Solutions to Homework Assignment 5
MATH 24901 and 02
Section 12.5, Page 829
1, 2, 3, 6, 9, 11, 13, 14, 1924, 26, 30, 37, 39, 40, 4550, 53, 54, 58, 6569
3. A vector equation should look like
r
(
t
) =
<

2
,
4
,
10
>
+
t <
3
,
1
,

8
> .
Comparing components gives
us the parametric equations
x
=

2 + 3
t,y
= 4 +
t,z
= 10

8
t.
6. A direction vector is just
<
1
,
2
,
3
>,
so our equations are
x
=
t,y
= 2
t,z
= 3
t.
Symmetric equations
are
x
=
y
2
=
z
3
.
9. A direction vector is
<
2

0
,
1

1
/
2
,

3

1
>
=
<
2
,
1
/
2
,

4
> .
Parametric equations are then
x
= 2+2
t,y
= 1+
t/
2
,z
=

3

4
t.
Solving for
t
gives the symmetric equations
x

2
2
= 2
y

2 =
z
+ 3

4
.
11. The given line has parametric equations
x
=
t

2
,y
= 2
t,z
=
t
+ 3
.
Its direction vector is thus
<
1
,
2
,
1
>
(from the coeﬃcients of
t
). We get
x
= 1 +
t,y
=

1 + 2
t,z
= 1 +
t.
19.
L
1
is in the direction
<

6
,
9
,

3
>,
and
L
2
is in the direction
<
2
,

3
,
1
> .
These are parallel, so the
lines are parallel.
21.