Hence, there do not exist scalars
c
1
,
c
2
, and
c
3
which satisfy the given equation.
5.
(c)
v
=
[
3
2
+ 4
2
+ 0
2
+ (–12)
2
]
1/2
=
169 = 13
6.
(a)
u
+
v
= (4, 4, 10, 1)
= 4
2
+ 4
2
+ 10
2
+ 1
2]
1/2
=
133
(c)
–2
u
+ 2
u
=
[
(–8)
2
+ (–2)
2
+ (–4)
2
+ (–6)
2
]
1/2
+ 2
[
4
2
+ 1
2
+ 2
2
+ 3
2
]
1/2
= 120
1/2
+ 2[30]
1/2
= 4
30
(e)
8.
Since
k
v
=
[
(–2
k
)
2
+ (3
k
)
2
+ 0
2
+ (6
k
)
2
]
1/2
=
[
49
k
2
]
1/2
= 7

k

, we have 
k
v
 = 5 if and only
if
k
= ±5/7
.
9.
(a)
(2,5)
•
(–4,3) = (2)(–4) + (5)(3) = 7
(c)
(3, 1, 4, –5)
•
(2, 2, –4, –3) = 6 + 2 –16 + 15 = 7
10.
(a)
Let
v
= (
x
,
y
) where
v
= 1. We are given that
v
•
(3, –1) = 0. Thus, 3
x
–
y
= 0 or
y
= 3
x
. But
v
= 1 implies that
x
2
+
y
2
=
x
2
+ 9
x
2
= 1 or
x
= ±1/
10. Thus, the only
possibilities are
v
= (1/
10, 3/
10) or
v
= (–1/
10, –3/
10).
You should graph these two vectors and the vector (3, –1) in an
xy
coordinate system.
(b)
Let
v
= (
x
,
y
,
z
) be a vector with norm 1 such that
x
– 3
y
+ 5
z
= 0
This equation represents a plane through (0,0,0) which is perpendicular to (1, –3, 5).
There are infinitely many vectors
v
which lie in this plane and have norm 1 and initial
point (0,0,0).
1
1
3
1
2
2
3 1 2 2
1
2
2
2
2
2
1 2
w
w
=
+
+
+
(
)
=
,
,
,
,
/
,
,
1
3
2
2
3
2
3
180
Exercise Set 4.1
11.
(a)
d
(
u
,
v
) =
[
(1 – 2)
2
+ (–2 – 1)
2
]
1/2
=
10
(c)
d
(
u
,
v
) =
[
(0 + 3)
2
+ (–2 – 2)
2
+ (–1 – 4)
2
+ (1 – 4)
2
]
1/2
=
59
14.
(e)
Since
u
•
v
= 0 + 6 + 2 + 0 = 8, the vectors are not orthogonal.
15.
(a)
We look for values of
k
such that
u
•
v
= 2 + 7 + 3
k
= 0
Clearly
k
= –3 is the only possiblity.
16.
We must find two vectors
x
= (
x
1
,
x
2
,
x
3
,
x
4
) such that
x
•
x
= 1 and
x
•
u
=
x
•
v
=
x
•
w
= 0. Thus
x
1
,
x
2
,
x
3
, and
x
4
must satisfy the equations
x
2
1
+
x
2
2
+
x
2
3
+
x
2
4
= 1
2
x
1
+
x
2
– 4
x
3
= 0
–x
1
– x
2
+ 2
x
3
+ 2
x
4
= 0
3
x
1
+ 2
x
+ 5
x
3
+ 4
x
4
= 0
The solution to the three linear equations is
x
1
= –34
t
,
x
2
= 44
t
,
x
3
= –6
t
, and
x
4
= 11
t
. If we
substitute these values into the quadratic equation, we get
[
(–34)
2
+ (44)
2
+ (–6)
2
+ (11)
2
]
t
2
= 1
or
Therefore, the two vectors are
±
1
57
(–34, 44, –6, 11)
t
= ±
= ±
1
3249
1
57
Exercise Set 4.1
181
17.
(a)
We have

u
•
v
 = 3(4) + 2(
–
1) = 10,
while
u
v
= [3
2
+ 2
2
]
1/2
[4
2
+ (–1)
2
]
1/2
=
221.
(d)
Here

u
•
v
 = 0 + 2 + 2 + 1 = 5,
while
u
v
= [0
2
+ (–2)
2
+ 2
2
+ 1
2
]
1/2
[(–1)
2
+ (–1)
2
+ 1
2
+ 1
2
]
1/2
= 6.
18.
(a)
In this case
and
20.
By Theorem 4.1.6, we have
u
•
v
=
1
4
u
+
v
2
–
1
4
u
–
v
2
=
1
4
(1)
2
–
1
4
(5
2
) = –6.
22.
Note that
u
•
a
= 4 and
a
2
= 15. Hence, by Theorem 3.3.3, we have
and
u
u
u
u a
a
a
−
=
−
⋅
=
−
(
)
−
−
proj
a
2
2 1 4
1
4
15
4
15
8
15
4
, , ,
,
,
,
5
34
15
11
15
52
15
9
5
=
−
,
,
,
proj
a
2
u
u a
a
a
=
⋅
=
−
(
)
4
15
1 1 2 3
, , ,
A
T
u . v
=
−
= −
9
1
2
6
12
u .
v
A
=
⋅
−
= −
3
1
10
18
12
u
v
.
A
T
=
⋅
−
−
3
1
2
3
1
4
2
6
=
⋅
=
3
1
14
26
68
Au . v
=
⋅
−
=
5
13
2
6
68
182
Exercise Set 4.1
23.
We must see if the system
3 + 4
t
=
s
2 + 6
t
= 3 – 3
s
3 + 4
t
= 5 – 4
s
–1 – 2
t
= 4 – 2
s
is consistent. Solving the first two equations yield