388
CHAPTER 13
VECTOR GEOMETRY
(ET CHAPTER 12)
SOLUTION
The trace of the plane
−
x
+
y
=
4onthe
xz
plane is the set of all points that satisfy both the equation of
the given plane and the equation
y
=
0ofthe
plane. That is, the set of all points
(
x
,
0
,
z
)
such that
−
x
+
0
=
4, or
x
=−
4. This is a vertical line in the
plane.
41.
−
x
+
y
=
4,
yz
The trace of the plane
−
x
+
y
=
plane is the set of all points that satisfy both the equation of
the plane and the equation
x
=
plane. That is, the set of all points
(
0
,
y
,
z
)
such that
−
0
+
y
=
4, or
y
=
4.
This is a vertical line in the
plane.
42.
Does the plane
x
=
5 have a trace in the
plane? Explain.
The
plane has the equation
x
=
0, hence the
x
coordinates of the points in this plane are zero, whereas
the
x
coordinates of the points in the plane
x
=
5 are 5. Thus, the two planes have no common points.
43.
Give equations for two distinct planes whose trace in the
xy
plane has equation 4
x
+
3
y
=
8.
The
plane has the equation
z
=
0, hence the trace of a plane
ax
+
by
+
cz
=
0inthe
plane is
obtained by substituting
z
=
0 in the equation of the plane. Therefore, the following two planes have trace 4
x
+
3
y
=
8
in the
plane:
4
x
+
3
y
+
z
=
8
;
4
x
+
3
y
−
5
z
=
8
44.
Find parametric equations for the line through
P
0
=
(
3
,
−
1
,
1
)
perpendicular to the plane 3
x
+
5
y
−
7
z
=
29.
We need to ±nd a direction vector for the line. Since the line is perpendicular to the plane 3
x
+
5
y
−
7
z
=
29, it is parallel to the vector
n
=h
3
,
5
,
−
7
i
normal to the plane. Hence,
n
is a direction vector for the line. The vector
parametrization of the line is, thus,
r
(
t
)
3
,
−
1
,
1
i+
t
h
3
,
5
,
−
7
i
This yields the parametric equations
x
=
3
+
3
t
,
y
1
+
5
t
,
z
=
1
−
7
t
45.
Find all planes in
R
3
whose intersection with the
plane is the line
r
(
t
)
=
t
h
2
,
1
,
0
i
.
The intersection of the plane
+
+
=
d
with the
plane is obtained by substituting
z
=
equation of the plane. This gives the line
+
=
d
,inthe
plane. We ±nd the equation of the line
l
(
t
)
=
t
h
2
,
1
,
0
i
.
On this line we have
x
=
2
t
y
=
t
⇒
y
=
1
2
x
⇒
x
−
2
y
=
0
We thus must have
d
=
0and
b
a
2,
a
±=
0. That is,
d
=
0,
b
2
a
,
a
±=
0. Notice that
c
can have any value. Hence,
the planes are
−
2
ay
+
=
0
,
a
±=
0
46.
Find all planes in
R
3
whose intersection with the
plane is the line with equation 3
x
+
2
z
=
5.
The intersection of the plane
+
+
=
d
with the
plane is obtained by substituting
y
=
equation of the plane. This gives the following line in the
plane:
+
=
d
This is the equation of the line 3
x
+
2
z
=
5 if and only if for some
λ
±=
0,
a
=
3
,
c
=
2
,
d
=
5
Notice that
b
can have any value. The planes are thus
(
3
)
x
+
+
(
2
)
z
=
5
,
±=
0
.
47.
Let
P
be the plane
n
· h
x
,
y
,
z
i =
d
,where
n
±=
0
,andlet
P
1
be the parallel plane
n
· h
x
,
y
,
z
i =
d
1
(Figure
10).
(a)
Show that the line through
n
intersects
P
at the terminal point of the vector
µ
d
k
n
k
¶
e
n
.
(b)
Show that the distance between
P
and
P
1
is

d
−
d
1

k
n
k
.