360
C H A P T E R
13
VECTOR GEOMETRY
(ET CHAPTER 12)
We now verify the equality:
v
2
w
2
−
(
v
·
w
)
2
=
17
·
21
−
18
2
=
33
=
v
×
w
2
40.
Find the volume of the parallelepiped spanned by
u
,
v
, and
w
in Figure 17.
SOLUTION
Using
u
=
1
,
0
,
4 ,
v
=
1
,
3
,
1
and
w
=
−
4
,
2
,
6 , the scalar triple product is the following 3
×
3
determinant:
u
·
(
v
×
w
)
=
1
0
4
1
3
1
−
4
2
6
=
1
·
(
18
−
2
)
−
0
+
4
(
2
+
12
)
=
16
+
56
=
72
41.
Find the area of the parallelogram spanned by
v
and
w
in Figure 17.
y
x
z
u
= ⟨
1, 0, 4
⟩
w
= ⟨−
4, 2, 6
⟩
v
= ⟨
1, 3, 1
⟩
FIGURE 17
SOLUTION
The area of the parallelogram equals the length of the cross product of the two vectors
v
=
1
,
3
,
1
and
w
= −
4
,
2
,
6 . We calculate the cross product as follows:
v
×
w
=
i
j
k
1
3
1
−
4
2
6
=
(
18
−
2
)
i
−
(
6
+
4
)
j
+
(
2
+
12
)
k
=
16
i
−
10
j
+
14
k
The length of this vector 16
i
−
10
j
+
14
k
is
16
2
+
10
2
+
14
2
=
2
√
138. Thus, the area of the parallelogram is 2
√
138.
42.
Calculate the volume of the parallelepiped spanned by
u
=
2
,
2
,
1
,
v
=
1
,
0
,
3
,
w
=
0
,
−
4
,
0
SOLUTION
Using
u
=
2
,
2
,
1 ,
v
=
1
,
0
,
3 , and
w
=
0
,
−
4
,
0 , the volume is given by the following scalar triple
product:
u
·
(
v
×
w
)
=
2
2
1
1
0
3
0
−
4
0
=
2
(
0
+
12
)
−
2
(
0
−
0
)
+
1
(
−
4
−
0
)
=
24
−
4
=
20
.
43.
Sketch and compute the volume of the parallelepiped spanned by
u
=
1
,
0
,
0
,
v
=
0
,
2
,
0
,
w
=
1
,
1
,
2
SOLUTION
Using
u
=
1
,
0
,
0 ,
v
=
0
,
2
,
0 , and
w
=
1
,
1
,
2 , the volume is given by the following scalar triple
product:
u
·
(
v
×
w
)
=
1
0
0
0
2
0
1
1
2
=
1
(
4
−
0
)
−
0
+
0
=
4
.
u
v
w
y
x
z