17.1 SOLUTIONS
1197
CHAPTER SEVENTEEN
Solutions for Section 17.1
Exercises
1.
One possible parameterization is
x
= 3 +
t,
y
= 2
t,
z
=

4

t.
2.
One possible parameterization is
x
= 1 + 3
t,
y
= 2

3
t,
z
= 3 +
t.
3.
One possible parameterization is
x
=

3 + 2
t,
y
= 4 + 2
t,
z
=

2

3
t.
4.
One possible parameterization is
x
= 5
,
y
=

1 + 5
t,
z
= 1 + 2
t.
5.
One possible parameterization is
x
=
t,
y
= 1
,
z
=

t.
6.
One possible parameterization is
x
= 1
,
y
= 0
,
z
=
t.
7.
The displacement vector from the Frst point to the second is
~v
= 4
~
i

5
~
j

3
~
k
. The line through point
(1
,
5
,
2)
and
with direction vector
~v
= 4
~
i

5
~
j

3
~
k
is given by parametric equations
x
= 1 + 4
t,
y
= 5

5
t,
z
= 2

3
t.
Other parameterizations of the same line are also possible.
8.
The vector connecting the two points is
3
~
i

~
j
+
~
k
. So a possible parameterization is
x
= 2 + 3
t,
y
= 3

t,
z
=

1 +
t.
9.
The displacement vector from the Frst point to the second is
~v
= (

1

(

3))
~
i
+ (

3

(

2))
~
j
+ (

1

1)
~
k
=
2
~
i

~
j

2
~
k
. The line through point
(

3
,

2
,
1)
and with direction vector
~v
= 2
~
i

~
j

2
~
k
is given by parametric
equations
x
=

3 + 2
t,
y
=

2

t,
z
= 1

2
t.
Other parameterizations of the same line are also possible.
10.
The line passes through
(3
,

2
,
2)
and
(0
,
2
,
0)
. The displacement vector from the Frst of these points to the second is
~v
= (0

3)
~
i
+ (2

(

2))
~
j
+ (0

2)
~
k
=

3
~
i
+ 4
~
j

2
~
k
. The line through point
(3
,

2
,
2)
and with direction
vector
~v
=

3
~
i
+ 4
~
j

2
~
k
is given by parametric equations
x
= 3

3
t,
y
=

2 + 4
t,
z
= 2

2
t.
Other parameterizations of the same line are also possible.