324
CHAPTER 13
VECTOR GEOMETRY
(ET CHAPTER 12)
We use the formula for the midpoint of a segment to Fnd the coordinates of the points
P
and
Q
.Thisgives
P
=
µ
1
+
0
2
,
0
+
1
2
,
1
+
1
2
¶
=
µ
1
2
,
1
2
,
1
¶
Q
=
µ
1
+
0
2
,
1
+
1
2
,
0
+
1
2
¶
=
µ
1
2
,
1
,
1
2
¶
Substituting in (1) yields the following vector:
v
=
−→
PQ
=
¿
1
2
−
1
2
,
1
−
1
2
,
1
2
−
1
À
=
¿
0
,
1
2
,
−
1
2
À
.
54.
±ind the components of the vector
w
whose tail is
C
and head is the midpoint of
AB
in ±igure 18.
B
=
(1, 1, 0)
C
=
(0, 1, 1)
A
=
(1, 0, 1)
y
x
z
FIGURE 18
SOLUTION
We denote the midpoint of
by
M
=
(
a
,
b
,
c
)
. To Fnd the coordinates of
M
we Frst parametrize the
line through
A
=
(
1
,
0
,
1
)
and
B
=
(
1
,
1
,
0
)
:
r
(
t
)
=
(
1
−
t
)
h
1
,
0
,
1
i+
t
h
1
,
1
,
0
i=h
1
−
t
,
0
,
1
−
t
i+h
t
,
t
,
0
1
,
t
,
1
−
t
i
The midpoint of
occurs at
t
=
1
2
, hence the vector
OM
is
h
1
,
1
2
,
1
2
i
.
B
=
(1, 1, 0)
C
=
(0, 1, 1)
A
=
(1, 0, 1)
y
x
z
The point
M
is the terminal point of
,thatis,
M
=
³
1
,
1
2
,
1
2
´
. We now Fnd the vector
w
=
CM
:
w
=
¿
1
−
0
,
1
2
−
1
,
1
2
−
1
À
=
¿
1
,
−
1
2
,
−
1
2
À
.
Further Insights and Challenges
In Exercises 55–59, we consider the equations of a line in
symmetric form
.