CHAPTER 1
1.1. Given the vectors
M
= −
10
a
x
+
4
a
y
−
8
a
z
and
N
=
8
a
x
+
7
a
y
−
2
a
z
, find:
a) a unit vector in the direction of
−
M
+
2
N
.
−
M
+
2
N
=
10
a
x
−
4
a
y
+
8
a
z
+
16
a
x
+
14
a
y
−
4
a
z
=
(
26
,
10
,
4
)
Thus
a
=
(
26
,
10
,
4
)

(
26
,
10
,
4
)

=
(
0
.
92
,
0
.
36
,
0
.
14
)
b) the magnitude of 5
a
x
+
N
−
3
M
:
(
5
,
0
,
0
)
+
(
8
,
7
,
−
2
)
−
(
−
30
,
12
,
−
24
)
=
(
43
,
−
5
,
22
)
, and

(
43
,
−
5
,
22
)
 =
48
.
6
.
c)

M

2
N

(
M
+
N
)
:

(
−
10
,
4
,
−
8
)

(
16
,
14
,
−
4
)

(
−
2
,
11
,
−
10
)
=
(
13
.
4
)(
21
.
6
)(
−
2
,
11
,
−
10
)
=
(
−
580
.
5
,
3193
,
−
2902
)
1.2. Given three points,
A(
4
,
3
,
2
)
,
B(
−
2
,
0
,
5
)
, and
C(
7
,
−
2
,
1
)
:
a) Specify the vector
A
extending from the origin to the point
A
.
A
=
(
4
,
3
,
2
)
=
4
a
x
+
3
a
y
+
2
a
z
b) Give a unit vector extending from the origin to the midpoint of line
AB
.
The vector from the origin to the midpoint is given by
M
=
(
1
/
2
)(
A
+
B
)
=
(
1
/
2
)(
4
−
2
,
3
+
0
,
2
+
5
)
=
(
1
,
1
.
5
,
3
.
5
)
The unit vector will be
m
=
(
1
,
1
.
5
,
3
.
5
)

(
1
,
1
.
5
,
3
.
5
)

=
(
0
.
25
,
0
.
38
,
0
.
89
)
c) Calculate the length of the perimeter of triangle
ABC
:
Begin with
AB
=
(
−
6
,
−
3
,
3
)
,
BC
=
(
9
,
−
2
,
−
4
)
,
CA
=
(
3
,
−
5
,
−
1
)
.
Then

AB
 + 
BC
 + 
CA
 =
7
.
35
+
10
.
05
+
5
.
91
=
23
.
32
1.3. The vector from the origin to the point
A
is given as
(
6
,
−
2
,
−
4
)
, and the unit vector directed from the
origin toward point
B
is
(
2
,
−
2
,
1
)/
3. If points
A
and
B
are ten units apart, find the coordinates of point
B
.
With
A
=
(
6
,
−
2
,
−
4
)
and
B
=
1
3
B(
2
,
−
2
,
1
)
, we use the fact that

B
−
A
 =
10, or

(
6
−
2
3
B)
a
x
−
(
2
−
2
3
B)
a
y
−
(
4
+
1
3
B)
a
z
 =
10
Expanding, obtain
36
−
8
B
+
4
9
B
2
+
4
−
8
3
B
+
4
9
B
2
+
16
+
8
3
B
+
1
9
B
2
=
100
or
B
2
−
8
B
−
44
=
0. Thus
B
=
8
±
√
64
−
176
2
=
11
.
75 (taking positive option) and so
B
=
2
3
(
11
.
75
)
a
x
−
2
3
(
11
.
75
)
a
y
+
1
3
(
11
.
75
)
a
z
=
7
.
83
a
x
−
7
.
83
a
y
+
3
.
92
a
z
1