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Section 12.3 – The Dot Product
Definition: If a  a1, a2 , a3 and b  b1, b2 , b3 , then the dot product of a and b ,
written a  b, is given by:
a  b  a1b1  a2b2  a3b3
In the case that a  a1 , a2 and b  b1 , b2 , a  b  a1b1  a2b2 .
Example:
2,4  3, 1 
Example:
1,7,4  6,2, 
Example:
1

2
i  2j  3k    2j  k  
Notice that the dot product of two vectors is a scalar! We can only take the dot product
of two vectors.
Properties of the Dot Product: If a,b,c V3 , and k  , then the following hold:
1) a  b  b  a
2) 0  a  0
 k a   b  k a  b  a  k b
4) a   b  c   a  b  a  c
3)
5) a  a  a
2
The dot product has an interesting geometric interpretation:
 is the angle between vectors a and b , such that 0     , then
ab
.
a  b  a b cos , or, equivalently, cos 
a b
Theorem: If
If we apply the Law of Cosines to triangle OAB here, we get:
|AB| 2 = |OA| 2 + |OB| 2 – 2|OA||OB| cos θ

Observe that
the Law of Cosines
still applies in
the limiting cases
when θ = 0 or π, or
a = 0 or b = 0
Example: Find a  b if a  4, b  10, and the angle between a and b is
2
.
3
Example: Find the angle between the vectors a  2i  j  k and b  3i  2 j  k .
Example: Find the three angles of the triangle with vertices at A  0,1,1 , B  2,4,3 ,
and C 1,2, 1 .
Definition: Two nonzero vectors are said to be perpendicular, or orthogonal, if the
angle between them is  

2
. (The zero vector 0 is considered to be perpendicular to all
vectors.)
Claim: Two nonzero vectors, a and b , are orthogonal if and only if a  b  0 .
The dot product is positive if a and b point in the same “general” direction, 0 if they are
perpendicular and negative if they point in “generally” opposite directions.
What are  and cos  when a and b point in exactly opposite directions?
What are  and cos  when a and b point in exactly the same direction?
Example: Tell whether each pair of vectors is parallel (what does that mean?),
orthogonal, or neither:
1) a  3,9,6 ,b  4, 12, 8
2) a  i  j  2k, b  2i  j  k
3) u  a, b, c , v  b, a,0
4) u  2,2, 1 , v  5, 4,2
Definition: The direction angles of a vector a are the angles
,  ,
and
 in the
interval  0,   that a makes with the positive x  axis, positive y  axis, and positive
z  axis, respectively. (on [0, π])
The cosines of these direction angles; (that is, cos , cos  , and cos  ) are called the
direction cosines of a .
How do we find the direction cosines and direction angles effectively?
Summarizing, we have:
cos 
cos  
cos  
a  a1, a2, a3  a cos  , a cos  , a cos   a cos  , cos  , cos 
therefore
a
 cos  , cos  , cos 
a
What this really says is that the direction cosines of a are the components of the unit
vector in the direction of a .
Example: Find the direction cosines and direction angles (in degrees) of the vector
1,2,3 .
Example: If a vector has direction angles
angle,
 . (There may be two answers.)


4
and


3
, find the third direction
Projection Vectors: Suppose we have two vectors, PQ and PR , which we will call a
and b, respectively. If S is the foot of the perpendicular from R to the line containing
PQ , then PS is called the vector projection of b onto a , and it is denoted proja b.
Then, the vector with representation PS is called the vector projection of b onto a and is
denoted by proja b.

You can think of it as a shadow of b.
The scalar projection of b onto a is defined as the signed magnitude of proja b.
compa b 
ab
a
This is the number |b| cos θ, where θ is the angle between a and b.

This is denoted by compa b.

Observe that it is negative if π/2 < θ ≤ π
.
The equation a ∙ b = |a||b| cos θ = |a|(|b| cos θ) shows that:

The dot product of a and b can be interpreted
as the length of a times the scalar projection of b onto a.
How do we compute proja b?


ab

a
Theorem: Vector projection of b onto a: proja b 
 2 
a


Proof: Step 1: Why is this a scalar multiple of vector a ?
Step 2: What is a vector, having length 1, in the direction of vector a ?
So, the vector projection is the scalar projection times the unit vector in the direction of
a
Step 3: Now, if we figure out the length of the projection vector, we’ll essentially be
done. We claim that the length of the projection vector is b cos . Why? How else can
the expression b cos be written?
In Summary:
Scalar projection of b onto a:
compa b 
ab
|a|
 a b  a a b
 2a

 |a| | a| |a|
Vector projection of b onto a: proja b  
Example: If a  1,1,2 and b  2,3,1 , find the vector and scalar projections of b
onto a.
Example: If a  2i  3j  k, and b  i  6 j  2k, find proja b and proja b .
A wagon is pulled a distance of 100 m along a horizontal path by a constant force of 70N.
The handle of the wagon is held at an angle of 35° above the horizontal.

Find the work
done by the force.
The work done by this force is defined to be the product of the component of the force
along D and the distance moved:
W = (|F| cos θ)|D|
W = |F||D| cos θ
=F ∙D
Therefore, the work done by a constant force F is:

The dot product F ∙ D, where D is
the displacement vector.
W = F ∙ D = |F||D| cos 35°
= (70)(100) cos 35°
≈ 5734 N∙m
= 5734 J