SOLUTIONS FOR HOMEWORK 1
Section 1.1
1. (a)
The vectors (3
,
1
,
2) and (6
,
4
,
2) are parallel iff (this is an
abbreviation for “if and only if”) there exists
t
∈
R
s.t.
t
(3
,
1
,
2) =
(6
,
4
,
2).
That is, for some
t
, three equalities must be satisfied:
(i)
3
t
= 6, (ii)
t
= 4, and (iii) 2
t
= 2. But there is no such
t
. Thus, the
vectors are
not parallel
.
(c)
The vectors (5
,
−
6
,
7) and (
−
5
,
6
,
−
7)
are parallel
, since the equal-
ity
t
(5
,
−
6
,
7) = (
−
5
,
6
,
−
7) is satisfied for
t
=
−
1.
2. (c)
The line in question is the set of all points of the form (3
,
7
,
2)+
t
(
(3
,
7
,
−
8)
−
(3
,
7
,
2)
)
= (3
,
7
,
−
10
t
), with
t
∈
R
.
3. (a)
The plane in question is the set of all points of the form
(2
,
−
5
,
−
1) +
t
(
(0
,
4
,
6)
−
(2
,
−
5
,
−
1)
)
+
s
(
(
−
3
,
7
,
1)
−
(2
,
−
5
,
−
1)
)
= (2
−
2
t
−
5
s,
−
5 +
t
−
2
s,
−
1 + 7
t
+ 2
s
)
,
with
t,s
∈
R
.
Section 1.2
1. (h) No
. Consider, for instance
p
(
x
) = 13 + 2
x
, and
q
(
x
) =
−
2
x
.
Then deg(
p
) = deg(
q
) = 1. However,
p
(
x
) +
q
(
x
) = 13, hence deg(
p
+
q
) = 0
<
1.
4. (a)
parenleftbigg
2
5
−
3
1
0
7
parenrightbigg
+
parenleftbigg
4
−
2
5
−
5
3
2
parenrightbigg
=
parenleftbigg
6
3
2
−
4
3
9
parenrightbigg
.
(e)
(2
x
4
−
7
x
3
+4
x
+3)+(8
x
3
+2
x
2
−
6
x
+7) = 2
x
4
+
x
3
+2
x
2
−
2
x
+10.
7.
We have to show that
f
(
t
) =
g
(
t
) for any
t
∈
S
=
{
0
,
1
}
. That is,
we need to show the equalities
f
(0) =
g
(0), and
f
(1) =
g
(1). But these
are easily verified. The equality
f
+
g
=
h
is handled in the same way.