2
KIAM HEONG KWA
where
r
0
=
h
x
0
, y
0
, z
0
i
,
a
=
h
a
1
, a
2
, a
3
i
, and
b
=
h
b
1
, b
2
, b
3
i
are vectors
in
R
3
such that (1)
a
6
=
0
, (2)
b
6
=
0
, and (3)
a
and
b
are nonparallel
(i.e.,
a
×
b
6
=
0
) represents a plane in
R
3
that contains the point
(
x
0
, y
0
, z
0
) and has
a
×
b
as a normal vector.
To see this, write the
vector equation as
r
(
u, v
)

r
0
=
u
a
+
v
b
. Then
(
r
(
u, v
)

r
0
)
·
a
×
b
=
u
a
·
a
×
b
+
v
b
·
a
×
b
= 0
because
a
·
a
×
b
=
b
·
a
×
b
=
0
.
Let us apply the general statement to the vector equation
r
(
u, v
) = (
u
+
v
)
i
+ (3

v
)
j
+ (1 + 4
u
+ 5
v
)
k
= 3
j
+
k
+
u
(
i
+ 4
k
) +
v
(
i

j
+ 5
k
)
.
Identifying the vectors
r
0
,
a
, and
b
, one has
r
0
= 3
j
+
k
,
a
=
i
+ 4
k
,
and
b
=
i

j
+ 5
k
.
In particular,
a
×
b
= 4
i

j

k
.
Denoting
r
(
u, v
) =
x
i
+
y
j
+
z
k
, one has
[
x
i
+ (
y

3)
j
+ (
z

1)
k
]
·
(4
i

j

k
) = 4
x

y

z
+ 4 = 0
.
Example 2
(Problem 6 in the text)
.
The vector equation
r
(
s, t
) =
h
s
sin 2
t, s
2
, s
cos 2
t
i
defines a circular paraboloid whose axis is the
y

axis. To see this, consider the equivalent parametric equations
x
=
s
sin 2
t, y
=
s
2
, z
=
s
cos 2
t.
Note that
x
2
+
z
2
=
s
2
(sin
2
2
t
+ cos
2
2
t
) =
s
2
=
y.
Traces parallel to the
xz
planes are circles, while traces parallel to the
xy
 and
yz
planes are parabolas.
Example 3
(Problem 20 in the text)
.
We would like to find a paramet
ric representation for the lower half of the ellipsoid
2
x
2
+ 4
y
2
+
z
2
= 1
.
Note that this can be described by
2
x
2
+ 4
y
2
+
z
2
= 1
and
z
≤
0
and thus more compactly by
z
=

p
1

2
x
2

4
y
2
.
Hence a parametric representation of the surface is
x
=
x, y
=
y, z
=

p
1

2
x
2

4
y
2
,