324
C H A P T E R
13
VECTOR GEOMETRY
(ET CHAPTER 12)
We use the formula for the midpoint of a segment to find the coordinates of the points
P
and
Q
. This gives
P
=
1
+
0
2
,
0
+
1
2
,
1
+
1
2
=
1
2
,
1
2
,
1
Q
=
1
+
0
2
,
1
+
1
2
,
0
+
1
2
=
1
2
,
1
,
1
2
Substituting in (1) yields the following vector:
v
=
−→
P Q
=
1
2
−
1
2
,
1
−
1
2
,
1
2
−
1
=
0
,
1
2
,
−
1
2
.
54.
Find the components of the vector
w
whose tail is
C
and head is the midpoint of
AB
in Figure 18.
B
=
(1, 1, 0)
C
=
(0, 1, 1)
A
=
(1, 0, 1)
y
x
z
FIGURE 18
SOLUTION
We denote the midpoint of
AB
by
M
=
(
a
,
b
,
c
)
. To find the coordinates of
M
we first parametrize the
line through
A
=
(
1
,
0
,
1
)
and
B
=
(
1
,
1
,
0
)
:
r
(
t
)
=
(
1
−
t
)
1
,
0
,
1
+
t
1
,
1
,
0
=
1
−
t
,
0
,
1
−
t
+
t
,
t
,
0
=
1
,
t
,
1
−
t
The midpoint of
AB
occurs at
t
=
1
2
, hence the vector
OM
is 1
,
1
2
,
1
2
.
B
=
(1, 1, 0)
C
=
(0, 1, 1)
A
=
(1, 0, 1)
y
x
z
The point
M
is the terminal point of
OM
, that is,
M
=
1
,
1
2
,
1
2
. We now find the vector
w
=
CM
:
w
=
1
−
0
,
1
2
−
1
,
1
2
−
1
=
1
,
−
1
2
,
−
1
2
.
Further Insights and Challenges
In Exercises 55–59, we consider the equations of a line in
symmetric form
.