Dot Product

Report
The definition of the product of two vectors is:
v w  a1a2  b1b2
where v  a1i  b1 j and w  a2i  b2 j
If v  2i  5 j and w  4i 1 j, find v  w
v  w    2  4   5 1
 8  5  3
This is called the dot product. Notice the answer is just
a number NOT a vector.
The dot product is useful for several things. One of
the important uses is in a formula for finding the angle
between two vectors that have the same initial point.
If u and v are two nonzero vectors, the angle
 , 0   <  , between u and v is determined
by the formula cos   u  v
u v
v

u
Technically there are two angles between
these vectors, one going the "shortest" way
and one going around the other way. We
are talking about the smaller of the two.
Find the angle  between u = 2i  j and v = 4i + 3j.
uv
cos  
u v
u  v   2 4    13  8  3  5
u 
2   1 
2
2
5
v  4  3  16  9  25  5
2
v  4i  3j

u  2i  j
2
uv
cos  

u v
  cos
1
5
5 5

1
5
1
 63.4
5
Find the angle between the vectors
v = 3i + 2j and w = 6i + 4j
vw
18

8
cos 


v w
13 52
  cos 1  0
1
w  6i  4 j
v  3i  2 j
26
676
1
What does it mean when the
angle between the vectors is 0?
The vectors have the same
direction. We say they are
parallel because remember
vectors can be moved around
as long as you don't change
magnitude or direction.

If the angle between 2 vectors is
, what would their dot
w
=
2i
+
8j
2
product be?

uv

cos  
u v
2
Since cos 2 is 0, the
dot product must be 0.
Vectors u and v in thisvcase
are- called
orthogonal.
=
4i
j
(similar to perpendicular but refers to vectors).
Determine whether the vectors v = 4i - j and
w = 2i + 8j are orthogonal.
compute their dot product
and see if it is 0
v  w   4 2    18  0
v w  0
The vectors v and w are orthogonal.
A use of the dot product is found in the formula below:
The work W done by a constant force F in
moving an object from A to B is defined
as

W  F  AB
This means the force is in
some direction given by the
vector F but the line of
motion of the object is along
a vector from A to B
Find the work done by a force of 50 kilograms acting in the
direction 3i + j in moving an object 20 metres from (0, 0) to
(20, 0).
Let's find a unit vector in the direction 3i + j
Remember to get a unit
vector, divide a vector by
it's magnitude

3i + j
W  F  AB
20i + 0j
(20, 0)
 3
1 
W  50 
i
j    20 i  0 j 
10 
 10
3i  j 
3  1
2
2
 10
3
1
u
i
j
10
10
Our force vector is in this
direction but has a
magnitude of 50 so we'll
multiply our unit vector by
50.

3000
W 
 948.68 m kgs F  50

10
3
1 
i
j
10
10 
Acknowledgement
I wish to thank Shawna Haider from Salt Lake Community College, Utah
USA for her hard work in creating this PowerPoint.
www.slcc.edu
Shawna has kindly given permission for this resource to be downloaded
from www.mathxtc.com and for it to be modified to suit the Western
Australian Mathematics Curriculum.
Stephen Corcoran
Head of Mathematics
St Stephen’s School – Carramar
www.ststephens.wa.edu.au

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