem a1 is -10, a2 is -10, b1 is 11, and b2 is 11.
v  v  ( 10)(10)  (11)(11)
v  v  100  121
v  v  221
Given the vectors u = 8i + 8j,
v = —10i + 11j, and w = 9i + 7j find
the following.
u  v  w 
We start by doing the part in the parentheses
v  w  ( 10i  11 j)  (9i  7 j)
v  w  10i  11 j  9i  7 j
v  w  10i  9i  11 j  7 j
v  w  1i  18 j
For this problem, the dot product formula to use is
v  w  a1a2  b1b2
continued on next slide
Given the vectors u = 8i + 8j,
v = —10i + 11j, and w = 9i + 7j find
the following.
u  v  w 
We will now do the dot product of our result from the previous slide
and the vector u.
For this problem, the dot product formula to use is
v  w  a1a2  b1b2
In our problem a1 is 8, a2 is -1, b1 is 8, and b2 is 18.
u  ( v  w)  (8i  8 j)  ( 1i  18 j)
u  ( v  w)  (8)(1)  (8)(18)
u  ( v  w)  8  144
u  ( v  w)  136
continued on next slide
Given the vectors u = 8i + 8j,
v = —10i + 11j, and w = 9i + 7j find
the following.
u v  uw
We start with doing the dot product of u and v and the dot product
of u and w.
For this problem, the dot product formula to use is
v  w  a1a2  b1b2
u  v  (8i  8 j)  ( 10i  11 j)
u  w  (8i  8 j)  (9i  7 j)
u  v  (8)(10)  (8)(11)
u  w  (8)(9)  (8)(7)
u  v  80  88
u  w  72  56
uv  8
u  w  128
continued on next slide
Given the vectors u = 8i + 8j,
v = —10i + 11j, and w = 9i + 7j find
the following.
u v  uw
Now we will add the dot products from the previous slide.
u v  8
u  w  128
u  v  u  w  8  128
u  v  u  w  136
Notice that this is the same as the answer that we got for u•(v + w).
This tells us that the dot product can be distributed over addition and
subtraction.
continued on next slide
Given the vectors u = 8i + 8j,
v = —10i + 11j, and w = 9i + 7j find
the following.
3v  u  12w  w 
We start with doing the dot product of v and u and the dot product
of w and w.
For this problem, the dot product formula to use is
v  w  a1a2  b1b2
v  u  ( 10i  11 j)  (8i  8 j) w  w  (9i  7 j)  (9i  7 j)
v  u  ( 10)(8)  (11)(8)
w  w  (9)(9)  (7)(7)
v  u  80  88
w  w  81  49
vu 8
u  w  130
continued on next slide
Given the vectors u = 8i + 8j,
v = —10i + 11j, and w = 9i + 7j find
the following.
3v  u  12w  w 
Now we multiply by the dot products we found by 3 and 12,
respectively, and subtract the two results.
3( v  u)  12( w  w)  3(8)  12(130)
3( v  u)  12( w  w)  24  1560
3( v  u)  12( w  w)  1536
Find the angle θ in degrees
measured between the vectors
u = 10i + 3j and v = 1i — 7j .
For this problem, we want to use the formula for the angle between two
vectors.


1 v  w
  cos 


v
w


For this we need the to find the magnitude of each vector and the
dot product of the two vectors.
For this problem, the dot product formula to use is v  w  a1a2  b1b2
In our problem a1 is 10, a2 is 1, b1 is 3, and b2 is -7.
u  v  (10)(1)  (3)(7 )
u  v  10  21
u  v  11
continued on next slide
Find the angle θ in degrees
measured between the vectors
u = 10i + 3j and v = 1i — 7j .
 vw 

  cos 

v
w


1
u  v  11
Now that we have the dot product, we need to find the magnitude of
each vector. We can use the alternate formula for magnitude.
v  a 2  b 
2
Magnitude of u
alternate magnitude formula.
Magnitude of v
u  (10)2  (3)2
v  (1)2  ( 7 )2
u  100  9
v  1  49
u  109
v  50
continued on next slide
Find the angle θ in degrees measured between
the vectors u = 10i + 3j and v = 1i — 7j .
 vw
 v w
  cos 1 
u  109




u  v  11
v  50
Now we are set to plug everything into the formula to find the angle
between the vectors u and v.




 uv 

 cos 

u
v


 11


 cos 1 

 109 50 
 11 
1 
 cos 

 5450 
 98.56914188
1
Determine if the pair of vectors
is orthogonal, parallel, or neither.
The first step necessary to answer this question is to find the dot
product of the two vectors. If the dot product is 0, then the vectors
are orthogonal and we can stop. If the dot product is not 0, then we
must go on to find the angle between the vectors.
16
u  8i  3 j and v  2i 
j
3
For this problem, the dot product formula to use is v  w  a a
1 2
 b1b2
In our problem a1 is -8, a2 is 2, b1 is 3, and b2 is 16/3.
16 

u  v  ( 8i  3j)   2i  j 
3 

 16 
u  v  ( 8)(2)  (3) 
3
u  v  16  16
uv  0
Since the dot product is 0, we
can say that the vectors u and v
are orthogonal.
continued on next slide
Determine if the pair of vectors
is orthogonal, parallel, or neither.
u  7i  3j and v  21i  9 j
For this problem, the dot product formula to use is v  w  a1a2  b1b2
In our problem a1 is 7, a2 is 21, b1 is -3, and b2 is -9.
16 

u  v  ( 8i  3j)   2i  j 
3 

u  v  (7 )(21)  ( 3)(9)
u  v  147  27
u  v  174
Since the dot product is not equal to 0, we know that the two vectors
are not orthogonal. This means we need to continue with the formula
for finding the angle between the two vectors by calculating the
magnitude of each vector.
continued on next slide
Determine if the pair of vectors
is orthogonal, parallel, or neither.
u  7i  3j and v  21i  9 j
 vw 

  cos 

v
w


1
u  v  174
Now that we have the dot product, we need to find the magnitude of
each vector. We can use the alternate formula for magnitude.
v  a 2  b 
2
alternate magnitude formula.
Magnitude of u
Magnitude of v
u  (7)2  ( 3)2
v  (21)2  ( 9)2
u  49  9
v  441  81
u  58
v  522
continued on next slide
Determine if the pair of vectors is orthogonal, parallel,
or neither. u  7i  3j and v  21i  9 j
 vw
 v w
  cos 1 
u  58
v  522




u  v  174
Now we are set to plug everything into the formula to find the angle
between the vectors u and v.
 uv
 u v




174


  cos 1 

 58 522 
 174 
  cos 1 

 30276 
  cos 1 
 174 

174


  cos 1 1  0 
  cos 1 
Since the angle between the
two vectors is 0 degrees, the
vectors are parallel.