328
C H A P T E R
13
VECTOR GEOMETRY
(ET CHAPTER 12)
13.3 Dot Product and the Angle Between Two Vectors
(ET Section 12.3)
Preliminary Questions
1.
Is the dot product of two vectors a scalar or a vector?
SOLUTION
The dot product of two vectors is the sum of products of scalars, hence it is a scalar.
2.
What can you say about the angle between
a
and
b
if
a
·
b
<
0?
SOLUTION
Since the cosine of the angle between
a
and
b
satisfies cos
θ
=
a
·
b
a
b
, also cos
θ
<
0. By definition
0
≤
θ
≤
π
, but since cos
θ
<
0 then
θ
is in
[
π
/
2
,
π
]
. In other words, the angle between
a
and
b
is obtuse.
3.
Suppose that
v
is orthogonal to both
u
and
w
. Which property of dot products allows us to conclude that
v
is
orthogonal to
u
+
w
?
SOLUTION
One property is that two vectors are orthogonal if and only if the dot product of the two vectors is zero. The
second property is the Distributive Law. Since
v
is orthogonal to
u
and
w
, we have
v
·
u
=
0 and
v
·
w
=
0. Therefore,
v
·
(
u
+
w
)
=
v
·
u
+
v
·
w
=
0
+
0
=
0
We conclude that
v
is orthogonal to
u
+
w
.
4.
What is proj
v
(
v
)
?
SOLUTION
The projection of
v
along itself is
v
, since
proj
v
(
v
)
=
v
·
v
v
·
v
v
=
v
5.
What is the difference, if any, between the projection of
u
along
v
and the projection along the unit vector
e
v
?
SOLUTION
The projection of
u
along
v
is the vector
proj
v
(
u
)
=
(
u
·
e
v
)
e
v
(1)
The projection of
u
along
e
v
is the vector
proj
e
v
(
u
)
=
(
u
·
e
e
v
)
e
e
v
(2)
Since
e
e
v
=
e
v
, the vectors in (1) and (2) are identical. That is, the projection of
u
along
v
is the projection of
u
along
e
v
.
6.
Suppose that proj
v
(
u
)
=
v
. Determine:
(a)
proj
2
v
(
u
)
(b)
proj
v
(
2
u
)
SOLUTION
(a)
The projection of
u
along 2
v
is the following vector:
proj
2
v
(
u
)
=
u
·
2
v
(
2
v
)
·
(
2
v
)
2
v
=
2
u
·
v
4
v
·
v
2
v
=
4
u
·
v
4
v
·
v
v
=
u
·
v
v
·
v
v
=
proj
v
(
u
)
=
v
(b)
The projection of 2
u
along
v
is the following vector:
proj
v
(
2
u
)
=
2
u
·
v
v
·
v
v
=
2
u
·
v
v
·
v
v
=
2proj
v
(
u
)
=
2
·
v
=
2
v
7.
Let
u

be the projection of
u
along
v
. Which of the following is the projection
u
along the vector 2
v
?
(a)
1
2
u

(b) u

(c)
2
u

SOLUTION
Since
u
is the projection of
u
along
v
, we have,
u
=
u
·
v
v
·
v
v
The projection of
u
along the vector 2
v
is
proj
2
v
(
u
)
=
u
·
2
v
2
v
·
2
v
2
v
=
2
u
·
v
4
v
·
v
2
v
=
4
u
·
v
4
v
·
v
v
=
u
·
v
v
·
v
v
=
u
That is,
u
is the projection of
u
along 2
v
. Notice that the projection of
u
along
v
is the projection of
u
along the unit
vector
e
v
, hence it depends on the direction of
v
rather than on the length of
v
. Therefore, the projection of
u
along
v
and
along 2
v
is the same vector.
8.
Let
θ
be the angle between
u
and
v
. Which of the following is equal to the cos
θ
?
(a) u
·
v
(b) u
·
e
v
(c) e
u
·
e
v
SOLUTION
By the Theorems on the Dot Product and the Angle Between Vectors, we have
cos
θ
=
u
·
v
u
v
=
u
u
·
v
v
=
e
u
·
e
v
The correct answer is (c).
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
S E C T I O N
13.3
Dot Product and the Angle Between Two Vectors
(ET Section 12.3)
329
Exercises
In Exercises 1–12, compute the dot product.
This is the end of the preview.
Sign up
to
access the rest of the document.
 Spring '08
 Hubscher
 Vectors, Scalar, Dot Product, Cos

Click to edit the document details