Chapter 5
ISM:
Linear Algebra
Chapter 5
5.1
1.
v
=
√
7
2
+ 11
2
=
√
49 + 121 =
√
170
≈
13
.
04
2.
v
=
√
2
2
+ 3
2
+ 4
2
=
√
4 + 9 + 16 =
√
29
≈
5
.
39
3.
v
=
√
2
2
+ 3
2
+ 4
2
+ 5
2
=
√
4 + 9 + 16 + 25 =
√
54
≈
7
.
35
4.
θ
= arccos
u
·
v
u
v
= arccos
7+11
√
2
√
170
= arccos
18
√
340
≈
0
.
219 (radians)
5.
θ
= arccos
u
·
v
u
v
= arccos
2+6+12
√
14
√
29
≈
0
.
122 (radians)
6.
θ
= arccos
u
·
v
u
v
= arccos
2

3+8

10
√
10
√
54
≈
1
.
700 (radians)
7. Use the fact that
u
·
v
=
u
v
cos
θ
, so that the angle is acute if
u
·
v >
0, and obtuse if
u
·
v <
0. Since
u
·
v
= 10

12 =

2, the angle is obtuse.
8. Since
u
·
v
= 4

24 + 20 = 0, the two vectors enclose a right angle.
9. Since
u
·
v
= 3

4 + 5

3 = 1, the angle is acute (see Exercise 7).
10.
u
·
v
= 2 + 3
k
+ 4 = 6 + 3
k
. The two vectors enclose a right angle if
u
·
v
= 6 + 3
k
= 0,
that is, if
k
=

2.
11. a.
θ
n
= arccos
u
·
v
u
v
= arccos
1
√
n
θ
2
= arccos
1
√
2
=
π
4
(= 45
◦
)
θ
3
= arccos
1
√
3
≈
0
.
955 (radians)
θ
4
= arccos
1
2
=
π
3
(= 60
◦
)
b. Since
y
= arccos(
x
) is a continuous function,
lim
n
→∞
θ
n
= arccos
lim
n
→∞
1
√
n
= arccos(0) =
π
2
(= 90
◦
)
12.
v
+
w
2
= (
v
+
w
)
·
(
v
+
w
) (by hint)
=
v
2
+
w
2
+ 2(
v
·
w
) (by definition of length)
≤
v
2
+
w
2
+ 2
v
w
(by CauchySchwarz)
234
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ISM:
Linear Algebra
Section 5.1
= (
v
+
w
)
2
, so that
v
+
w
2
≤
(
v
+
w
)
2
Taking square roots of both sides, we find that
v
+
w
≤
v
+
w
, as claimed.
13. Figure 5.1 shows that
F
2
+
F
3
= 2 cos
(
θ
2
)
F
2
= 20 cos
(
θ
2
)
.
It is required that
F
2
+
F
3
= 16, so that 20 cos
(
θ
2
)
= 16, or
θ
= 2 arccos(0
.
8)
≈
74
◦
.
Figure 5.1: for Problem 5.1.13.
14. The horizontal components of
F
1
and
F
2
are

F
1
sin
β
and
F
2
sin
α
, respectively (the
horizontal component of
F
3
is zero).
Since the system is at rest, the horizontal components must add up to 0, so that

F
1
sin
β
+
F
2
sin
α
= 0 or
F
1
sin
β
=
F
2
sin
α
or
F
1
F
2
=
sin
α
sin
β
.
To find
EA
EB
, note that
EA
=
ED
tan
α
and
EB
=
ED
tan
β
so that
EA
EB
=
tan
α
tan
β
=
sin
α
sin
β
·
cos
β
cos
α
=
F
1
F
2
cos
β
cos
α
.
Since
α
and
β
are two distinct acute angles, it follows that
EA
EB
=
F
1
F
2
, so that Leonardo was mistaken.
15. The subspace consists of all vectors
x
in
R
4
such that
x
·
v
=
x
1
x
2
x
3
x
4
·
1
2
3
4
=
x
1
+ 2
x
2
+ 3
x
3
+ 4
x
4
= 0.
These are vectors of the form

2
r

3
s

4
t
r
s
t
=
r

2
1
0
0
+
s

3
0
1
0
+
t

4
0
0
1
.
The three vectors to the right form a basis.
235
Chapter 5
ISM:
Linear Algebra
16. You may be able to find the solutions by educated guessing.
Here is the systematic
approach: we first find all vectors
x
that are orthogonal to
v
1
, v
2
, and
v
3
, then we identify
the unit vectors among them.
Finding the vectors
x
with
x
·
v
1
=
x
·
v
2
=
x
·
v
3
= 0 amounts to solving the system
x
1
+
x
2
+
x
3
+
x
4
= 0
x
1
+
x
2

x
3

x
4
= 0
x
1

x
2
+
x
3

x
4
= 0
(
we can omit all the coe
ffi
cients
1
2
)
.
The solutions are of the form
x
=
x
1
x
2
x
3
x
4
=
t

t

t
t
.
Since
x
= 2

t

, we have a unit vector if
t
=
1
2
or
t
=

1
2
. Thus there are two possible
choices for
v
4
:
1
2

1
2

1
2
1
2
and

1
2
1
2
1
2

1
2
.
17. The orthogonal complement
W
⊥
of
W
consists of the vectors
x
in
R
4
such that
x
1
x
2
x
3
x
4
·
1
2
3
4
= 0 and
x
1
x
2
x
3
x
4
·
5
6
7
8
= 0.