Math 601 Solutions to Homework 8
1. Find the distance between the following two lines:
x
y
z
=
0
1
2
+
t
1
1
0
x
y
z
=
1

1
3
+
t
1
1
0
Answer:
First, note that the two lines are parallel, since both lines
are in the direction given by the vector
a
=
1
1
0
.
A point on the ﬁrst line is
P
0
= (0
,
1
,
2) and a point on the second line
P
1
= (1
,

1
,
3). Thus, the vector
→
P
0
P
1
=
1

2
1
goes from one line to
the other line. We can project this vector onto one of the lines, as in
the following picture:
T
#
T
"
We ﬁnd the vector projection of
→
P
0
P
1
onto the vector
a
:
ˆ
→
P
0
P
1
·
a
k
a
k
!
a
k
a
k
=

1
/
2

1
/
2
0
Then, if we subtract this result from
→
P
0
P
1
, we will get a vector perpen
dicular to the two lines, which goes between the two lines. The length
of this vector will be distance between the two lines.
1

2
1


1
/
2

1
/
2
0
=
3
/
2

3
/
2
1
The length of this vector is
r
9
4
+
9
4
+ 1 =
√
22
2
. Thus, the distance
between the two lines is
√
22
2
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View Full Document2. Consider the four points
P
1
= (5
,
2
,
0),
P
2
= (1
,
5
,

1),
P
3
= (1
,
7
,
3),
and
P
4
= (4
,
6
,
3) in
R
3
.
(a) Find the equation for the plane consisting of all points equidistant
from
P
1
and
P
2
.
(b) Do the same for
P
1
and
P
3
and for
P
1
and
P
4
.
(c) Suppose that all four points lie on the surface of a sphere
S
. De
termine the center point of
S
.
Answer:
(a) The plane equidistant to the two points is the plane that lies di
rectly in between the two points. This plane will be perpendicular
to the line segment between the two points, and it will contain the
midpoint of that line segment.
Thus, the vector
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 Spring '08
 alndy
 Math, Derivative, Sin, Cos, p1

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