ANSWERS PAMPHLET
51
b
)
x
(
s, t
)
5
0
3
2
1
3
0
2
1
t
1
2
2
2
2
s
,
x
1
2
y
1
3
z
5
12.
10.37
a
) Rewriting the symmetric equations,
y
5
y
0
2
(
b
6
a
)
x
0
1
(
b
6
a
)
x
and
z
5
z
0
2
(
c
6
a
)
x
0
1
(
c
6
a
)
x
. Then
x
(
t
)
y
(
t
)
z
(
t
)
5
x
0
y
0
z
0
1
1
b
6
a
c
6
a
t
or
x
(
t
)
y
(
t
)
z
(
t
)
5
x
0
y
0
z
0
1
a
b
c
t.
b
) The two planes are described by any two distinct equalities in system
(20). For example,
x
2
x
0
a
5
y
2
y
0
b
and
y
2
y
0
b
5
z
2
z
0
c
.
In other words,
2
bx
1
ay
5 2
bx
0
1
ay
0
and
cy
2
bz
5
cy
0
2
bz
0
.
c
)
i
)
x
1
2
2
2
1
5
x
2
2
3
4
5
x
3
2
1
5
,
ii
)
x
1
2
1
4
5
x
2
2
2
5
5
x
3
2
3
6
.
d
)
i
) 4
x
1
1
x
2
5
11 and 5
x
2
2
4
x
3
5
11.
ii
) 5
x
1
2
4
x
2
5 2
3 and 6
x
2
2
5
x
3
5 2
3
.
10.38
a
) Normals (1
,
2
,
2
3) and (1
,
3
,
2
2) do not line up, so planes intersect.
b
) Normals (1
,
2
,
2
3) and (
2
2
,
2
4
,
6) do line up, so planes do not intersect.
10.39
a
) (
x
2
1
, y
2
2
, z
2
3)
?
(
2
1
,
1
,
0)
5
0, so
x
2
y
5 2
1.
b
) The line runs through the points (4
,
2
,
6) and (1
,
3
,
11), so the plane must
be orthogonal to the difference vector (
2
3
,
1
,
5). Thus (
x
2
1
, y
2
1
,
z
1
1)
?
(
2
3
,
1
,
5)
5
0, or
2
3
x
1
y
1
5
z
5 2
7.
c
) The general equation for the plane is
a
x
1
b
y
1
g
z
5
d
. The equations
to be satisfied are
a
5
d
6
a
,
b
5
d
6
b
, and
c
5
d
6
g
. A solution is
a
5
1
6
a
,
b
5
1
6
b
.
g
5
1
6
c
and
d
5
1, so (1
6
a
)
x
1
(1
6
b
)
y
1
(1
6
c
)
z
5
1.
10.40
Plug
x
5
3
1
t
,
y
5
1
2
7
t
, and
z
5
3
2
3
t
into the equation
x
1
y
1
z
5
1
of the plane and solve for
t
: (3
1
t
)
1
(1
2
7
t
)
1
(3
2
3
t
)
5
1
=
⇒
t
5
2
6
3
.
The point is
(
11
6
3
,
2
11
6
3
,
1
)
.
10.41
1
1
2
1
.
.
.
4
1
2
1
.
.
.
3
=
⇒
1
0
2
3
.
.
.
5
0
1
2
.
.
.
2
1
=
⇒
x
5
5
1
3
z
y
5 2
1
2
2
z.
Taking
z
5
t
, write the line as
x
y
z
5
5
2
1
0
1
t
3
2
2
1
.